# Clipping and V supply and Voltage at collector of a transistor

1. Apr 14, 2010

### morrisshokr

Hi, im new here, I'm still in high school here in Melbourne Australia, so please excuse my stupidity.

Theres a question come up on a massive project that has a large bearing on our end of year results, its got to do with clipping of an output signal, and relating it to the Supply voltage and the Voltage detected at the collector of a transistor. It asks us to comment on the clipping and relate it to Vs and Vc. Our whole year level, at a large leading high school here, has no idea.

Im relatively ignorant in relation to this so please keep it as simple as you can. You might need more info, please tell me.

Thanks!!

2. Apr 14, 2010

### Studiot

So what do you think happens if you set a transistor up with a gain of 100 and a supply voltage of 10 volts and an input signal of 1 volt?

3. Apr 14, 2010

### MATLABdude

Welcome to PhysicsForums!

Generally, homework goes in the homework section of this forum, but this question may be general enough that a mod will leave this here.

Since you're in highschool, I don't know if you've studied various BJT amplifier topologies (presumably, you're talking about a common emitter configuration--look it up in Wikipedia, if you're unfamilliar, but the specific details are not too important to the discussion at hand here).

So, let's generalize this to the point where you have not some BJT amplifier, but just a black box amplifier that provides 10x gain, and uses +/-10V supplies.

Now, you have a 1.5 V max (3 V peak-to-peak) sine wave input. What is the output going to be through this 10x amplifier?

EDIT: Both Studiot and I are trying to get at the same point, but his/her example is much simpler.

Last edited: Apr 14, 2010
4. Apr 14, 2010

### morrisshokr

ok guys thanks for your replies. It's just that the values we got are confusing. For the supply we got 5.05 with a gain of 50. We had 3 different graphs. Ill take the first one as an example. There was clipping of the output on both positive and negative. The input for that graph was 2.34V. And the clipping for that occurs at 1.2V and -3.5V. This is why I am confused. How do you get clipping at those values, with that supply, gain and input?

Am i missing something blatantly obvious here?
Can you guide me through the calculations or the thought behind it?

Last edited: Apr 14, 2010
5. Apr 14, 2010

### Studiot

I do believe that in your high school physics/electronics you study a transistor circuit arrangement like the attached.

Please confirm that you understand this sketch as it gives us a working basis for discussion.

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6. Apr 14, 2010

### morrisshokr

Yes I do understand

7. Apr 14, 2010

### Studiot

OK so I have redrawn the sketch a little differently.

R1 and R2 have become part of a slider and R4 is zero.

Let us start with the slider at the bottom end so the base is at zero volts (Vb=0).
The transistor is off so the collector passes no current (Ic=0)
So there is zero current through R3, so the voltage across R3 = 0xR3 = 0 (Ohms law)
So the collector is at the supply voltage Vs. (Vc=Vs)

Now let us slowly raise the slider so the base voltage increases.
At around Vb=0.7 the transistor turns on and starts to conduct collector currrent.
I don't know if your course tells you that Ic is beta times the base current or it is a function of Vb.
Both are true but neither are relevant here yet.
The point is that as we raise Vb the base current increase and therefore so does the collector current.

As Ic increases it passes through R3 and the voltage across R3 increases from zero.

So the collector voltage Vc must fall so Vc=Vs-IcR3.

The more we push up the slider the more the collector voltage falls.
Vc cannot fall all the way to zero because of the mechanics of the transistor, but it gets quite close.
It actually falls down to a voltage called Vce saturated, which is the minimum possible voltage across the collector - emitter.
We say the transistor is saturated or bottomed, because it cannot fall any further no matter how hard we drive the base.

If we now reintroduce R4, can you see that the collector current passes through the transistor and 'out' of the emitter and thus through R4?

This current through R4 has the effect of raising the voltage at the emitter.
Since there has to be at least the saturation voltage across the transistor this reduces how far the collector voltage can drop.