# Calculating capacitance required for a pulse duration of time T using a class AB amplifier

Firstly I apologise if this question is similar to another one I asked a while ago. But, as you will see, this is question is slightly different.

I have an amplifier which is a common emitter driving an AB push pull stage using MJE340/350 transistors.

I am using it to amplify a signal of 250kHz. The (unloaded) output voltage at 250kHz is 145.6 Vpp. This amplifier must be able to drive a signal for a time T.

I want to use it to drive a load with impedance Z = 669.2Ohms.

From V=IR, I can see that the current requirement is 0.21757 A. The power supplys I am using are R12-100B's which, from the data sheet, provide 50 mA each, So the total current provided is 100 mA (I added them because I am working with the peak to peak voltage, so both halves of the wave).

This leaves me with a current deficit of 0.1176 mA. To supply this excess current there are two capacitors. The value of the capacitors is 3.3 uF rated to take 450V. These are decoupling the supply rails which are at +/- 120 V. So essentially I have one 6.6uF capacitor charged to 240 V.

From Q = CV I know there are 0.001584 coulombs of charge stored. I also know that 1 Amp is defined as is 1 Coulomb of charge passing a point in 1 second. For this I can say that if A = C/s then C/A = s.

So to calculate the maximum time T which I can transmit this signal for I divide the amount of charge by the current deficit. This gives me a value of 13.47 mS which the signal can be transmitted for.

Now I know that as a capacitor discharges the voltage across the capacitor decreases so this method is not quite correct as this is not taken in to account.

So my questions are:

1) How much (if any) of my calculations are valid?

2) How should I take into account the voltage drop as the caps discharge (I assume I just a bigger value...)?

3) Should I use a few smaller caps because of the high frequency being used?

4) The voltage output was measured (using a pico scope with a probe on *10) with no load applied, will this value still be valid when the amplifier is loaded?

• This really does need a block diagram or circuit diagram showing decoupling capacitors to the load. You shouldn't use p-p to calc current into the load - that makes no sense. Also, numerically what is time, T? – Andy aka Apr 11 '14 at 15:32
• Its just an AB push pull, with a +120V rail, with one cap to ground, and a -120 rail with another cap to ground. Thanks, I had wondered if Vpp was right. The time T is the length of time the signal (sinewave) is transmitting for, so I can calculate the time provided by the 3.3uF caps, and I want to go backwards and calc the value of a cap required for a signal duration of a time T. – Tim Mottram Apr 11 '14 at 15:36
• Sorry if my question was unclear, I'll try and get a circuit diagram up. Once I figure out how to do that!! – Tim Mottram Apr 11 '14 at 15:44
• Listen, this is why I asked for a circuit because asking questions about what you have doesn't work. Read my previous comment. – Andy aka Apr 11 '14 at 15:44

This answer is not based on seeing a circuit diagram. A circuit would help a lot.

If the load is to mid-rail of the power supply and you are driving a sinewave at 145.6 Vp-p, then the current calculation is this.

1. Calculate the RMS voltage - 145.6 Vp-p is 51.5 Vrms
2. Calculate the load power - $\frac{V^2}{R}$ = 3.96 watts
3. Estimate input power - input power is going to be no-worse than twice output power so say 8 watts max.
4. Estimate dc current drawn from each 120V - each half cycle occupies 50% of the time of the sine period therefore each power supply is supplying 4 watts or 33 mA.

Sanity check, 240V x 33mA = 8 watts. OK I think I got it right.

Your power supplies can supply 50mA and this might be OK without adding capacitance (there'll be some inside the power supplies anyway) and the peak current taken by the load is

$\sqrt2 \times 33mA$ = 47mA.

I think you should be OK without adding capacitors and will run indefinitely as far as I can see.

• Hi Andy, thanks for your answer, you are, of course, correct. I have a few more questions, but I'll try and find the answers out on my own before bothering you guys with them. – Tim Mottram Apr 16 '14 at 14:03