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Newly assembled QMX, SWR=1.5 with dummy load

 

Hey,
?
I just assembled this a few days ago, made a couple of contacts with an end-fed. What I found odd is that when I attached a 50 ohm dummy load to the output and measured SWR, it was around 1.5 on 20m and <2 on other bands. Surely it shouldn't be that? What may have I done wrong? Should I worry?
?
73 de HF2J
 

Jan, First I would check the dummy load with a digital ohm meter. Is it actually 50 ohms? Next double check whatever cable or jumper you are using to attach the QMX to the dummy load. Check for shorts and/or opens. Double check your solder connections on the RF output of your QMX for cold solder joints or bridges.
Dean - KC9REN
 

Hi Jan,
Check the winding count of the transformer T507. Check for its proper soldering to the PCB.
?
 

I concur with the previous recommendations.
SWR into a dummy load from the QMX should show 1.0x:1 on the lower bands, increasing to perhaps 1.1x on the highest band.
 

The resistor is 50.7 ohm (if you trust the $20 multimeter), and 50.9 ohm with the cable attached. Solder joints appear to be solid, but I'll inspect it once again when I find my magnifying glass. The number of windings is correct. The L401 transformer has some looseness on one of the windings (1mm) - would that matter?

Also, it appears that the SWR is a function of power. It is greater when I measure it at 10/20% and lower when it is 50%. Is this normal?

Thanks,
J

On Mon, Apr 21, 2025 at 6:45?PM Stan Dye via <standye=[email protected]> wrote:
I concur with the previous recommendations.
SWR into a dummy load from the QMX should show 1.0x:1 on the lower bands, increasing to perhaps 1.1x on the highest band.
 

Are you using a dummy load or a resistor for a dummy load? If using a resistor what type? It would make a difference as some types of resistor are not suitable for RF.
 

I'm using an SMA terminator?which came with my NanoVNA.


On Mon, Apr 21, 2025 at 9:04?PM Richard Dyer via <rdyer39=[email protected]> wrote:
Are you using a dummy load or a resistor for a dummy load? If using a resistor what type? It would make a difference as some types of resistor are not suitable for RF.
 

On Mon, Apr 21, 2025 at 11:45 AM, Jan Szumiec wrote:
The L401 transformer has some looseness on one of the windings (1mm) - would that matter?
No, that won't hurt anything, and does not affect transmit (T401 is only used on receive).
Also, it appears that the SWR is a function of power. It is greater when I measure it at 10/20% and lower when it is 50%. Is this normal?
No, SWR is not a function of power, if everything is operating correctly.? The SMA terminator from your nanovna kit should work fine at the lower powers, but is not built for 5W - so don't use it with full power, or you may burn it out.
?
The SWR and power displayed by the QMX are both read from the bridge circuit, of which T507 is the principle part.? What the CPU actually reads from that bridge is the peak forward and reverse voltage (divided by 10), from which it calculates power and swr.? So if one of the solder joints to T507 is a bit 'flaky' or 'cold', it could cause either the forward or reverse voltage to be incorrect or 'noisy', causing incorrect measurements.? So check those and re-flow them, leaving the iron on the connection for several seconds in case there is still some enamel that needs burned away from the wires.? [There are also SMD resistors, 2 diodes and 2 capacitors in that bridge circuit - it is less likely, but if one of those have been disturbed, it could also cause this issue.]
 

Thank you. I'll try that tomorrow.


On Mon, Apr 21, 2025, 22:10 Stan Dye via <standye=[email protected]> wrote:
On Mon, Apr 21, 2025 at 11:45 AM, Jan Szumiec wrote:
The L401 transformer has some looseness on one of the windings (1mm) - would that matter?
No, that won't hurt anything, and does not affect transmit (T401 is only used on receive).
Also, it appears that the SWR is a function of power. It is greater when I measure it at 10/20% and lower when it is 50%. Is this normal?
No, SWR is not a function of power, if everything is operating correctly.? The SMA terminator from your nanovna kit should work fine at the lower powers, but is not built for 5W - so don't use it with full power, or you may burn it out.
?
The SWR and power displayed by the QMX are both read from the bridge circuit, of which T507 is the principle part.? What the CPU actually reads from that bridge is the peak forward and reverse voltage (divided by 10), from which it calculates power and swr.? So if one of the solder joints to T507 is a bit 'flaky' or 'cold', it could cause either the forward or reverse voltage to be incorrect or 'noisy', causing incorrect measurements.? So check those and re-flow them, leaving the iron on the connection for several seconds in case there is still some enamel that needs burned away from the wires.? [There are also SMD resistors, 2 diodes and 2 capacitors in that bridge circuit - it is less likely, but if one of those have been disturbed, it could also cause this issue.]
 

"I'm using an SMA terminator which came with my NanoVNA."
?
Are you shure that this dummy-load can withstand the output-power of the QMX (up to 5W!)?
?
73, Wolfgang

 

¿ªÔÆÌåÓý

The SMA terminator which comes with the NanoVNA can only withstand 1W, maybe 2W Max.?

From: [email protected] <[email protected]> on behalf of Wolfgang, DF1KZ via groups.io <df1kz@...>
Sent: Monday, April 21, 2025 10:58:54 PM
To: [email protected] <[email protected]>
Subject: Re: [QRPLabs] Newly assembled QMX, SWR=1.5 with dummy load
?

I'm using an SMA terminator which came with my NanoVNA.
?
Are you shure that this dummy-load can withstand the output-power of the QMX (up to 5W!)
 

¿ªÔÆÌåÓý

For a few seconds maybe. ?It is a small chip capacitor in a tiny metal case. Not designed to dissipate any power.

Real Dummy Loads are easy to make. ?Twenty 1k quarter watt resistors in parallel make 50 Ohms and will safely dissipate five watts if they all have good air separation.

Dave

On Apr 21, 2025, at 17:06, bobfox via groups.io <oe7oer@...> wrote:

?
The SMA terminator which comes with the NanoVNA can only withstand 1W, maybe 2W Max.?

From: [email protected] <[email protected]> on behalf of Wolfgang, DF1KZ via groups.io <df1kz@...>
Sent: Monday, April 21, 2025 10:58:54 PM
To: [email protected] <[email protected]>
Subject: Re: [QRPLabs] Newly assembled QMX, SWR=1.5 with dummy load
?

I'm using an SMA terminator which came with my NanoVNA.
?
Are you shure that this dummy-load can withstand the output-power of the QMX (up to 5W!)
 

On Mon, Apr 21, 2025 at 05:04 PM, Wolfgang, DF1KZ wrote:
Are you shure that this dummy-load can withstand the output-power of the QMX (up to 5W!)?
It is also poor lab practice to use precision calibration kit standards as adapters or dummy loads.? They are precision instruments and should be given requisite care.??
10 watt 30dB attenuators are less than $20 on AMZN and likely stupidly cheap from China, they will work as a dummy load or attenuator.??
?
No ham station should be without a suitable dummy load, anyway, IMO.? Suitable meaning able to handle full output power for long enough to make meaningful measurements.
73, Don N2VGU
?
 

I agree.
Never use those dinky 50 ohm calibration references as a transmitter dummy load.
Unless your transmitter is sending out a max of a tenth Watt or so.
?
Hans has a nice 20W dummy load for $8.50,? perhaps include that with your next order.
https://qrp-labs.com/dummy.html
?
A cheap dummy load can be made with a bunch of 3 Watt metal film resistors in parallel,?
For example, four 200 ohm parts in parallel can deal with 5 Watts all day long, and cost about $0.25 each.
Mouser part number 603-RSF3WSJR-73-200R? is a good choice.
For 50 Watts, perhaps use twenty 1000 ohm 3 Watt resistors, Mouser 603-RSF3WSJT-52-1K.
?
If you already have a dummy load and want to use it as a 30 dB attenuator, you can do something like this:
?

in ---+--------R------+------- out to coax
? ? ? | ? ? ? ? ? ? ? |
? ? 50ohms? ? ? ? ? 50ohms
? ? ? |? ? ? ? ? ? ? ?|
? ? ?GND ? ? ? ? ? ? GND
?
The first 50 ohm resistor is your dummy load, it must be high power to deal with the transmitter.
R should be 1/2 Watt for a 30dB attenuator, can be 1/8 Watt for a 40dB attenuator, value of R calculated below.
The second 50 ohm resistor at the output can be 1/8 Watt.
I assume you are sending the sample out to something like a tinySA over coax.
The second 50 ohm resistor is the source termination to absorb any reflections coming back from the coax
There is a third 50 ohm resistor inside the tinySA that is in parallel with the second 50 ohm resistor,
together they make a load of 25 ohms.
For 30dB of attenuation we have a power ratio of 1000, so a voltage ratio of sqrt(1000)=31.62.
Those two parallel 50 ohm resistors in series with resistor R form a voltage divider.
To get a voltage ratio of 31.62, the value of R should be around (31.62 - 1) * 25 = 765.5 ohms for 30dB of attenuation.
For 40dB of attenuation, the value of R should be (sqrt(10000) - 1) * 25 = 2475 ohms.
Anything within 5% or so should be fine for our needs.
?
On the 30dB attenuator, having (765.5+25) ohms in parallel with the 50 ohm dummy load brings it down to? 47 ohms.
So in that case we might be better off with four 220 ohm resistors instead of 200 ohm resistors.
The transmitter now sees 55 ohms in parallel with 765.5+25, or 51.3 ohms, which is good enough for most of us.
If we then add a 2000 ohm resistor in parallel with that 51.3 ohms, we then have the transmitter seeing 50.02 ohms.
On the 40dB attenuator that additional parallel load of (2475+25) is so small it won't matter, so use four 200 ohm resistors.
At 40dB, you need to start being careful with how it's built to avoid unwanted coupling from input to output.
?
Check out that attenuator before using, as you don't want to burn out your tinySA.
I'd just hook the transmitter input port to a 12v battery, check for the a voltage of 12v/31.62 at the output
with a voltmeter.? Add that third 50 ohm resistor before measuring.
?
?
For US residents buying stuff direct from China:
In summary, the de-minimus exemption from tariffs for packages worth less than $800 expires on May 2,
at which point the minimum customs fee will be $100.? On June 2 it rises further to $200.
So you could be spending $210 for that $10 attenuator.
If you place an order now, it will probably arrive at US customs around May 2, so place your bets.
?
Jerry, KE7ER
?
?
On Mon, Apr 21, 2025 at 08:26 PM, Donald S Brant Jr wrote:

On Mon, Apr 21, 2025 at 05:04 PM, Wolfgang, DF1KZ wrote:
Are you shure that this dummy-load can withstand the output-power of the QMX (up to 5W!)?
It is also poor lab practice to use precision calibration kit standards as adapters or dummy loads.? They are precision instruments and should be given requisite care.??
10 watt 30dB attenuators are less than $20 on AMZN and likely stupidly cheap from China, they will work as a dummy load or attenuator.??
?
No ham station should be without a suitable dummy load, anyway, IMO.? Suitable meaning able to handle full output power for long enough to make meaningful measurements.
73, Don N2VGU
 

Thank you all - this is sound advice. I own a dummy load, but am in a different QTH so I used what I had at hand.


On Tue, Apr 22, 2025, 07:46 Jerry Gaffke via <jgaffke=[email protected]> wrote:
I agree.
Never use those dinky 50 ohm calibration references as a transmitter dummy load.
Unless your transmitter is sending out a max of a tenth Watt or so.
?
Hans has a nice 20W dummy load for $8.50,? perhaps include that with your next order.
?
A cheap dummy load can be made with a bunch of 3 Watt metal film resistors in parallel,?
For example, four 200 ohm parts in parallel can deal with 5 Watts all day long, and cost about $0.25 each.
Mouser part number 603-RSF3WSJR-73-200R? is a good choice.
For 50 Watts, perhaps use twenty 1000 ohm 3 Watt resistors, Mouser 603-RSF3WSJT-52-1K.
?
If you already have a dummy load and want to use it as a 30 dB attenuator, you can do something like this:
?

in ---+--------R------+------- out to coax
? ? ? | ? ? ? ? ? ? ? |
? ? 50ohms? ? ? ? ? 50ohms
? ? ? |? ? ? ? ? ? ? ?|
? ? ?GND ? ? ? ? ? ? GND
?
The first 50 ohm resistor is your dummy load, it must be high power to deal with the transmitter.
R should be 1/2 Watt for a 30dB attenuator, can be 1/8 Watt for a 40dB attenuator, value of R calculated below.
The second 50 ohm resistor at the output can be 1/8 Watt.
I assume you are sending the sample out to something like a tinySA over coax.
The second 50 ohm resistor is the source termination to absorb any reflections coming back from the coax
There is a third 50 ohm resistor inside the tinySA that is in parallel with the second 50 ohm resistor,
together they make a load of 25 ohms.
For 30dB of attenuation we have a power ratio of 1000, so a voltage ratio of sqrt(1000)=31.62.
Those two parallel 50 ohm resistors in series with resistor R form a voltage divider.
To get a voltage ratio of 31.62, the value of R should be around (31.62 - 1) * 25 = 765.5 ohms for 30dB of attenuation.
For 40dB of attenuation, the value of R should be (sqrt(10000) - 1) * 25 = 2475 ohms.
Anything within 5% or so should be fine for our needs.
?
On the 30dB attenuator, having (765.5+25) ohms in parallel with the 50 ohm dummy load brings it down to? 47 ohms.
So in that case we might be better off with four 220 ohm resistors instead of 200 ohm resistors.
The transmitter now sees 55 ohms in parallel with 765.5+25, or 51.3 ohms, which is good enough for most of us.
If we then add a 2000 ohm resistor in parallel with that 51.3 ohms, we then have the transmitter seeing 50.02 ohms.
On the 40dB attenuator that additional parallel load of (2475+25) is so small it won't matter, so use four 200 ohm resistors.
At 40dB, you need to start being careful with how it's built to avoid unwanted coupling from input to output.
?
Check out that attenuator before using, as you don't want to burn out your tinySA.
I'd just hook the transmitter input port to a 12v battery, check for the a voltage of 12v/31.62 at the output
with a voltmeter.? Add that third 50 ohm resistor before measuring.
?
?
For US residents buying stuff direct from China:
In summary, the de-minimus exemption from tariffs for packages worth less than $800 expires on May 2,
at which point the minimum customs fee will be $100.? On June 2 it rises further to $200.
So you could be spending $210 for that $10 attenuator.
If you place an order now, it will probably arrive at US customs around May 2, so place your bets.
?
Jerry, KE7ER
?
?
On Mon, Apr 21, 2025 at 08:26 PM, Donald S Brant Jr wrote:
On Mon, Apr 21, 2025 at 05:04 PM, Wolfgang, DF1KZ wrote:
Are you shure that this dummy-load can withstand the output-power of the QMX (up to 5W!)?
It is also poor lab practice to use precision calibration kit standards as adapters or dummy loads.? They are precision instruments and should be given requisite care.??
10 watt 30dB attenuators are less than $20 on AMZN and likely stupidly cheap from China, they will work as a dummy load or attenuator.??
?
No ham station should be without a suitable dummy load, anyway, IMO.? Suitable meaning able to handle full output power for long enough to make meaningful measurements.
73, Don N2VGU
 

Hello all,
?
My best bet is that the 50 ohm SMA load has already been overheated, changing the value of the terminating resistor, which is the cause of the 1:1.5 SWR after all.
?
73,
Roelof, pa0rdt
 

My dummy load uses 4 x 51R 2W SMD resistors connected in series parallel, then 2k5 to 2 x 100R to give a dummy load come 40dB attenuator.? It also has a BAT41 Schottky diode and 0.1 uF capacitor to give the peak reading.? AIUI SMD resistors tend to be lower inductance than through hole ones.
?
A dummy load is about third on my list of essential test gear, first is a multimeter, then a controllable voltage and current power supply with voltage and current display, then the dummy load.
 

Adding Chris's BAT41 and 0.1uF cap to read the peak voltage with a DVM is a good idea, will work great for a QMX.
Han's 20W dummy load does the same, but uses a 1n4004 diode, good for 400 Volts, assembly instructions have the schematic.
The BAT41 is a shottky diode so will be more accurate, but 20W would be close to blowing through it's 100 Volt max spec.
The DVM will see the peak RF voltage from ground, so multiply by 0.707 to get Vrms (the rms voltage of your transmitted signal).
The power out in Watts is then equal to Vrms*Vrms/50, and that should be fairly accurate assuming the RF is a sine wave.
?
Chris has two 100 ohm resistors in parallel to form the 50 ohm on the source end of the coax that goes to
some measuring device such as a tinySA, which we assume has an internal 50 ohm termination resistor.
So his circuit is the same as what I was suggesting for the 50 dB attenuator
His 2.5k resistor is very close to the ideal value of 2475 ohms that I had calculated.
?
At 10 Watts out into 50 ohms, the voltage across the 50 ohms is sqrt(10*50) = 22.4 Vrms.
And the voltage across Chris's two 100 ohm parallel resistors is? ?22.4v/(2500+50) * 50 = 0.44 Vrms.
Power dissipated by that 50 ohms is? 0.44*0.44/50 = 0.004 Watts.
Even ridiculously small 0402 (imperial size) chip resistors are rated for 0.062 Watts.
Two 100 ohm resistors in parallel present half the inductance so will be more accurate at 50 MHz,
but if using a surface mount chip resistor that shouldn't matter much.
?
Jerry, KE7ER

?
?
On Tue, Apr 22, 2025 at 04:38 AM, Chris, G5CTH wrote:

My dummy load uses 4 x 51R 2W SMD resistors connected in series parallel, then 2k5 to 2 x 100R to give a dummy load come 40dB attenuator.? It also has a BAT41 Schottky diode and 0.1 uF capacitor to give the peak reading.? AIUI SMD resistors tend to be lower inductance than through hole ones.
?
A dummy load is about third on my list of essential test gear, first is a multimeter, then a controllable voltage and current power supply with voltage and current display, then the dummy load.
 

If you are worried about series inductance of resistors, best to build a 50 ohm dummy load
of all parallel resistors.? ?Not a series combination of two sets of two 50 ohm resistors in parallel.
?
Jerry, KE7ER
?
On Tue, Apr 22, 2025 at 07:00 AM, Jerry Gaffke wrote:

Adding Chris's BAT41 and 0.1uF cap to read the peak voltage with a DVM is a good idea, will work great for a QMX.
Han's 20W dummy load does the same, but uses a 1n4004 diode, good for 400 Volts, assembly instructions have the schematic.
The BAT41 is a shottky diode so will be more accurate, but 20W would be close to blowing through it's 100 Volt max spec.
The DVM will see the peak RF voltage from ground, so multiply by 0.707 to get Vrms (the rms voltage of your transmitted signal).
The power out in Watts is then equal to Vrms*Vrms/50, and that should be fairly accurate assuming the RF is a sine wave.
?
Chris has two 100 ohm resistors in parallel to form the 50 ohm on the source end of the coax that goes to
some measuring device such as a tinySA, which we assume has an internal 50 ohm termination resistor.
So his circuit is the same as what I was suggesting for the 50 dB attenuator
His 2.5k resistor is very close to the ideal value of 2475 ohms that I had calculated.
?
At 10 Watts out into 50 ohms, the voltage across the 50 ohms is sqrt(10*50) = 22.4 Vrms.
And the voltage across Chris's two 100 ohm parallel resistors is? ?22.4v/(2500+50) * 50 = 0.44 Vrms.
Power dissipated by that 50 ohms is? 0.44*0.44/50 = 0.004 Watts.
Even ridiculously small 0402 (imperial size) chip resistors are rated for 0.062 Watts.
Two 100 ohm resistors in parallel present half the inductance so will be more accurate at 50 MHz,
but if using a surface mount chip resistor that shouldn't matter much.
?
Jerry, KE7ER
 

Thanks Jerry,
I had a load of 2W 51R SMD resistors so using 4 in a series parallel arrangement gave me 8W power handling.
?
Another way to do the power calculation is Vpk * Vpk / 100, that's practically mental arithmetic, or if you have a scope that measures the peak to peak voltage Vpp * Vpp / 400.
?
This simple little dummy load come peak V measurement come attenuator was the best thing I made to help with testing a QMX.
?
Incidentally the 2K5 resistor is 1k0 + 1k5.

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