Amplifier circuit questions

[jackorocko]

1. The gain of the op-amp is 10? 47k/4.7k = 10
2. This is a non-inverting amp?

3. The transistors in this setup kind of confuse me. The PNP transistor connected to the +15V side will only turn on when the base is more negative then the emitter. The 180Ohm resistor will have a voltage drop across it which I believe this is what turns on the transistor and sets the base current. But, what I don't understand is when this transistor will turn on. I assume it will depend on when the output of the op-amp is positive. But how exactly does the op-amp effect the transistor, is it by drawing more current which causes a larger voltage drop across the 180Ohm resistor? How do you figure the voltage drop across the resistor?

Always seems to be the same with me, figuring out the math. I see how it works, but for the life of me I can not figure out the math simply because I don't understand I guess how an op-amp works. Frustrating, can anyone explain to me what I am missing to get the answer I want?

 

[Harald Kapp]

A positive input on "In" will drive the output of the OpAmp positive (gain=10). This will cause current to flow from the output of the OpAmp through the load to GND. Where does the current come from? From the positive supply of the OpAmp (7). This current will increase the voltage drop across the 180 Ohm resistor, thus niasing the PNP into conduction mode. Current through the PNP will flow through the load, too, thus increasing the output voltage. Thanks to the feedback resistors the OpAmp will regulate the output such that Vout=10*Vin.

The same goes for negative input voltage, polarities reversed.


Harald

 

[jackorocko]

So pretty much what I thought. This is also a simple push-pull class AB amp correct? Now how do they know what value to select for the resistors connected to the base and Vcc of the op-amp? There has to be a fundamental way to mathematically calculate this value. What is it?

 

[Harald Kapp]

The idle current of the OpAmp is known from the datasheet. On idle, the transistors should not yet conduct. So I(idle)*R<=V(be).
The exact value is probably not that important because some inaccuracy can be nulled by the feedback action ofthe OpAmp.

Harald

 

[jackorocko]

Thanks harald, I thought the idle current had something to do with the output. But now I understand.

I got another question if you don't mind. How do they figure out the power output of this amplifier? I know the TIP125 has a Vce drop of -2v@saturation. So that means I will get a max of 13V out of the transistor. Does this voltage add to the voltage coming from the op-amp output?

Working backwards, I know the power output is close to 12.5W from the designers comments.
sqrt(12.5W * 8Ohms) = 10V max output?
10V/8Ohms = 1.25A max output?

 

[Harald Kapp]

How could it?
Voltages can be added only by stacking the voltage sources atop each other. But here both OpAmp and Transistors are powered by the same supply. So the max. output voltage is +-13 V (@2 V Ucesat).
What the OpAmp does is add some mA to the output current (that is the current providing bias for the transistors via the supply wires). But that should be only a small part of the output current, depending on the current gain of the transistors.

So the output power can be effectively calculated by
P=(13 V)^2 /2*Rload
(divide by 2 since 13 V is peak voltage, RMS is 13 V / sqr(2).)

Harald

 

[Resqueline]

The gain of that amplifier is actually 11.
If you have an input amplitude of 1, it also means there's 1 across the 4.7k and 10 across the 47k. Since these are in series the output amplitude becomes 11.

The output power estimate is here just a function of available peak voltage and nominal speaker impedance. 13V peak = 9.2V rms = 1.15A in 8 Ohms = 10.6W
There's no voltage or current coming from the op-amp that can be added to anything on the output here.

 

[jackorocko]

Yep, so I am still with you. Working backwards from the designers comments seem to prove what you two have said (edited last post). I am a noob when it comes to this math relationship stuff with electronics.

The gain of that amplifier is actually 11.

Right because the formula should be (1 + R2/R1) What would be a peak input voltage? Let us just assume, peak 1V.

1V/1.141 = .876Vrms
.876Vrms * 11 = 9.64Vrms

If you are right and my max output voltage is 9.2Vrms, then that would mean @ 1V peak input I would get clipping on the output?

 

[(*steve*)]

The gain of that amplifier is actually 11.

The voltage gain is 11.

The power gain is probably going to be much higher.

(This comment may be superfluous depending on the OP's knowledge and/or need for this )

 

[jackorocko]

What knowledge?

An ideal opamp will have power gain of infinity. Again, in principal I understand this. The idea that you can actually amplify nothing in a perfect opamp. What that means in real world I am not exactly sure, but I know it will still be a huge gain.