Measuring the power in an AC circuit - Page 1
Forum section: electronics / general

author: raff5184
author_rank: Regular Contributor
timestamp: March 08, 2017, 01:56:26 am
content: Hi, I have a simple circuit, basically a voltage generator (working at 700kHz) powering a piezoelectric transducer.I don't know what the correct way is for calculating the power transferred to the load.This is what I have done so far:I know the amplitude of the voltage signal V, hence I can calculate its rms value V/sqrt(2)As for the current, I only have an oscilloscope so I have to calculate it indirectly. I put a 100resistor in series with the load and calculate the voltage drop across it by measuring the voltages at the endsof the resistor and then taking the difference. So, by using Ohms's law, I know the current in the circuit.Now how do I find the power absorbed by the transducer? Do I have to consider only the active power?

author: acbern
author_rank: Frequent Contributor
timestamp: March 08, 2017, 07:34:00 am
content: I am not the expert on piezos, however, generally:A piezo is a highly capacitive load. So there will be a considerable amount of peactive power. If you multiply the momentary voltage with the momentary current, you get the momentary effective power. The integral will bring you to the effective power. Some scopes will do that math for you.

author: Le_Bassiste
author_rank: Frequent Contributor
timestamp: March 08, 2017, 11:08:15 am
content: Quote from: raff5184 on March 08, 2017, 01:56:26 amyes. the active (= real) power tells you the sum of ohmic losses inside the transducer plus the power that the transducer converts into acoustical power.hence you would calculate the real power from your measurements byPreal = VRMS x IRMS x cos(phi),where phi is the phase angle between voltage and current. note that you want to use a low shunt resistance for your current measurement in order to not falsify the phase angle too much.

author: Conrad Hoffman
author_rank: Super Contributor
timestamp: March 11, 2017, 12:32:47 am
content: Big question- is your pzt operating anywhere near a mechanical resonance? If it is, you may get some unusual and unstable measurements as temperature and operating conditions change. The higher the Q and the softer the ceramic, the worse it will be and at some point physical measurements become almost impossible. Are you using an inductor to cancel the capacitance? Those are imperfect and will dissipate power too.For simple off-resonance, things are much easier. A pure capacitance can't dissipate any power- it's all dissipated by the losses. As said above, use a small resistance to measure the current, since you don't want to influence the operating conditions too much. Get the voltage and phase angle- it will be negative for a capacitative load. You have to get both waveforms on the scope to determine the phase angle- just amplitudes won't do it. Use the formula given above.I've got a conversion utility on my site that will give you a lot of parameters, though not power. You'll still have to figure that manually. Use the IVP button. Get the parallel losses from the utility and you can just do e^2/R (I think). First item on this list-http://conradhoffman.com/chsw.htm