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I am working on a lab for school and I am having a problem. I am attempting to build a compensated class b power amplifier.

I understand the common emitter configuration (to boost voltage) and the class b push pull amplifier stage (to boost current). I understand that diodes in the circuit are used as diode current mirror bias and they SHOULD eliminate crossover distortion... my first problem is that my output is still showing cross over distortion.

Knowing that the current mirror bias should have eliminated it, I went back into my circuit to check to see if I connected anything incorrectly (by checking voltages at various points in the circuit) and then I realized my second problem... Due to the -Vcc, I actually didn't know how to calculate values by hand (and therefore, did not know how to double check myself). I tried my textbook and some online sites, but they mostly reference ground and +VCC. Which I know how to calculate (finding the base current using a voltage divider formula, -0.7V for the FWD bias drop to emitter for VE, using the rest of the drop across Re and Ohms law to find Ie, then using Ie=Ic to find VC and VCE.). But, with a negative supply I am clueless. I know my total voltage (from +9 to -9) is 18V. I just dont know how to use that number correctly.

The circuit I am trying to build looks like the one found here, except my CE amplifier has a voltage divider bias

My base of Q1 is voltage divider bias. The voltage divider consisting of a 68kOhm resistor connected +9V Vcc and the other portion of the voltage divider ("lower" section) consists of a potentiometer 5k Ohm and a 10k Ohm resistor, in series, connected to -9Vcc. I am not sure what the purpose of the 2W potentiometer here is supposed to do. I am not sure why we couldn't just use a 15 k Ohm (rather than a 5k potentiometer and 10k resistor in series), though I know that a potentiometer is used to regulate voltage.

How do I calculate the voltage at the base of Q1? The emitter of Q1 is connected to a 2.7kOhm resistor, which is then connected to -9V.

I am just not sure how to calculate any values due to the double voltage source. Do I use a superimposed DC technique and pretend one supply is not there, calculate all currents then do the same for the other? I hope there is an easier, less tedious way!

Thanks in advance for your help.

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# Calculating voltage class ab power amp

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