# Figuring out outputs

Discussion in 'Homework Help' started by Velocity, Apr 19, 2009.

1. ### Velocity Thread Starter Member

Feb 27, 2009
14
0
Hello all,

I've got an assignment in which I have to sketch out the outputs of various circuit designs. The circuits are displayed below. Each input (point A) is being driven by a sine wave with a period of one millisecond, centered at zero volts. Circuit #1's input is a sine wave with a period of one millisecond, oscillating between +1 and -1 volt. Circuits #2 and #3 have sine wave inputs with a period of one millisecond, oscillating between +5 and -5 volts.

The purpose is to sketch the outputs at each of the marked points (B, C, D) in the three corresponding circuits.

Now, I have been able to figure some things out for this assignment...

Circuit #1: Point B experiences a gain of -10; the voltage swings between -10V and +10V. Point C experiences a gain of -400, assuming a 25Ω resistance in the transistor. However, this gain is not constant, and that's where I'm having trouble figuring out the output.

Circuit #2: From the output of the op-amp around to ground is a voltage divider with part C in the middle... the divider has a gain of 5.

Circuit #3: The capacitor-resistor is a differentiator. It's also a high-pass filter with an f-3dB point at about 159 Hz. If I'm correct, at part B we should have an output graph of what looks like a cosine wave oscillating between +5V and -5V (centered about zero) with a period of 1 ms.

That's all I have at the moment. In summary, point B of Circuit #2, point C of all three circuits, and part D of Circuit #3 are the ones on which I'm not sure. Any help?

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Last edited: Apr 20, 2009
2. ### PRS Well-Known Member

Aug 24, 2008
989
36
vxsd nCircuit 1 is an amplifier with gain of 10K/1K=-10 at the collector, point b. The negative just means it is 180 degrees out of phase with the input. And, you're right, we have +/- 10 volts there at the collector. But don't forget this signal is at 1 kHZ and riding on a DC voltage at the collector. Ththe emitter voltage, given Vb=0, is -15 -(-.6) volts = -14.4 volts. The current, Ie =-(.6-(-15))/1000= 14.4mA. Since Ic is approximately Ie we have 15-10k*14.4mA=.6volts. So draw the .6 volts (straight horizontal line) on a graph then draw a +/_10 volt sinewave with a frequency of 1000 Hz. Remember it should be drawn 180 degrees out of phase with the input.

At point c the output is in phase with the input with a gain of approximately 1. The dc voltage there is -14.4 volts. Draw that as a straight horizontal line on the graph.

Notice that the collector voltage just touches on that line without getting clipped. So it's a good design. The emitter signal is +/- 1 volt. Draw that on the -14.4 bias line.

At point A we simply have the signal +/-1volt, 1 kHz sinewave on with a bias of 0 volts.

Last edited: Apr 19, 2009
3. ### Velocity Thread Starter Member

Feb 27, 2009
14
0
Thanks PRS... I think.

I should have mentioned which graphs I already drew up. At point B in Circuit #1 I drew a sine wave ranging from +10V to -10V, 180º out of phase with the input, centered about 0 volts. For point B in Circuit #3 I drew a sine wave ranging from +5V to -5V, 90º out of phase with the input (i.e. a cosine wave).

I understand the diode effect on the voltage in the first circuit, but I want to make sure of things before I put the graphs on paper. For part B, it's a +10/-10V swing, but is that centered at +.6V or zero? Same for part C; it's not centered at -14.4V is it? That wouldn't make sense...

Does anybody else have anything to add to help me out a bit here?

4. ### Velocity Thread Starter Member

Feb 27, 2009
14
0
All right, so thus far I have figured out four of the graphs... I think. I'm still stuck on #2 though as far as drawing the output graphs.

Anyway, for point B on circuit #1 I've got the +/- 10V sine wave centered at .6V with a period of 1 ms, and it's inverted. For point C I've got a +/- 1V sine wave centered at -.6V, in phase with the input. Is this correct?

For Circuit #3, I have at point B a cosine wave with +/- 5V swing, centered at 0V, with a period of 1 ms. (From what I saw, the capacitor/resistor is a differentiator.) At point C, I have a cosine wave which is clipped; its peak is 4.4V (taking into account the diode drop from the transistor) and it clips at the zero point due to the ground at the emitter. It too has a period of 1 ms and is in phase with point B.

Am I on the right track?

5. ### PRS Well-Known Member

Aug 24, 2008
989
36
LOL

I'm pretty sure it's .6 volts. Calculate the current at the emitter, Ie, (remember Vbe=.6 volts and Vb=0 volts). The dc collector voltage is given by Vcc-RcIc and Ic is about the same as Ie. Furthermore, consider the graph. If Vc were not .6 volts the signal would get clipped.

I didn't evaluate circuit #3 or #2. I thought I could help you with the first one and you'd get the idea. But I'll look at them.

Right, just think of that be junction as a diode which has a value of .6 volts for silicon and .2 volts for germanium.

Yes it is a sine wave centered on a bias of .6 volts.

Sorry, you're right. It rides on a dc voltage of 0-.6 = -.6volts. Go to the head of the class, Velocity. Now graph this stuff and you will see this circuit clips the negative going sinewave at -.6 volts. This is true because the sinewave must rise and fall through the availabe Vce. Draw the load line and you will also see this.

6. ### PRS Well-Known Member

Aug 24, 2008
989
36
It's been years since I learned this stuff, Velocity, so give me a little room to make mistakes. Circuit 2 is that of a non-inverting op amp. The gain is given by vo=10k/2k+1=6 volts per volt. It is in phase with the input. If the input is +-5 volts then the gain results in +/- 30 volts and clipping occurs at +/- 15 volts. This wave rides on approximately 0 volts dc. The result is sort of like a square wave having rounded rise and fall times.

But what about point C in circuit 2? That's the hard part. Feedback from an in phase output signal is going to the inverting terminal of an op amp. Remember differential amplifiers are used at the input of op amps. By feeding positive feedback to the inverting terminal it is made thereby negative feedback as it subtracts from the input signal. But how much does it subtract? I believe this is given as Vout*(2k/(2K+10K) This, due to voltage division.

Correct, but it is important to show the bottom of the sinewave getting clipped at -.6 volts.

I need to take another look at it....

Last edited: Apr 20, 2009
7. ### PRS Well-Known Member

Aug 24, 2008
989
36
This part of the circuit is a high pass filter. The capacitor blocks dc. As the frequency increases so does the gain. The 3dB point is calculated as 1/(2*pi*R*C). At that point the gain is .707 times the midband gain.

As far as circuit 3 goes, we don't know Rc or Re, right? So we have no idea whether it is biased right or not. But we do know that at that 3dB point the gain decreases with decreasing frequency and will not pass dc.

You are doing good, Velocity! Hope I am?

8. ### PRS Well-Known Member

Aug 24, 2008
989
36
One last thought. Your teacher really socked it to you! LOL, in fun, not malice.

9. ### Velocity Thread Starter Member

Feb 27, 2009
14
0
Actually after asking my professor, Rc and Re on Circuit #3 are the same as Circuit #1; i.e., Rc = 10kΩ and Re = 1kΩ.

Thanks for the help. I'm starting to understand it more, I think. I know the purpose of the various circuits in general, but my biggest trouble is calculating everything and understanding it fully. Go figure, it makes sense when the professor talks about it in class but as soon as I'm out of there... I get lost.

10. ### PRS Well-Known Member

Aug 24, 2008
989
36
In that case use the calculated results from 1 and project them onto 2 with the exception of the frequency roll on at 1/(2*pi*RC)

11. ### Velocity Thread Starter Member

Feb 27, 2009
14
0
3dB point is about 159 Hz from what I calculated. So for a 1kHz sine wave everything should pass through, yes?

I also got the right gain for Circuit 2, it's the graphing I can never seem to get. So it ranges ±30V and looks like a sine wave, but it clips at ±15V. If I'm thinking correctly, it'll have the rounding at the top/bottom, go horizontal at ±15V, but at the "zero" point of a regular sine wave, this graph would switch immediately to the opposite side... I'll have to draw one and upload a pic to be sure you know what I mean.

12. ### PRS Well-Known Member

Aug 24, 2008
989
36
Well, I appreciate the fact that you're putting some confidence in me now, Velocity. But we are yet to see if I can be trusted. In exchange for my time I'd appreciate it if you'd let me know what the right answers are.

As for frequency response just remember that there is no output at dc, infinitesimal output at .ooo1 volt and you don't approach the midband gain -- if Re and Rc are the same as circuit 1 then that is 10 volts per volt -- until 159 Hz. At that point the gain is .707*10=7.07. But are you expected to know this or only infer the amp won't have much of a response until 159 Hz? I don't know where you are at exactly in your studies.

As far as circuit 2 goes you have a gain of 6 with +/-5*6 = +/- 30volts. But your supply limits you to +/-15 volts, therefore the clipping, and yes it just flattens the waveform at that clipping point. I think you have the right idea.

13. ### Velocity Thread Starter Member

Feb 27, 2009
14
0
Okay, I've got the idea of the graph of the second one. I think we're on the same page there too. As for the impact of the frequency response... I'm not sure. We didn't really get too far into it. All I know is that at the 3dB point, the power drops by half and output voltage has dropped by the .707 ratio you mentioned.

The fact that the input is at 1kHz makes me think that the 3dB point of the C-R duo won't have much of an effect. Again, our professor didn't go too in-depth in that regard.

And of course I'll repost later with the correct answers... that is, assuming that my professor returns the homework at all. He hasn't been that good with returning our homework on time this semester.

14. ### Velocity Thread Starter Member

Feb 27, 2009
14
0
Okay, I'm nearly done with this, but I've got just another question on the third circuit... the C-R does act as a high-pass filter, but based on what I read in my text it also looks like a differentiator. Secondly, the voltage difference - it differs from #1 in that the applied voltages to the C and E are +15V and ground, respectively. This would imply a centering voltage of 7.5V. Thus for part D, would this and the gain of -10 correspond to a +/- 10V wave centered at 7.5V?

EDIT: From the calculations I made similar to Circuit #1, the current through the collector region is .6 mA and results in a 6V drop across Rc. This leaves 9V at point D. Would this, then, be the center of the graph of a +/- 10V output...?

15. ### PRS Well-Known Member

Aug 24, 2008
989
36
You're right again. Seems I need a refresher.

16. ### PRS Well-Known Member

Aug 24, 2008
989
36
Here's the actual output from the circuit. The capacitor is there merely to block any dc voltage that might come from the source. That way the circuit is self-biasing. The edges of the Vc signal are merely tapered, not curved. I can't draw well with Paint.

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17. ### Velocity Thread Starter Member

Feb 27, 2009
14
0
I'm assuming the top graph is what occurs at point B of the circuit and the bottom graph is what happens at point D?

EDIT: Never mind, I see where it says Ve and Vc now. That'd be C and D in the circuits I drew, respectively.

Last edited: Apr 21, 2009