|
Hello all,
My QDX, which has been working perfectly, failed on transmit yesterday while completing a WSPR transmission. ?It was entirely my own doing - the QDX is built for 12V but I get 5W out on 10.4V and I¡¯d been accidentally running the QDX on 13.8V. ?EEK! ?Anyway, it died mid-transmission and gave out that heart-sink-inducing burnt aroma. ?
I had some spare BS170¡¯s, so replaced all four, making a statistical assumption it was those were the toasted part. But, after replacement, the output voltage measured across a 50ohm load is under 1V on all bands. ?So now I need to get methodical and do some diagnostics.
The rig is otherwise functioning well - receives perfectly and CAT control is working well.
I have a basic digital oscilloscope here to help with troubleshooting.
Can anyone point me in the direction of a flow chart to allow me to chase the signal through the transmit chain - what are the spec voltages along the way that I should expect to see. ?How do I test whether the driver IC5 is also fried (or functioning normally)? ?I¡¯m quite patient with diagnostics, so I¡¯m determined to find the fault and fix it, but would value your help.
Many thanks and 73,
John
VK7JB?
|
No substitute for careful fault finding but the driver IC (or one part at least) frequently fails if the BS170¡¯s do.?
I would replace it as a precaution then fault-find further if it¡¯s still not working?
|
Hi John,
A first step for me would be to verify the 5-volt squarewave on the gates of the BS170s.? This is a quick test to see if IC5 failed.? A high current on receiving and a hot IC2 is another indication after BS170 failures.
Hans does a good job of describing how the QDX works in the manuals.? There is also a troubleshooting page on the QRP-Labs web pages:
73
Evan
AC9TU
|
John VK7JB,
I have TWO QDXs that both suffered the same fate as yours -- working perfectly, then mid-transmission death resulting in high current on receive (400 mA) and a hot IC2.? I removed the four BS170s but the high current remains.? IC5 may be the problem, or perhaps Q6 (the T/R switch).? But nothing other than IC2 seems hot.
Have you solved your problem and/or do you or others have suggestions for me?
Thanks so much and 73,
Kirk
W1FA
|
Hi John,
I am sure that IC5 has failed.? This is common when the BS170s fail.? Continuing to try to run the QDX could cause IC2 to fail and possibly other issues.? I would remove IC5 and verify that the receive current goes down and IC2 does not get overly warm.
73
Evan
AC9TU
|
Hi Kirk and Evan,
I¡¯ve ordered some replacement IC5¡¯s and am waiting for them to arrive down under. ?This weekend, I¡¯ll put the oscilloscope onto the gates of the BS120s while the old IC5 is still in place and see what I find. ?I¡¯ll also see if the high current state persists with the (probably) damaged IC5 in place. ?I didn¡¯t notice IC2 getting hot. ?I¡¯d better check that¡¯s still doing its job.
73,
John
VK7JB
|
On 14/09/2023 07:12, John Farmer VK7JB via groups.io wrote: I¡¯ll put the oscilloscope onto the gates of the BS120s while the old IC5 is still in place and see what I find John, Just be aware that in at least one case, I think a QCX, the Si5351 and processor were damaged by 12 volts flowing back through the driver IC from a BS170 with, presumably a drain/gate short. As Evan says testing with a likely damaged part still present is somewhat risky even if the 12V is not there. 73 Alan G4ZFQ
|
Which Final FET Fails?
My low band QDX has just failed for the fourth time. ?Each time only Q9 has failed (shorted D-S). ?No other parts have been replaced. ??
My high band QDX also failed once, also Q9. ??
Both are built as 9 volt version, power is via a linear LM317 regulator with 1 amp current limiting.
Any ideas why this one part is more stressed than others?
I have just added 1N4148 diodes as in the QMX schematic.?
|
"I have just added 1N4148 diodes as in the QMX schematic."
Wut?
JZ
toggle quoted message
Show quoted text
On Thu, Sep 14, 2023 at 8:13?AM Jerry <Jerryh47@...> wrote:
Which Final FET Fails?
My low band QDX has just failed for the fourth time. Each time only Q9 has failed (shorted D-S). No other parts have been replaced.
My high band QDX also failed once, also Q9.
Both are built as 9 volt version, power is via a linear LM317 regulator with 1 amp current limiting.
Any ideas why this one part is more stressed than others?
I have just added 1N4148 diodes as in the QMX schematic.
|
On Thu, Sep 14, 2023, at 02:12 AM, John Farmer VK7JB wrote:
Hi Kirk and Evan,
I¡¯ve ordered some replacement IC5¡¯s and am waiting for them to arrive down under. ?This weekend, I¡¯ll put the oscilloscope onto the gates of the BS120s while the old IC5 is still in place and see what I find. ?I¡¯ll also see if the high current state persists with the (probably) damaged IC5 in place. ?I didn¡¯t notice IC2 getting hot. ?I¡¯d better check that¡¯s still doing its job.
73,
John
VK7JB
John, If you decide to test IC5, I would strongly suggest that you remove the BS170s and reduce the supplied voltage to 7 volts.? While trying to do that diagnostic, I destroyed the processor on a failed QDX that I got from another Ham.? I did not remove the BS170s when I should have.? The safe route is to remove all 4 BS170s and replace IC5, then do the IC5 testing.? After removing the BS170s, you can verify that IC5 works as expected (0 volts on the BS170 gates when receiving and 5 volts square wave-2.5 volts measured with DMM).? Also, limit the current supplied to 200ma if you have a current limiting supply.? Without the BS170s, the QDX will not need a lot of current. Suggestions from experience ;-) 73 Evan AC9TU
|
On Thu, Sep 14, 2023, at 07:12 AM, Jerry wrote:
Which Final FET Fails?
My low band QDX has just failed for the fourth time. ?Each time only Q9 has failed (shorted D-S). ?No other parts have been replaced. ??
My high band QDX also failed once, also Q9. ??
Both are built as 9 volt version, power is via a linear LM317 regulator with 1 amp current limiting.
Any ideas why this one part is more stressed than others?
I have just added 1N4148 diodes as in the QMX schematic.?
Jerry, The diodes in the QMX schematic are Zeners, NOT 1n4148.? Also, the single 1n4148 for the QDX protection is across L14, the RF choke that feeds power to the finals.? My simulations have shown that the Zeners OR the commutating diode across L14 are good choices.? You do not need both.? The simpler solution is John's commutating diode.? The commutating diode could be a 1n4148.? I have only bench-tested the commutating diode. 73 Evan AC9TU
|
On 14/09/2023 12:49, Jerry wrote:
Which?BS120 Fails?
My low band QDX has just failed for the fourth time. ?Each time
only Q9 has failed (shorted D-S). ?No other parts have been
replaced. ??
My high band QDX also failed once, also Q9. ??
Both are built as 9 volt version, power is via a linear LM317
regulator with 1 amp current limiting.
Any ideas why this one part is more stressed than others?
I have just added 1N4148 diodes as in the QMX schematic.
You do realise that the diodes between the source and drain of
the BS170s in the QMX schematic (D503 and D504) are place holders
for Zener diodes?
You need something like 1N4756 47V Zener diodes there.
The place for an 1N4148 is across the inductor that drives them,
L502.
Chris, G5CTH
|
Chris, & all here,
Here is a fun exercise:
Assume that no commutating diode is present at L14 in QDX, and that,
for convenience, the PA is drawing 1A through L14. Then at the moment
the last BS170 transistors switch off, a large V=Ldi/dt spike will
take the transistors to avalanche at 60 Volts. At that moment, 60
watts of instantaneous power (60V x 1A) is being dumped into the four
of them. It will last about half a microsecond in duration.
If 47 volt Zeners are intended to absorb the spike, the calculation
results in half an amp flowing to each zener, and 47 watts being
delivered to the pair. The Zener resistance will allow for
considerable elevation of the drain voltage above 47 volts, and the 60
volt avalanche spec may still be violated in spite of the presence of
the Zeners, assuming they can even handle the instantaneous power
dump.
Hans has shown through bench testing that high voltages are extremely
unlikely as the result of high SWR. Simulations done by Evan and I
agree with his bench testing. It seems the Zener diodes offer no real
value here, and that the L14 commutating diode is a better overall
protective solution. It can handle the L14 current easily, which
becomes more important under low Z / high SWR conditions.
More musings from one of those pesky nerds...
JZ KJ4A
toggle quoted message
Show quoted text
On Thu, Sep 14, 2023 at 10:56?AM Chris <chris.rowland@...> wrote:
On 14/09/2023 12:49, Jerry wrote:
Which BS120 Fails?
My low band QDX has just failed for the fourth time. Each time only Q9 has failed (shorted D-S). No other parts have been replaced.
My high band QDX also failed once, also Q9.
Both are built as 9 volt version, power is via a linear LM317 regulator with 1 amp current limiting.
Any ideas why this one part is more stressed than others?
I have just added 1N4148 diodes as in the QMX schematic.
You do realise that the diodes between the source and drain of the BS170s in the QMX schematic (D503 and D504) are place holders for Zener diodes?
You need something like 1N4756 47V Zener diodes there.
The place for an 1N4148 is across the inductor that drives them, L502.
Chris, G5CTH
|
Hello John
Isn't the calculation fraught with difficulties? Where did the 0.5 microseconds assumption come from? Wouldn't that depend on the effective resistance to ground of the parallel BS170s during breakdown? Which is something I'm not sure of.
The spike occurs because the inductor tries to keep the same 1A current going. But now there's nowhere for it to go so the voltage rises. Not infinitely as it would leak somepleace else presumably. But let's say it rises above 60V and something rising about 60V causes avalanche breakdown.?
So then consider another way to calculate it - during breakdown if the resistance falls to zero (assume) then the power dissipation in the transistor is zero since current x voltage is zero. The energy is then dissipated in the effective internal resistance of the power supply rails or everything else attached to the power supply rails? Or??
So the resistance in breakdown isn't zero. Then equivalently the voltage won't fall to zero either. But neither will it stay at 60V. The voltage during breakdown (and dissipation of the inductor energy) will be determined by the current flow and the resistance of the transistor.?
Still another way to think of it... the energy storage in an inductor is 1/2 L I^2. Since 10 turns on an FT37-43 is 34uA the energy storage for a 1A current flow is 17uJ (micro joules). If this really was all dissipated as heat in the transistors in 1/2 us then the power dissipation would be 17 uJ / 0.5 us = 34W. But practically speaking I see no way to estimate how much of the energy is dissipated as heat in the transistor, compared to other places such as the power supply impedance or even the load on that side of the broadband transformer.?
A further difficulty is that avalanche breakdown is, I believe, a somewhat random process; this combined with the device variations means that all four transistors are unlikely to avalanche together. So most likely only one of the transistors avalanches.
Yet another way to think of it is just a current spike into the transistor... if the transistors can take pulse currents of 1.2A for 300us it seems unlikely 1A for 0.5us would cause damage. Unless avalanche conditions are substantially different from regular pulse current.
Where does the 0.5us come from anyway?
An interesting related observation. At some point in my QCX 50W PA kit development during TX as well as the side tone I could hear this random crackling. In the end I realized the BS170 transistor I had switching the receive bypass PIN diode switch, was seeing the full DC generated from the rectified/doubled RF and that was in excess of it's breakdown. That was when I realized I need a transistor with a higher Vds. Attached are some scope screenshots I took at the time. The actual voltage at which breakdown occurred was a lot more than 60V (vertical resolution is 50V/Div). The BS170s were fine, the crackling audio was too annoying through.
Then again what does the datasheet even mean here? It says Drain Source breakdown is 60V, with conditions Vgs = 0 and Id = 100uA. But how to we relate this to the real world and what happens in other conditions? How could there be 100uA Id anyway if Vgs = 0?
Overall it seems to me there are rather many unknowns and it is difficult to make any predictive calculations. But intuively, which may be a totally wrong feeling, it seems unlikely to me that this avalanche breakdown, if it occurs, would damage or increase the probability of damage of the transistor.?
I don't know. It's not a simple case. Maybe this is why I like the suck-it-and-see approach so much. Because with so many unknowns and complexities how can we be sure the model even approximates the reality? Particularly in situations like this where it's hard to make an experimental observation that could confirm the model under specific conditions and give some confidence. Without being able to tie it other than loosely to observations it seems to me we end up rather uncomfortably deep into speculation territory.
Chris, & all here,
Here is a fun exercise:
Assume that no commutating diode is present at L14 in QDX, and that,
for convenience, the PA is drawing 1A through L14. Then at the moment
the last BS170 transistors switch off, a large V=Ldi/dt spike will
take the transistors to avalanche at 60 Volts. At that moment, 60
watts of instantaneous power (60V x? 1A) is being dumped into the four
of them. It will last about half a microsecond in duration.
If 47 volt Zeners are intended to absorb the spike, the calculation
results in half an amp flowing to each zener, and 47 watts being
delivered to the pair. The Zener resistance will allow for
considerable elevation of the drain voltage above 47 volts, and the 60
volt avalanche spec may still be violated in spite of the presence of
the Zeners, assuming they can even handle the instantaneous power
dump.
Hans has shown through bench testing that high voltages are extremely
unlikely as the result of high SWR. Simulations done by Evan and I
agree with his bench testing. It seems the Zener diodes offer no real
value here, and that the L14 commutating diode is a better overall
protective solution. It can handle the L14 current easily, which
becomes more important under low Z / high SWR conditions.
More musings from one of those pesky nerds...
JZ KJ4A
On Thu, Sep 14, 2023 at 10:56?AM Chris <chris.rowland@...> wrote:
>
> On 14/09/2023 12:49, Jerry wrote:
>
> Which BS120 Fails?
>
> My low band QDX has just failed for the fourth time.? Each time only Q9 has failed (shorted D-S).? No other parts have been replaced.
>
> My high band QDX also failed once, also Q9.
>
> Both are built as 9 volt version, power is via a linear LM317 regulator with 1 amp current limiting.
>
> Any ideas why this one part is more stressed than others?
>
> I have just added 1N4148 diodes as in the QMX schematic.
>
> You do realise that the diodes between the source and drain of the BS170s in the QMX schematic (D503 and D504) are place holders for Zener diodes?
>
> You need something like 1N4756 47V Zener diodes there.
>
> The place for an 1N4148 is across the inductor that drives them, L502.
>
> Chris, G5CTH
>
>
>
|
Hans,
For any dissipative non-reactive component, linear or not, the power being dissipated at any instant is V(across) x I(through), period. It's that simple.
There are not many places for the stored inductive energy to go other than the dissipative nature of the avalanching transistor or the Zener. Everything else fades to insignificance in comparison.
The Zener dissipation is actually underestimated in my discussion, as I wanted to keep things simple. Here. V(across) will be the Zener potential plus the IR drop of the Zener internal resistance. It is that IR drop that renders the Zener diode potentially unable to limit the Vds at the transistor to safe values. Simulation using the 1N4756 library model agrees. Note also? the peak current spec for the 1N4756: a mere 95 mA. It will not appreciate having half an amp coming it's way!
My discussion is augmented by simulation experience. The 0.5 microsecond time comes from simulation. The instantaneous Pd is also confirmed in simulation.
?Your calculation of Pd is in the ballpark with mine and with simulation, too. I think you meant to call the inductor value 34uH (not uA), but OK. I use 35 uH in my simulation. It does not enter into the mind experiment calculation, though, as the current I remaining unchanged at the moment of transistor shutdown is what is important.
One important factor that the mind experiment does not capture, but which simulation does, is the effect of non-instantaneous transistor turn-off. That definitely takes some of the punch out of V=Ldi/dt by reducing di/dt.
Is there enough energy present here to fry a transistor in avalanche? That is a great question, but it is probably the wrong question.
?As I have said on previous occasions when we have discussed this, the right question is whether there is sufficient voltage present to break down the gate oxide of the BS170. I believe the answer is solidly Yes. That breakdown can take place in an instant.
It is unfortunate that the worst case for the voltage spike scenario is also the worst case for the high SWR scenario, ie. very low load impedance. It complicates the conversation and provides a fig leaf to hide behind.
Your friendly hand-wringing nerd :-) JZ
Hello John
Isn't the calculation fraught with difficulties? Where did the 0.5 microseconds assumption come from? Wouldn't that depend on the effective resistance to ground of the parallel BS170s during breakdown? Which is something I'm not sure of.
The spike occurs because the inductor tries to keep the same 1A current going. But now there's nowhere for it to go so the voltage rises. Not infinitely as it would leak somepleace else presumably. But let's say it rises above 60V and something rising about 60V causes avalanche breakdown.?
So then consider another way to calculate it - during breakdown if the resistance falls to zero (assume) then the power dissipation in the transistor is zero since current x voltage is zero. The energy is then dissipated in the effective internal resistance of the power supply rails or everything else attached to the power supply rails? Or??
So the resistance in breakdown isn't zero. Then equivalently the voltage won't fall to zero either. But neither will it stay at 60V. The voltage during breakdown (and dissipation of the inductor energy) will be determined by the current flow and the resistance of the transistor.?
Still another way to think of it... the energy storage in an inductor is 1/2 L I^2. Since 10 turns on an FT37-43 is 34uA the energy storage for a 1A current flow is 17uJ (micro joules). If this really was all dissipated as heat in the transistors in 1/2 us then the power dissipation would be 17 uJ / 0.5 us = 34W. But practically speaking I see no way to estimate how much of the energy is dissipated as heat in the transistor, compared to other places such as the power supply impedance or even the load on that side of the broadband transformer.?
A further difficulty is that avalanche breakdown is, I believe, a somewhat random process; this combined with the device variations means that all four transistors are unlikely to avalanche together. So most likely only one of the transistors avalanches.
Yet another way to think of it is just a current spike into the transistor... if the transistors can take pulse currents of 1.2A for 300us it seems unlikely 1A for 0.5us would cause damage. Unless avalanche conditions are substantially different from regular pulse current.
Where does the 0.5us come from anyway?
An interesting related observation. At some point in my QCX 50W PA kit development during TX as well as the side tone I could hear this random crackling. In the end I realized the BS170 transistor I had switching the receive bypass PIN diode switch, was seeing the full DC generated from the rectified/doubled RF and that was in excess of it's breakdown. That was when I realized I need a transistor with a higher Vds. Attached are some scope screenshots I took at the time. The actual voltage at which breakdown occurred was a lot more than 60V (vertical resolution is 50V/Div). The BS170s were fine, the crackling audio was too annoying through.
Then again what does the datasheet even mean here? It says Drain Source breakdown is 60V, with conditions Vgs = 0 and Id = 100uA. But how to we relate this to the real world and what happens in other conditions? How could there be 100uA Id anyway if Vgs = 0?
Overall it seems to me there are rather many unknowns and it is difficult to make any predictive calculations. But intuively, which may be a totally wrong feeling, it seems unlikely to me that this avalanche breakdown, if it occurs, would damage or increase the probability of damage of the transistor.?
I don't know. It's not a simple case. Maybe this is why I like the suck-it-and-see approach so much. Because with so many unknowns and complexities how can we be sure the model even approximates the reality? Particularly in situations like this where it's hard to make an experimental observation that could confirm the model under specific conditions and give some confidence. Without being able to tie it other than loosely to observations it seems to me we end up rather uncomfortably deep into speculation territory. Chris, & all here,
Here is a fun exercise:
Assume that no commutating diode is present at L14 in QDX, and that,
for convenience, the PA is drawing 1A through L14. Then at the moment
the last BS170 transistors switch off, a large V=Ldi/dt spike will
take the transistors to avalanche at 60 Volts. At that moment, 60
watts of instantaneous power (60V x? 1A) is being dumped into the four
of them. It will last about half a microsecond in duration.
If 47 volt Zeners are intended to absorb the spike, the calculation
results in half an amp flowing to each zener, and 47 watts being
delivered to the pair. The Zener resistance will allow for
considerable elevation of the drain voltage above 47 volts, and the 60
volt avalanche spec may still be violated in spite of the presence of
the Zeners, assuming they can even handle the instantaneous power
dump.
Hans has shown through bench testing that high voltages are extremely
unlikely as the result of high SWR. Simulations done by Evan and I
agree with his bench testing. It seems the Zener diodes offer no real
value here, and that the L14 commutating diode is a better overall
protective solution. It can handle the L14 current easily, which
becomes more important under low Z / high SWR conditions.
More musings from one of those pesky nerds...
JZ KJ4A
On Thu, Sep 14, 2023 at 10:56?AM Chris <chris.rowland@...> wrote:
>
> On 14/09/2023 12:49, Jerry wrote:
>
> Which BS120 Fails?
>
> My low band QDX has just failed for the fourth time.? Each time only Q9 has failed (shorted D-S).? No other parts have been replaced.
>
> My high band QDX also failed once, also Q9.
>
> Both are built as 9 volt version, power is via a linear LM317 regulator with 1 amp current limiting.
>
> Any ideas why this one part is more stressed than others?
>
> I have just added 1N4148 diodes as in the QMX schematic.
>
> You do realise that the diodes between the source and drain of the BS170s in the QMX schematic (D503 and D504) are place holders for Zener diodes?
>
> You need something like 1N4756 47V Zener diodes there.
>
> The place for an 1N4148 is across the inductor that drives them, L502.
>
> Chris, G5CTH
>
>
>
|
Not sure if I'm the Chris but...
I was pointing out that the diodes
mentioned needed to be Zeners, where they were positioned.
What I don't understand is how people
claim to be continually blowing BS170s, fit diodes, and stop doing
so. I've seen no explanation not even any acknowledgement.
Hans' experiments are designed to
demonstrate no failure and so do so.
This may not be consciously
intentional, systems are very good at teaching people to avoid
problems. We saw this at work, somebody new would find all sorts
of bugs that the development, sales, marketing, and service people
had all missed. We would check and indeed there was a problem,
which we fixed of course.? We would keep people away from a new
product, then introduce it at the late beta phase and pay
attention to what they found and how they reacted to it. A red
light was "It's OK once you get used to it".? This is the sort of
professionalism I'm used to.
An experiment I'd be interested in
would be a length of unterminated coax and a frequency sweep from
the 1/4 to 1/2 wave positions at least, including both.? At what
point does it fail and why?
Chris, G5CTH
On 14/09/2023 16:17, John Z wrote:
toggle quoted message
Show quoted text
Chris, & all here,
Here is a fun exercise:
Assume that no commutating diode is present at L14 in QDX, and that,
for convenience, the PA is drawing 1A through L14. Then at the moment
the last BS170 transistors switch off, a large V=Ldi/dt spike will
take the transistors to avalanche at 60 Volts. At that moment, 60
watts of instantaneous power (60V x 1A) is being dumped into the four
of them. It will last about half a microsecond in duration.
If 47 volt Zeners are intended to absorb the spike, the calculation
results in half an amp flowing to each zener, and 47 watts being
delivered to the pair. The Zener resistance will allow for
considerable elevation of the drain voltage above 47 volts, and the 60
volt avalanche spec may still be violated in spite of the presence of
the Zeners, assuming they can even handle the instantaneous power
dump.
Hans has shown through bench testing that high voltages are extremely
unlikely as the result of high SWR. Simulations done by Evan and I
agree with his bench testing. It seems the Zener diodes offer no real
value here, and that the L14 commutating diode is a better overall
protective solution. It can handle the L14 current easily, which
becomes more important under low Z / high SWR conditions.
More musings from one of those pesky nerds...
JZ KJ4A
On Thu, Sep 14, 2023 at 10:56?AM Chris <chris.rowland@...> wrote:
On 14/09/2023 12:49, Jerry wrote:
Which BS120 Fails?
My low band QDX has just failed for the fourth time. Each time only Q9 has failed (shorted D-S). No other parts have been replaced.
My high band QDX also failed once, also Q9.
Both are built as 9 volt version, power is via a linear LM317 regulator with 1 amp current limiting.
Any ideas why this one part is more stressed than others?
I have just added 1N4148 diodes as in the QMX schematic.
You do realise that the diodes between the source and drain of the BS170s in the QMX schematic (D503 and D504) are place holders for Zener diodes?
You need something like 1N4756 47V Zener diodes there.
The place for an 1N4148 is across the inductor that drives them, L502.
Chris, G5CTH
|
Hans said:
¡°Then again what does the datasheet even mean here? It says Drain Source breakdown is 60V, with conditions Vgs = 0 and Id = 100uA. But how to we relate this to the real world and what happens in other conditions? How could there be 100uA Id anyway if Vgs = 0?¡±
Hans, Vgs=0 doesn¡¯t mean that no drain current can flow. All devices exhibit leakage current, even when they¡¯re supposedly ¡°off¡±. This spec is clear - it means that as Vds is increased, leakage current increases. The breakdown voltage is where the drain current begins to increase dramatically, aka, the knee of the curve. Further increases in Vds will cause a great deal of current to flow because the field effect block due to zero gate voltage has been overwhelmed.
Tony - AC9QY
Hello John
Isn't the calculation fraught with difficulties? Where did the 0.5 microseconds assumption come from? Wouldn't that depend on the effective resistance to ground of the parallel BS170s during breakdown? Which is something I'm not sure of.
The spike occurs because the inductor tries to keep the same 1A current going. But now there's nowhere for it to go so the voltage rises. Not infinitely as it would leak somepleace else presumably. But let's say it rises above 60V and something rising about 60V causes avalanche breakdown.?
So then consider another way to calculate it - during breakdown if the resistance falls to zero (assume) then the power dissipation in the transistor is zero since current x voltage is zero. The energy is then dissipated in the effective internal resistance of the power supply rails or everything else attached to the power supply rails? Or??
So the resistance in breakdown isn't zero. Then equivalently the voltage won't fall to zero either. But neither will it stay at 60V. The voltage during breakdown (and dissipation of the inductor energy) will be determined by the current flow and the resistance of the transistor.?
Still another way to think of it... the energy storage in an inductor is 1/2 L I^2. Since 10 turns on an FT37-43 is 34uA the energy storage for a 1A current flow is 17uJ (micro joules). If this really was all dissipated as heat in the transistors in 1/2 us then the power dissipation would be 17 uJ / 0.5 us = 34W. But practically speaking I see no way to estimate how much of the energy is dissipated as heat in the transistor, compared to other places such as the power supply impedance or even the load on that side of the broadband transformer.?
A further difficulty is that avalanche breakdown is, I believe, a somewhat random process; this combined with the device variations means that all four transistors are unlikely to avalanche together. So most likely only one of the transistors avalanches.
Yet another way to think of it is just a current spike into the transistor... if the transistors can take pulse currents of 1.2A for 300us it seems unlikely 1A for 0.5us would cause damage. Unless avalanche conditions are substantially different from regular pulse current.
Where does the 0.5us come from anyway?
An interesting related observation. At some point in my QCX 50W PA kit development during TX as well as the side tone I could hear this random crackling. In the end I realized the BS170 transistor I had switching the receive bypass PIN diode switch, was seeing the full DC generated from the rectified/doubled RF and that was in excess of it's breakdown. That was when I realized I need a transistor with a higher Vds. Attached are some scope screenshots I took at the time. The actual voltage at which breakdown occurred was a lot more than 60V (vertical resolution is 50V/Div). The BS170s were fine, the crackling audio was too annoying through.
Then again what does the datasheet even mean here? It says Drain Source breakdown is 60V, with conditions Vgs = 0 and Id = 100uA. But how to we relate this to the real world and what happens in other conditions? How could there be 100uA Id anyway if Vgs = 0?
Overall it seems to me there are rather many unknowns and it is difficult to make any predictive calculations. But intuively, which may be a totally wrong feeling, it seems unlikely to me that this avalanche breakdown, if it occurs, would damage or increase the probability of damage of the transistor.?
I don't know. It's not a simple case. Maybe this is why I like the suck-it-and-see approach so much. Because with so many unknowns and complexities how can we be sure the model even approximates the reality? Particularly in situations like this where it's hard to make an experimental observation that could confirm the model under specific conditions and give some confidence. Without being able to tie it other than loosely to observations it seems to me we end up rather uncomfortably deep into speculation territory. Chris, & all here,
Here is a fun exercise:
Assume that no commutating diode is present at L14 in QDX, and that,
for convenience, the PA is drawing 1A through L14. Then at the moment
the last BS170 transistors switch off, a large V=Ldi/dt spike will
take the transistors to avalanche at 60 Volts. At that moment, 60
watts of instantaneous power (60V x? 1A) is being dumped into the four
of them. It will last about half a microsecond in duration.
If 47 volt Zeners are intended to absorb the spike, the calculation
results in half an amp flowing to each zener, and 47 watts being
delivered to the pair. The Zener resistance will allow for
considerable elevation of the drain voltage above 47 volts, and the 60
volt avalanche spec may still be violated in spite of the presence of
the Zeners, assuming they can even handle the instantaneous power
dump.
Hans has shown through bench testing that high voltages are extremely
unlikely as the result of high SWR. Simulations done by Evan and I
agree with his bench testing. It seems the Zener diodes offer no real
value here, and that the L14 commutating diode is a better overall
protective solution. It can handle the L14 current easily, which
becomes more important under low Z / high SWR conditions.
More musings from one of those pesky nerds...
JZ KJ4A
On Thu, Sep 14, 2023 at 10:56?AM Chris <chris.rowland@...> wrote:
>
> On 14/09/2023 12:49, Jerry wrote:
>
> Which BS120 Fails?
>
> My low band QDX has just failed for the fourth time.? Each time only Q9 has failed (shorted D-S).? No other parts have been replaced.
>
> My high band QDX also failed once, also Q9.
>
> Both are built as 9 volt version, power is via a linear LM317 regulator with 1 amp current limiting.
>
> Any ideas why this one part is more stressed than others?
>
> I have just added 1N4148 diodes as in the QMX schematic.
>
> You do realise that the diodes between the source and drain of the BS170s in the QMX schematic (D503 and D504) are place holders for Zener diodes?
>
> You need something like 1N4756 47V Zener diodes there.
>
> The place for an 1N4148 is across the inductor that drives them, L502.
>
> Chris, G5CTH
>
>
>
|
Thanks Tony
Very useful into to know. 73 Hans G0UPL
On Thu, Sep 14, 2023, 9:30 PM Tony Scaminaci < tonyscam@...> wrote: Hans said:
¡°Then again what does the datasheet even mean here? It says Drain Source breakdown is 60V, with conditions Vgs = 0 and Id = 100uA. But how to we relate this to the real world and what happens in other conditions? How could there be 100uA Id anyway if Vgs = 0?¡±
Hans, Vgs=0 doesn¡¯t mean that no drain current can flow. All devices exhibit leakage current, even when they¡¯re supposedly ¡°off¡±. This spec is clear - it means that as Vds is increased, leakage current increases. The breakdown voltage is where the drain current begins to increase dramatically, aka, the knee of the curve. Further increases in Vds will cause a great deal of current to flow because the field effect block due to zero gate voltage has been overwhelmed.
Tony - AC9QY Hello John
Isn't the calculation fraught with difficulties? Where did the 0.5 microseconds assumption come from? Wouldn't that depend on the effective resistance to ground of the parallel BS170s during breakdown? Which is something I'm not sure of.
The spike occurs because the inductor tries to keep the same 1A current going. But now there's nowhere for it to go so the voltage rises. Not infinitely as it would leak somepleace else presumably. But let's say it rises above 60V and something rising about 60V causes avalanche breakdown.?
So then consider another way to calculate it - during breakdown if the resistance falls to zero (assume) then the power dissipation in the transistor is zero since current x voltage is zero. The energy is then dissipated in the effective internal resistance of the power supply rails or everything else attached to the power supply rails? Or??
So the resistance in breakdown isn't zero. Then equivalently the voltage won't fall to zero either. But neither will it stay at 60V. The voltage during breakdown (and dissipation of the inductor energy) will be determined by the current flow and the resistance of the transistor.?
Still another way to think of it... the energy storage in an inductor is 1/2 L I^2. Since 10 turns on an FT37-43 is 34uA the energy storage for a 1A current flow is 17uJ (micro joules). If this really was all dissipated as heat in the transistors in 1/2 us then the power dissipation would be 17 uJ / 0.5 us = 34W. But practically speaking I see no way to estimate how much of the energy is dissipated as heat in the transistor, compared to other places such as the power supply impedance or even the load on that side of the broadband transformer.?
A further difficulty is that avalanche breakdown is, I believe, a somewhat random process; this combined with the device variations means that all four transistors are unlikely to avalanche together. So most likely only one of the transistors avalanches.
Yet another way to think of it is just a current spike into the transistor... if the transistors can take pulse currents of 1.2A for 300us it seems unlikely 1A for 0.5us would cause damage. Unless avalanche conditions are substantially different from regular pulse current.
Where does the 0.5us come from anyway?
An interesting related observation. At some point in my QCX 50W PA kit development during TX as well as the side tone I could hear this random crackling. In the end I realized the BS170 transistor I had switching the receive bypass PIN diode switch, was seeing the full DC generated from the rectified/doubled RF and that was in excess of it's breakdown. That was when I realized I need a transistor with a higher Vds. Attached are some scope screenshots I took at the time. The actual voltage at which breakdown occurred was a lot more than 60V (vertical resolution is 50V/Div). The BS170s were fine, the crackling audio was too annoying through.
Then again what does the datasheet even mean here? It says Drain Source breakdown is 60V, with conditions Vgs = 0 and Id = 100uA. But how to we relate this to the real world and what happens in other conditions? How could there be 100uA Id anyway if Vgs = 0?
Overall it seems to me there are rather many unknowns and it is difficult to make any predictive calculations. But intuively, which may be a totally wrong feeling, it seems unlikely to me that this avalanche breakdown, if it occurs, would damage or increase the probability of damage of the transistor.?
I don't know. It's not a simple case. Maybe this is why I like the suck-it-and-see approach so much. Because with so many unknowns and complexities how can we be sure the model even approximates the reality? Particularly in situations like this where it's hard to make an experimental observation that could confirm the model under specific conditions and give some confidence. Without being able to tie it other than loosely to observations it seems to me we end up rather uncomfortably deep into speculation territory. Chris, & all here,
Here is a fun exercise:
Assume that no commutating diode is present at L14 in QDX, and that,
for convenience, the PA is drawing 1A through L14. Then at the moment
the last BS170 transistors switch off, a large V=Ldi/dt spike will
take the transistors to avalanche at 60 Volts. At that moment, 60
watts of instantaneous power (60V x? 1A) is being dumped into the four
of them. It will last about half a microsecond in duration.
If 47 volt Zeners are intended to absorb the spike, the calculation
results in half an amp flowing to each zener, and 47 watts being
delivered to the pair. The Zener resistance will allow for
considerable elevation of the drain voltage above 47 volts, and the 60
volt avalanche spec may still be violated in spite of the presence of
the Zeners, assuming they can even handle the instantaneous power
dump.
Hans has shown through bench testing that high voltages are extremely
unlikely as the result of high SWR. Simulations done by Evan and I
agree with his bench testing. It seems the Zener diodes offer no real
value here, and that the L14 commutating diode is a better overall
protective solution. It can handle the L14 current easily, which
becomes more important under low Z / high SWR conditions.
More musings from one of those pesky nerds...
JZ KJ4A
On Thu, Sep 14, 2023 at 10:56?AM Chris <chris.rowland@...> wrote:
>
> On 14/09/2023 12:49, Jerry wrote:
>
> Which BS120 Fails?
>
> My low band QDX has just failed for the fourth time.? Each time only Q9 has failed (shorted D-S).? No other parts have been replaced.
>
> My high band QDX also failed once, also Q9.
>
> Both are built as 9 volt version, power is via a linear LM317 regulator with 1 amp current limiting.
>
> Any ideas why this one part is more stressed than others?
>
> I have just added 1N4148 diodes as in the QMX schematic.
>
> You do realise that the diodes between the source and drain of the BS170s in the QMX schematic (D503 and D504) are place holders for Zener diodes?
>
> You need something like 1N4756 47V Zener diodes there.
>
> The place for an 1N4148 is across the inductor that drives them, L502.
>
> Chris, G5CTH
>
>
>
|
Hans' experiments are designed to
demonstrate no failure and so do so.
I'm nowhere near so easily offended or anything, and I do very well know and understand exactly the bias tendency you mention. Nevertheless I think it might still be a little harsh.?
It was the same in my former Software Dev career as it is here. The very best types of problem reports are when there's a very clear definition of the problem and how to reproduce it. Whether hardware or firmware. Then I can reproduce it here and try and resolve it and re-test.?
I remember an Iambic keying problem early in the QCX days when two Israeli hams as I recall, provided me with the most concise description I've ever seen. Complete with oscilloscope traces, and a comparison with a real Curtis keyer chip, which is kind of the golden standard of Iambicism. It wasn't an easy problem to solve but it was very well defined and that made it 100x easier than it otherwise would have been.?
A more recent example, relating to QMX, was Gunnar's sudden supply voltage switch from 6 to 12V and again it's so well defined that I could set up a test jig, reproduce it, and find firmware improvements to largely mitigate it.?
Same when you go to the doctor. The more info you can provide the better chance the quack has of correctly diagnosing your condition and fixing it.?
But what we have here instead is a very nebulous problem. If indeed a problem at all. Some people with BS170 failures. Others use them for years without trouble (me included). Some talk about BS170 drain voltage and how bad SWRs of certain types could make the voltages, which find their way back through the filters and switches and output transformer to the BS170 drains and fry them. Specifically the high voltage case is important because it is very hard to protect against fast enough, whereas high current, if reasonably moderate, is a slower death by overheating and therefore in QMX it would be possible to act against it fast enough.
So far the only precisely defined scenarios are the open load and short load. This is why I tested them. Other recommendations largely boil down to twiddling LC matches on some uncool antenna etc. Which I can do but it isn't as well defined and likely to take more time, which is why I haven't got to it yet.?
I do sometimes wonder if I'm the only one here with a QDX and an average oscilloscope and an average skill to wield it... at the least the number of QDX + oscilloscope owners seems low in comparison to simulators... maybe I'm in the wrong business hi hi.?
The issue of the dI/dt inflicted spike caused by the sudden key-up of QDX is also interesting, and I can reproduce it and take scope shots of it... it kind of has a bearing on the other BS170 problems if we speculate that an already suffering transistor could be even more vulnerable to SWR problems. It's for such a short duration it isn't clear it's really a threat. Or a contributing factor...
Be clear also again, that this spike isn't an issue on QMX because of its envelope shaping.?
There seem so many unknowns. It's not a nice problem at all. It's not even clear if it's actually a real problem. The BS170 issue could simply be a straightforward matter of too much current in low impedance SWR mismatch situations causing overheating, for example. Throughout amateur radio history, damaging or blowing finals isn't totally an uncommon thing.?
73 Hans G0UPL
L
|
Thanks for the correction. Yes, I did put in 47 v zeners and not 1N4148 as I typed. My error. ?
I was more interested in responses to the main Question of my post ¡ª why should only Q9 be the failure multiple times? ?
Jerry N6JH
|
Bruce Akhurst