Using a resistor to messure power consumption of a circuit.

[tinman]

I was testing some very simple circuit's today,and found one that has my up most attention. As the frequency is around 127KHz,i don't trust my DMMs to give me an accurate amp average reading-although all 3 show the same current draw. I have drawn a basic circuit below,which we will use to answer my question.

Using the 3 ohm resistor in the circuit,can i use my scope across that 3 ohm resistor to work out the circuit's power draw,or will the rms voltage across that resistor only tell me what the resistor is dissipating?. (I tried a 1 ohm resistor in it's place,but i get a very noisy signal on my scope-the 3 ohm is much smoother on the scope.)

SG setting's.
Square wave at tuned frequency (around 127KHz)
Duty cycle-38%
VPP-8.2
Off set-4.1v-so as we have 0 volts at the bottom of the wave(62% off time)

[gyulasun]

Hi Brad,

The rms voltage across the resistor can indeed tell you the dissipation across the 3 Ohm:  P=V*V/R.  But you can learn about the current that flows in the circuit if you use Ohm's law for the rms voltage and the resistor values,  I=V/R

Here I assume the rms voltage value across the 3 Ohm is correct i.e. the scope considers duty cycle.

Finally the circuit power consumption could be calculated by taking the rms value of the 8.2Vpp square wave, 38% duty cycle and divide it by the rms current. 

I hope I am correct...

Gyula

[mscoffman]

tinman,

You should do what our lab instructor in school said to do: always draw the signal generator as a circle with the AC signal
for the picture *then* draw the generator's internal impedance as a resistor in series with it. So really your diagram should
have two resistors. Those coaxial cables that connect to
the circuit from the generator are supposed to match the impedance of the generator so as to not reflect any power back.
The manufacturer of the generator does quite a good job making the generator emit a near perfect signal, but he can't
create the impossible - a zero ohm impedance output. The generators output impedance is usually marked near that coaxial spigot.
Probably 50ohms. That step will help you analyzing what is happening. The reason is what you want your current shunt resistor
value to be is a trade off. The other thing is to not forget the AC power information; A two channel scope will let you determine
the phase shift between the voltage signal across the circuit and the current signal through it.  You can check whether your circuit
has a PF power factor that is inductive, capacitive, or resistive at the given frequency. Any inductor will have a self capacitance and
some frequency will cause parallel resonance and will allow you to tune to parallel frequency resonance of the inductor + self capacitor.

By the way a sinewave output RF frequency generator is easiest to understand, a pulse generator is somewhat different because
digital circuits require +/- unbalanced signals of fixed voltages creating asymmetric wave signals.

Also gyulason while you power numbers will be close to accurate, you really need to integrate (summate) the powers of the individual
Fast Forrier decomposition (sinewaves) of the voltage waveform with the Fast Forrier decomposition (sinewave) of the current waveform
then use that to calculate the power of each decomposition then add them back together to get total power. One of TB's capacitor
overunity proofs did this wrong and showed overunity in a capacitor where there really was none.

This is why it is better to measure the power after rectification then to try to calculate it.


:S: MarkSCoffman

[tinman]

Looking at the scope shot below(across the SG),which voltage do i use to calculate my P/in-as i already have the I/in.
My DMM is showing a very accurate I/in,as i checked it against my scope using a known value resistor across my SG at the same frequency.

In regards to the scope shot,it dosnt seem right that we should be using the RMS voltage to calculate power. As the duty cycle is 38%,and the vpp is 8.2,how is it that the RMS is 4.72?-which is more than half the 8.2 vpp,and yet only a 38% duty cycle.
The mean voltage is the same as the average voltage,and my DMM also shows 2.8 volts across the SG-same as means and average voltage on the scope.

My I/in is 2.75mA-can you work out the P/in from the scope shot and current(2.75mA)

[TinselKoala]

You have a situation where your input voltage is pulsed. So what you need to do to find the _average_ input power over many cycles, is to compute the VxI value during the pulse, then multiply that by the 0.38 duty cycle. Your voltage varies a little during the pulse, going from 8.2 down to perhaps 7.5 volts. So really you want an "average" here, take 7.8 volts for example. Your scope is computing the "average" and "mean" (here the same thing for this waveform) by doing something like this, taking the "average peak" and multiplying by the duty cycle. So you take the average voltage during the pulse, multiply by the average current during the pulse, and then multiply the result by the duty cycle, and this will be the average power. This won't be as strictly accurate as performing the instantaneous multiplication of the V and I traces and then integrating that resulting trace over time but it will be close.
I think.