# Calculating voltage class ab power amp

• Physixs
In summary: I am currently reading.In summary, the emitter of Q1 is connected to a 2.7kOhm resistor, which is then connected to -9V. 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
Physixs
Hello,

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 don't 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!

First the circuit shown provides a bias to help minimize crossover distortion, but it does not eliminate it. Minimizing this distortion comes at a cost of leaving the "off" transistor in a conducting state at a current level set by the bias levels. At far swings the "off" transistor switches off to a further degree and this transition between the on and off state contributes to the distortion.
Don't let the negative voltage throw you. Think of the minus rail as 0 and the positive rail as 18v then the idle output voltage is at 9v (instead of 0, when you use +9 & -9). You logic as .7 BE conduction state value is correct to evaluate a quick sketch of the voltages. The pot is likely used to "center" your output to the rail midpoint. Remember that the voltage divider still has some current through the base that must be accounted in your analysis.

1 person
Thanks Mjhilger. I really, really appreciate it.

So in this case, VBT3 would simply be 3.25V? Therefore my Ve would be 2.55V, Ie would be 945uA, Ic = Ie but this part I am a bit confused on...

So VBT1 would be 8.54V and VBT2 would be 7.145V? If so, I must have connected something wrong because my reading for VBT2 was close to zero and my PNP was burning up.

Lastly, I am not sure what you meant by "the potentiometer is likely used to center the output to the rail midpoint" Most likely because I don't understand how the potentiometer works. I did notice that even if I adjusted the adjustment knob on the potentiometer, it didn't change my output signal on my oscilloscope. So I was confused as to what the potentiometer even did in the circuit. So I pulled it out and put it on my ohmmeter. It was a 502 pot (5k, 2W) and it came in at 4.8k. No matter how I turned the knob it barely changed. On one extreme it went to 4.7k, on the other it was at 4.88k. I was under the impression that the potentiometer acted as a variable resistance able to range from 0 to 5k... which is clearly very wrong from what I learned. I never learned about potentiometers in class (aside a definition) and I never had to use one until now... so I am a bit confused.

Thanks so much for your help.

Sounds like your pot is broken.

What is the output swing? A large output swing will create a lot more distortion in that circuit than a small swing.

1 person
Remember Ic is approximately Ie; Ie = Ib + Ic, Ib is Ie/Hfe or the Beta of the transistor (for a 2n2222 about 75 - 250- look it up).
A pot is a variable resistor from around 0 to the max of the value. See the wiki:
http://en.wikipedia.org/wiki/Potentiometer
Your transistor is hot because the bias is forcing that one into full conduction while the other may be conducting in saturation. Heat is the dissipation of the current and voltage Pd = Ic * Vce. That pot needs to be connected correctly and adjusted to balance the quiescent point of your amp. You're close, I think you probably had the pot connected wrong. Look at the wiki and try again. Don't let the transistor get too hot and if it continues hot, turn off the supply and let it cool down. If you still have problems, get the lab instructor/monitor to help you.

1 person

## 1. How do you calculate the voltage class of an AB power amp?

To calculate the voltage class of an AB power amp, you need to know the supply voltage and the output power of the amplifier. You can then use the formula Vcc/√(2P) where Vcc is the supply voltage and P is the output power. The result will determine the voltage class, with Class A being <1, Class AB being slightly above 1, and Class B being close to 2.

## 2. What is the difference between Class A, Class AB, and Class B amplifiers?

Class A amplifiers have a single active device that conducts for the entire input signal, resulting in low distortion but high power consumption. Class AB amplifiers have two active devices that alternate conducting for different parts of the input signal, providing a balance between low distortion and moderate power consumption. Class B amplifiers have two active devices that only conduct for half of the input signal, resulting in higher distortion but lower power consumption.

## 3. How does the voltage class affect the performance of an amplifier?

The voltage class of an amplifier determines its efficiency, distortion level, and power consumption. Class A amplifiers have the lowest efficiency and highest distortion, while Class B amplifiers have the highest efficiency and moderate distortion. Class AB amplifiers provide a balance between the two, with slightly lower efficiency and distortion levels compared to Class B amplifiers.

## 4. What factors can affect the voltage class of an AB power amp?

The main factors that can affect the voltage class of an AB power amp are the supply voltage, output power, and biasing. A higher supply voltage or output power will result in a higher voltage class, while proper biasing can help maintain a consistent voltage class and improve amplifier performance.

## 5. How can I optimize the voltage class of an AB power amp for my specific needs?

To optimize the voltage class of an AB power amp, you can adjust the supply voltage, output power, and biasing. Lowering the supply voltage can decrease the voltage class, while adjusting the output power and biasing can help fine-tune the amplifier for your specific needs. It is also important to consider the load impedance and matching it to the amplifier's output impedance for optimal performance.

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