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"The Electronics of Radio" Problem

 

I started to work through The Electronics of Radio book. I can't figure out how to calculate the combined internal resistance of the capacitor and inductor in problem 8 A.

For that problem a 7MHz sine wave signal is fed across the variable capacitor C1 (9-50pF) and inductor L1 (15uH). Amplitude of the signal is 1V peak-to-peak. The assignment states to tune capacitor for resonance, the spot where the voltage measured across both is lowest. My test showed a voltage of about 260mV when tuned to resonance. Measured capacitance is 34pF. Calculations agree that the capacitance should be 34.5pF.

Now the problem instructs "Use the voltage to calculate the total resistance of the capacitor and inductor." I'm confused on how to do that without knowing/using the current. Am I missing there just a simple formula?

Thanks,
Claus
 

This problem should reduce to a simple series circuit, since the reactances
have cancelled. So you have a 1 V source in series with the 50 ohm source
resistance and an unknown loss resistance Rloss which has 0.260 volts
dropped across it. This knowledge should allow you to solve for Rloss in
the simple voltage divider circuit. If I didn't make any errors it's 17.6
ohms.

73-

Nick, WA5BDU


On Sat, Sep 7, 2019 at 10:02 PM Claus (K0TET) <claus-groupsio@...>
wrote:

I started to work through The Electronics of Radio book. I can't figure
out how to calculate the combined internal resistance of the capacitor and
inductor in problem 8 A.

For that problem a 7MHz sine wave signal is fed across the variable
capacitor C1 (9-50pF) and inductor L1 (15uH). Amplitude of the signal is
1V peak-to-peak. The assignment states to tune capacitor for resonance,
the spot where the voltage measured across both is lowest. My test showed
a voltage of about 260mV when tuned to resonance. Measured capacitance is
34pF. Calculations agree that the capacitance should be 34.5pF.

Now the problem instructs "Use the voltage to calculate the total
resistance of the capacitor and inductor." I'm confused on how to do that
without knowing/using the current. Am I missing there just a simple
formula?

Thanks,
Claus
 

Hi Nick,

I still don't understand how to solve the problem. Please show this math-challenged dummy all the steps needed to solve the problem.

Thanks.

73,

Steve AA7U

On 9/8/2019 6:23 AM, Nick Kennedy wrote:
This problem should reduce to a simple series circuit, since the reactances
have cancelled. So you have a 1 V source in series with the 50 ohm source
resistance and an unknown loss resistance Rloss which has 0.260 volts
dropped across it. This knowledge should allow you to solve for Rloss in
the simple voltage divider circuit. If I didn't make any errors it's 17.6
ohms.

73-

Nick, WA5BDU


On Sat, Sep 7, 2019 at 10:02 PM Claus (K0TET) <claus-groupsio@...>
wrote:

I started to work through The Electronics of Radio book. I can't figure
out how to calculate the combined internal resistance of the capacitor and
inductor in problem 8 A.

For that problem a 7MHz sine wave signal is fed across the variable
capacitor C1 (9-50pF) and inductor L1 (15uH). Amplitude of the signal is
1V peak-to-peak. The assignment states to tune capacitor for resonance,
the spot where the voltage measured across both is lowest. My test showed
a voltage of about 260mV when tuned to resonance. Measured capacitance is
34pF. Calculations agree that the capacitance should be 34.5pF.

Now the problem instructs "Use the voltage to calculate the total
resistance of the capacitor and inductor." I'm confused on how to do that
without knowing/using the current. Am I missing there just a simple
formula?

Thanks,
Claus
 

OK. I think a key is understanding that your voltage source is an ideal
voltage source plus its internal resistance, which with test equipment
would generally be a well defined 50 ohms. I had to bang this into my head
when I started measuring crystal parameters.

Now, if you've accepted that you don't even need to draw the L and C in
your circuit (because their reactances cancel), you can just leave them out
and draw a series circuit of: V source = 1 v, Rg = 50 and Rloss which is
unknown. (Rg means R of generator.)

Write the equation for current using Ohm's law. I = E/R where I is 1 V and
R is 50 + Rloss.

So I = 1/(50 + Rloss)

You already know that voltage across Rloss is 0.26 volts and it's also
equal to Rloss times the current. So we can say,

0.260 = Rloss * 1/(50 + Rloss)

Manipulate that equation with a little algebra. My steps would look like
this:

Rloss = 0.26*(50+Rloss), then

0.74*Rloss = 13

Rloss = 13/0.74 = 17.57 ohms

73-

Nick, WA5BDU


On Sun, Sep 8, 2019 at 10:50 AM Steve Ratzlaff <ratzlaffsteve@...>
wrote:

Hi Nick,

I still don't understand how to solve the problem. Please show this
math-challenged dummy all the steps needed to solve the problem.

Thanks.

73,

Steve AA7U
Chuck Carpenter
 

Nick,

The voltage is 1 Vpp. Should that be converted to RMS (0.707)?



Chuck, W5USJ (ex K2OFN)
ARCI 5422, SKCC 19956
EM22cv, Rains Co., Texas
Chuck Carpenter
 

Nick,

The voltage is 1 Vpp. Should that be converted to RMS (0.707)?



Chuck, W5USJ (ex K2OFN)
ARCI 5422, SKCC 19956
EM22cv, Rains Co., Texas
 

Hi Nick,

Thank you! I understand how to do it now.

73,

Steve AA7U

On 9/8/2019 9:24 AM, Nick Kennedy wrote:
OK. I think a key is understanding that your voltage source is an ideal
voltage source plus its internal resistance, which with test equipment
would generally be a well defined 50 ohms. I had to bang this into my head
when I started measuring crystal parameters.

Now, if you've accepted that you don't even need to draw the L and C in
your circuit (because their reactances cancel), you can just leave them out
and draw a series circuit of: V source = 1 v, Rg = 50 and Rloss which is
unknown. (Rg means R of generator.)

Write the equation for current using Ohm's law. I = E/R where I is 1 V and
R is 50 + Rloss.

So I = 1/(50 + Rloss)

You already know that voltage across Rloss is 0.26 volts and it's also
equal to Rloss times the current. So we can say,

0.260 = Rloss * 1/(50 + Rloss)

Manipulate that equation with a little algebra. My steps would look like
this:

Rloss = 0.26*(50+Rloss), then

0.74*Rloss = 13

Rloss = 13/0.74 = 17.57 ohms

73-

Nick, WA5BDU


On Sun, Sep 8, 2019 at 10:50 AM Steve Ratzlaff <ratzlaffsteve@...>
wrote:

Hi Nick,

I still don't understand how to solve the problem. Please show this
math-challenged dummy all the steps needed to solve the problem.

Thanks.

73,

Steve AA7U
 

Well, it can be in any units if kept consistent. I think it said pp in one
instance and didn¡¯t specify in the other. So I figured we were staying with
p-p.

I do normally think though, that when someone just says ¡®volts¡¯ that RMS
should be assumed.

73

Nick

On Sun, Sep 8, 2019 at 11:55 AM Chuck Carpenter <w5usj@...> wrote:

Nick,

The voltage is 1 Vpp. Should that be converted to RMS (0.707)?



Chuck, W5USJ (ex K2OFN)
ARCI 5422, SKCC 19956
EM22cv, Rains Co., Texas
 

Yes, both measurements are peak-to-peak voltages.

Thank you very much Nick for the great explanation. It's the internal 50 ohm resistance from the power source (function generator) that I was missing.

Next step is to measure this calculated number. Of course DC measurements are pointless. But I thought the antenna analyzer should be able to do it. Changed the antenna analyzer's frequency until reactance (X) was zero which append at 7.192MHz and got a resistance (R) of 14.1 Ohm. If you do a reverse calculation, the voltage across the LC at resonance would be 220mV. Knowing how hard it was to adjust the variable capacitor it's quite in the tolerance of my ability to adjust the capacitor for resonance.

Lab work successful. I learned a lot. Thank again Nick. Also thanks to Steve and Chuck for the great additional questions.

Claus

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