How to I calculate max current draw for this circuit?

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

Assumptions:

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

Questions:

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.

Photos:

• 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. – MarkU Mar 28 '18 at 0:20
• 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? – Smeboo Mar 28 '18 at 0:27
• 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. – MarkU Mar 28 '18 at 0:28
• 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. – MarkU Mar 28 '18 at 0:30
• Okay. I will give that a try then! – Smeboo Mar 28 '18 at 0:32