I’m trying to calculate the max current pulled from my 15V and -15V supplies. My OPAMPS also use the supplies for current.

All of my numbers seem really whack.

The part I am most confused about is the part with the transistors... I have struggled with transistor circuits a lot and am hoping someone could help walk me through what I need to do.

IC Datasheet


  • Caps are shorted because that is when we have max current
  • Both the speaker and mic do not draw current (?)


  1. What equations will I need for the PNP and NPN transistors? How do I implement them

  2. Do I need to worry about the mic or speaker in this situation?

  3. Should my voltages seriously be that large? I feel like they are too large but I cannot justify why since my gain is so large.


  1. Circuit Circuit diagram

  2. Work Work for circuit diagram Transistor Part

  • \$\begingroup\$ Va isn't 10V, it's approximately 0V, because the 0.1uF capacitor blocks DC current flow. If this is a practical op-amp (not an ideal op-amp), you should edit to link to its datasheet. The Electrical Characteristics table will list the typical and min/max quiescent current that the op-amp draws from the supply voltage, this quiescent bias point is an important part of estimating how much power the circuit needs. An ideal op-amp would draw no power (that's why the supply rails are not shown on an ideal op-amp). Good that you realized quickly that the estimate was way unreasonable. \$\endgroup\$
    – MarkU
    Mar 28, 2018 at 0:20
  • \$\begingroup\$ Okay. I will link to that. When you say that the capacitor is blocking the DC current flow that’s only after they are charged though, right? Wouldn’t I need to assume the caps act as a short to calculate max current? \$\endgroup\$
    – Smeboo
    Mar 28, 2018 at 0:27
  • \$\begingroup\$ I would approach this as three sub-circuits; the electret microphone (mic + R2), the 101V/V gain section (R3, opamp, R1, Rf, and the 10k trimpot), and the output stage. I think of the 0.1uF capacitor as the AC coupling linkage between the mic and the 101V/V gain stage. The current through the 0.1uF capacitor is negligible. Also, the current through the wiper of the 10k trimpot is negligible because it only drives the non-inverting input of the final stage. So these three sub-circuits can be analyzed separately. \$\endgroup\$
    – MarkU
    Mar 28, 2018 at 0:28
  • \$\begingroup\$ No, capacitors always behave the same way, but for DC analysis we treat capacitor as an open circuit. For AC analysis capacitor's impedance depends on frequency. For transient analysis we use an exponential function. For calculating the load current to power this circuit, we use DC analysis. \$\endgroup\$
    – MarkU
    Mar 28, 2018 at 0:30
  • \$\begingroup\$ Okay. I will give that a try then! \$\endgroup\$
    – Smeboo
    Mar 28, 2018 at 0:32

1 Answer 1


I'm not going to finish the problem for you, but i can point out a couple of places you went wrong.

  1. Why did you assume that the capacitor is shorted for max current draw? Instead, assume that max draw happens when you are saturating the amplifier - id assume a waveform that is 50% saturated high and 50% saturated low.

  2. Your gain of 101 only applies to small signal waveforms. In reality, if the output voltage of the op amp is greater than it's max output swing (typically +/- 13V or so for a +/- 15V op amp, it will rail high or low.

  3. The current through the base of the transistors will be probably 2 orders of magnitude lower than the current thorough the collector, so unless you're trying to be really precise, you can likely ignore it.

  4. The most current you could ever get through the transistor is what you'd get if one or the other transistor (Not both at the same time) was assumed to be fully on, and is limited by the 100 ohm resistor - as well as any resistance of the speaker.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.