Use of Capacitors in this circuit?

1. Mar 2, 2012

Xyius

In my lab course, we constructed a circuit that would amplify signals. Here is an image of the circuit.

http://desmond.imageshack.us/Himg442/scaled.php?server=442&filename=imgfw.jpg&res=medium [Broken]

My question is, what is the purpose of the capacitors? In other words, if I was making a signal amplification circuit without this as aid, why would I use capacitors? I asked my professor and he didn't have a concrete answer.

Also, is this the type of circuit that OpAmps Use?

Thanks!

Last edited by a moderator: May 5, 2017
2. Mar 2, 2012

schip666!

The circuit has (two) single ended DC power supplies -- batteries -- and you want to amplify an AC signal. The caps isolate the signal input and output from the DC offset in the circuit itself. So basically input and output swing +/- 0 volts, but the transistor amplifier operates between 0 and +15. If my professor couldn't explain this I think I would try a different class...

Since it's a common-emitter circuit it's mostly making square waves out of your input, so it might not matter in this particular case. But if you put a resistor in the emitter leg to ground you can reduce the gain and make it sort-of linear. In that case your input and output might be sine waves and the output can swing around ground such that no-input gives you 0v output. This might be of interest if the output was driving a loud speaker because you don't want the speaker to be powered (say pushed all the way in one direction or the other) when there's no signal.

Op-amps usually have push-pull stages rather than single ended like this. Spec sheets (used to have) schematics where you can see how the output stages work. They also tend to run on dual power supplies so as to avoid needing the AC-coupling-capacitors.

3. Mar 2, 2012

Xyius

I don't think I fully understand. I DID have a sine wave output. (Looked at the signal with a scope.)

Are you saying the capacitors offset the signal so it does not cross zero volts?

4. Mar 2, 2012

schip666!

The circuit shown has the transistor running at it's full gain, maybe 100 or something. If you had a very small input signal it would still retain it's shape without clipping.

What I was trying to say is the capacitors offset the input and output so they DO cross zero volts. The amplifier itself only runs between 0 and +15 volts. The 150K resistor on the base biases the input to the positive so the signal generator can "pull" it up or down.

5. Mar 2, 2012

Xyius

Okay so let me see if I understand. The first capacitor offsets the signal to be only positive because the transistor only works between 0 and 15 volts.

As for the 150K resistor, why must we use such a high resistor? If all we want to do is bias the input to be more positive, can't we just remove the resistor completely? Also, could we not achieve the same result by removing the capacitor as well and simply having the battery of 5 volts? The superposition of the signal wave and the DC voltage would make it above 0 volts would it not?

Also, how does the second capacitor help the signal?

Doesn't the capacitor "distort" or "mess up" the sine wave? Wouldn't it start to discharge when the voltage went negative and thus, change the look of the wave?

I am getting closer but I am not quite there yet! :(

6. Mar 2, 2012

Hassan2

Hello Xyius,

Note: Without the 150K resistor, your transistor breaks instantly!

Beside the useful and clear explanations of schip666!, the following comment may also help you.

1. The 150K resistor controls the base current , in a common emitter configuration, the gain is proportional to this current. A smaller resistor can yield a high base current. The collector current is β( ≈100) times larger than the base current, so the voltage drop along the 1K resistor can saturate the transistor. A high base current can damage the transistor too.

2. Without the 33μF capacitor , DC current would flow into the ac generator which has a very low impedance. Some related problems would arise as well.

3. As for the 100μF capacitor, if the transistor is working in its linear region ( this is possible for very small input AC voltages), we have an steady state sinusoidal wave with a DC offset at the collector. In this case, the capacitor does not distort the signal but it may attenuate it slightly. If the transistor get saturated or if negative half-cycle of the input signal turns the transistor off, then the output signal may deviate from being sinusoidal as you mentioned.

By the way, I think it's better to use a high resistor in series with the function generator.
I prevents saturation of the transistor in positive half cycles, Of course you will get a lower voltage gain for the whole circuit.

7. Mar 2, 2012

tamtam402

In my electrical engineering classes we learned to solve this kind of circuit by separating AC and DC signals, and using the superposition principle to find the "real" output of the circuit. DC current "sees" the capacitors as open circuits. If you're familiar with impedance, you can convince yourself by using the impedance formula for a capacitor, which is Z=1/jωc. For DC currents ω=0, so Z tends to ∞.

8. Mar 3, 2012

Xyius

Thanks a lot guys, I believe I understand now. :D

I just have one question. How did you guys know the transistor is rated to have a beta value of around 100?

9. Mar 3, 2012

Delta Kilo

10. Mar 3, 2012

likephysics

Why 33uf and 100n at the output?
Shouldn't they be equal since the input freq is equal to output freq.

11. Mar 3, 2012

Hassan2

ِDifferent impedance. The load resistance is not fixed and could be low. A larger capacitance is needed to make sure that the cut of frequency is lower than the working frequency.

12. Mar 3, 2012

Neandethal00

From now on, when you see this type of circuits with capacitors, remember only 2 properties of capacitors.
a) A capacitor passes AC (Sine wave type) signals. [picky people will bring in reactance here].
b) A capacitor blocks DC signals (Battery or DC power).

13. Mar 7, 2012

Xyius

Thank you! This helps tremendously! :)

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