Need help understanding Current Draw for a Power Amplifier
Hi,
I'm working on a project with a colleague involving amplifying some simple sine waves (20kHz) using an APEX Microtech Power Amplifier PA50 Two power supplies INSTEK SPS3610 are connected in series and are supplying the amplifier with +/ 20 volts each. Each power supply can provide up to 10 amps (i think that is continuous not peak). The APEX amplifier can output up to 40 amps continuous. The colleague I am working with has done some measurements with the amplifier and has measured 23.4 amps RMS or 33.4 amps Peak on the output (before the amplifier saturates). He used a hall effect sensor to measure the current going to the load (which is an electromagnet device immersed in a conductive fluid). I'm confused on how you can draw that much current when your supplies can provide only up to 10 amps maximum. With two power supplies in series, the maximum current provided to the amplifier is 10 amps. I always thought that for a typical op amp power amplifier, the current on the output of the amplifier is equivalent to the current draw from the supplies. Can anyone help me understand this? I am thinking my argument is wrong or my colleague has not calibrated the hall effect sensor properly. 
Try doing the calculation with power first, to reassure yourself that the power out is less than the power in. Then take a look at the output voltage swing, and see that it is less than the 40V DC input supply voltage. The current out can be more than the current in, as long as the amp is using a switching power supply, and the output voltage is less than the input voltage.

Hi,
The output power of the amplifier was 248 watts (10.5 volts RMS, 23.4 amps RMS), and the power from the supplies was 332 watts (8.3 amps at +/ 20 volts) The power supplies are both DC power supplies, and the amplifier itself doesnt have any switching power supplies. Is it still possible to have more current flow on the output of the amplifer than from the power supplies (with DC power supplies)? 
Since you state that your load is an electromagnet (inductor) you can't just read voltage and current.
You need to determine the phase angle (or power factor) involved as well. 
Hi,
Certainly, for power you have to include a phase angle. I believe that my colleague used a phase angle correction to get the correct output power. When I had worked on that project while I was a grad student I always accounted for the phase angle. I'm still lost on whether the current draw from the DC power supplies should be the same at the output of the amplifier. What confuses me is that he is drawing 8.3 amps from the power supplies in series, and getting 23.4 amps RMS on the output of the Amp Also, some of the connectors (to the load) he is using are rated at 10 amps with 16 gauge wiring, so i would have expected him to see some failures with those. 
Anytime you get to measuring power in big inductors there are lots of gotchas.
How about the definition of RMS itself? A lot of test equipment assumes a sine wave and just applies square root of 2 to peak voltage. If the waveform is not a sine then the number is meaningless. Since the load is varying the 8.3A shown by the DC power supply meter is some sort of average current. It can not represent a constant DC current. I'm thinking a crude current foldover sum of the dual 8.3A DC input should get about 16.6A RMS on the amp output (RMS uses peak voltage not peak to peak). The 23.4A rms isn't that far off from 16A. 
Here is what cold be happening. You say that the 2 power supplies are wired in series. That may be the case, but I would bet that the +/ node between the 2 supplies is wired to the amplifier. That being the case, one supply provides current for the positive portion of the cycle and the other supply provides current for the negative portion of the cycle. So each supply provides current for half the signal. If we assume the power supply is decoupled from the amplifier well enough to smooth out the current pulses, each half cycle probably draws 16.6 amps. When each half cycle is combined into a sine wave we can assume that there is a total current draw RMS of 16.6 amps. 16.6/23.4 is .709 which happens to be very close to the conversion of RMS to peak. I would guess the hall sensor is calibrated in peak in your case. If so, the numbers come out exactly as they should.

All times are GMT 5. The time now is 08:46 AM. 
Powered by vBulletin Copyright ©2000  2014, Jelsoft Enterprises Ltd.
© 2014 Physics Forums