## Transistor Amplifier

My year 12 exams are in 1 week exactly and I'm worried about a question (from last year's exam) that my teacher can't even answer. Each year they tend to repeat the difficult questions, and this was one of those.

I'll just give a summary of the question, and maybe you'll be able to answer my question without having to look at it.
- An AC input signal of 0.3V peak to peak is applied at the base.
- There is also a DC voltage of 4 V going through the base.
- The voltage gain is 200.
- The DC supply voltage is 20 V.
- They don't give a linear transfer characteristic.

To answer the question, you have to draw a graph of output voltage against time. In their answer, the graph is clipped at +8 and -12. I don't understand where they got those values from. What I thought I had to do was add the AC voltage to the DC across the base (ie. sinusoidal from 3.85 to 4.15, varying about 4V) and then normally you just read off the V in against V out graph.

The question is in this pdf: http://www.vcaa.vic.edu.au/vce/studi...hysics1w06.pdf

It's in the electronics section, question 3.

This post is so messy - hopefully you'll be able to understand it. I so desperately need the answer to this! Thanks for your help

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 In their answer, the graph is clipped at +8 and -12. I don't understand where they got those values from. What I thought I had to do was add the AC voltage to the DC across the base (ie. sinusoidal from 3.85 to 4.15, varying about 4V) and then normally you just read off the V in against V out graph.
To find the range of the output you need to find the compliance of the amplifier. The first step is to find the maximum peak-to-peak value that the output voltage can swing. This is found by: 2VCEQ.
VCEQ=Vcc-ICQ(Rc+Re)
As you know, ICQ is approximately equal to IE

If you're unclear as to how we got that equation, let us know.

 Quote by ranger To find the range of the output you need to find the compliance of the amplifier. The first step is to find the maximum peak-to-peak value that the output voltage can swing. This is found by: 2VCEQ. VCEQ=Vcc-ICQ(Rc+Re) As you know, ICQ is approximately equal to IE If you're unclear as to how we got that equation, let us know.
I know I didn't ask this question, but as someone who's interested in physics instrumentation (which requires a lot of electronics), I would like to know how you got that equation.

Recognitions:
Gold Member

## Transistor Amplifier

Based on the schematic (Figure 1, page 11) on the sample test, the amplifier is configured in the common-emitter configuration. The first step is to find the quiescent (at rest; no input signal) voltage and current. Quiescent states gives us an idea as to how the transistor is biased and dictates the compliace of the amplifier. I wont get into the details on how to do this, as the discussion will become very lengthly. For a very gentle intro to transistor amplifiers, I suggest Robert T. Paynter.

But we arrived at that equation by using KVL and the fact that collector and emitter at-rest currents are approximately equal. Since the 20V DC supply is connected the collector, it means that:
20V = ICRc+VCE+ICRe
Since we're finding these values when the amplifier is at "rest", it becomes
20V = ICQRc+VCEQ+ICQRe
Solving for VCEQ we get:
VCEQ=Vcc-ICQ(Rc+Re)

The value of quiescent collector current can be found by:
ICQ=IE=VE/RE

Since the input signal will oscillate about the quiescent state, its pretty obvious why the maximum output (peak-to-peak) swing is 2VCEQ.
 I really appreciate your help, but I still don't quite understand. I've read your explanation through a few times, but my knowledge is just so basic. What determines the point at which they clip?
 Recognitions: Gold Member Hi freswood, When we dc bias a transistor, we need to choose two quiescent points. These points are quiescent collector current and quiescent collector-emitter voltage. When we apply an input ac signal to be amplified, the ac signal "rides" on top of the quiescent points. In order for us to have the maximum possible output swings for the amplified input signal, the amplifier must be biased so that collector current and collector-emitter voltage are half there maximum possible value. Actually, I just glimpsed at my previous post where I gave you 2VCEQ to find the max swing. We need to also do 2ICQ*rC. rC is the parallel combination of the collector resistor and load resistance. The overall compliance is found to be the lower of the products (sorry I forgot to mention this). Once you've found the compliance, you need to know how the input ac signal will respond using the given voltage gain. If the output voltage exceeds the compliance, we have clipping.
 I talked to my teacher about it, and the answer she was told was this: -The clipping at the top is the same as the voltage drop across Rc (8V) -Then you know what the bottom clipping is because the value from 0 to the top clipping + 0 to the bottom clipping is the supply voltage (20V) which makes the bottom clipping -12V. Is this correct in any way. It's ok if I don't understand the science behind it, but is that the right way to go about doing the question?
 Note that Vout is taken after the Capacitor, this means that the DC part of the Voltage will be Removed.

Recognitions:
Gold Member
 Quote by freswood I talked to my teacher about it, and the answer she was told was this: -The clipping at the top is the same as the voltage drop across Rc (8V) -Then you know what the bottom clipping is because the value from 0 to the top clipping + 0 to the bottom clipping is the supply voltage (20V) which makes the bottom clipping -12V. Is this correct in any way. It's ok if I don't understand the science behind it, but is that the right way to go about doing the question?
I don't like that approach at all.

Here is an easier way to do it. You have to draw the DC loadline. You do know how to draw one right? On the line you'll of course plot Icq and Vceq. SO we have have Vin peak as 0.15V with a gain of 200. This means that the amplifier will be producing a 30V peak output wave form. But becuase of clipping, the positive and negative portions are distorted. To find the exact points of clipping, you'll that DC loadline I told you to draw.

Looking at the DC loadline, the range of [output] swing is basically from 0 <= Vce <= Vce(off). Vce(off) in this case is 20V. Like I said before, the output (Vce) will oscillate about its quiescent point (Vceq). This basically means that Vceq will be the horizontal axis when analyzing the graph.

Note: when I analyzed it, I got the answer you gave only when ignoring Vbe (0.7V). And I got -8V and +12V as the clipping points and not +8V and -12V.

 Quote by ranger I don't like that approach at all. Here is an easier way to do it. You have to draw the DC loadline. You do know how to draw one right? On the line you'll of course plot Icq and Vceq. SO we have have Vin peak as 0.15V with a gain of 200. This means that the amplifier will be producing a 30V peak output wave form. But becuase of clipping, the positive and negative portions are distorted. To find the exact points of clipping, you'll that DC loadline I told you to draw. Looking at the DC loadline, the range of [output] swing is basically from 0 <= Vce <= Vce(off). Vce(off) in this case is 20V. Like I said before, the output (Vce) will oscillate about its quiescent point (Vceq). This basically means that Vceq will be the horizontal axis when analyzing the graph. Note: when I analyzed it, I got the answer you gave only when ignoring Vbe (0.7V). And I got -8V and +12V as the clipping points and not +8V and -12V.
Check out

http://www.vcaa.vic.edu.au/vce/studi...ssrpjune06.pdf

page 4 tells you the answer to this question. Not the same signs as your clipping this is so frustrating.

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Gold Member
 Quote by poala999 Check out http://www.vcaa.vic.edu.au/vce/studi...ssrpjune06.pdf page 4 tells you the answer to this question. Not the same signs as your clipping this is so frustrating.
I usually don't do this, but you guys need it for a test on Monday, I'll post some work:
Attached Thumbnails

 thanks. great work. I know the site doesn't like full answer but i like to see full worked examples. helps so much, especially how you have related it with the graph. keep up the work. exam on tuesday, can't wait, i'm strange, i like exams.

Mentor
 Quote by ranger I usually don't do this, but you guys need it for a test on Monday, I'll post some work:
That's a pretty detailed post by ranger. I would have left out most of the numbers and used symbolic hints at what is going on, in order to avoid the interpretation of posting a complete solution (which is not allowed here on the PF, as has been mentioned earlier in this thread).

I think it would have been possible to teach almost as much with the graphs and explanations, without plugging in as many numbers. Make the students do at least that much, even if they do have a test tomorrow.

EDIT -- I do appreciate, BTW, ranger's patience in trying to help out you folks. ranger put a lot of effort into being tutorial in this thread.
 disagree with your teaching method. you can clearly tell the people who have tried and failed, and the people who simply want you to do there assignment. Had the exam, went great. Went all out this year and had two of those graphs. Vcc=6V and q point was 3V, with voltage gain of 80.... clipping occured at 3v when 240mv was put in, so happy. fairly easy. thanks.