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This *should* be easy ... simple transistor amp stage

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carbonzit

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I swear: I know a bit about electronics, but it appears I can't design a simple working circuit to save my life!

Circuit in this case is just a 1-stage transistor amplifier. Tried sim-ing it in LTspice, but I get zero output. Here it is:
Single-stage transistor amp 1.gif


I'm feeding it a low-frequency (500 Hz) 300 mV signal. Why don't I see anything at the output (on either side of the coupling cap)?

I'm not trying to claim this is supposed to be a particularly good amplifier; I'm just trying to get the damn thing to work so I can proceed with the other parts of a project.

Seems to me my problems might be with LTspice, not necessarily the circuit itself. I guess I need to separate the two (LTspice and design problems).

300 mV is not a huge voltage, but it certainly should be amplifiable, right? Is it possible I have this thing biased so that the transistor isn't anywhere near the active region? I know one wants to get the quiescent point somewhere near 1/2 Vcc; am I way off here? And 20K is just a guess at a light output load.

Any help would be appreciated here.
 

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  • Amp experiment 1.asc
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How do we even know the source impedance? It's one of LTspice's voltage sources: aren't these essentially "perfect" voltage sources with near-infinite impedance? Is that a parameter I can change in the simulation?

And besides, wouldn't a low source impedance work in our favor? I thought the problem was when one has a high-impedance source with a low-impedance load, not the other way around.
 
The input source varies from 0 to 300 mV, which is below the .6V+Ie*R2 needed to turn on the transistor. Its zero impedance means that the biasing network is pretty much ignored.

Impedance of an ideal voltage source is zero. LTSpice lets you set a series resistance in the properties of the source.

A DC-blocking capacitor will help.
 
Blocking cap (on the input, just after signal source, assuming that's where you were suggesting putting it) did nothing. But changing the source impedance (Rser) of the voltage source got me some output (~800 mV). Optimal value seemed to be about 600Ω, oddly enough.

But now I need more gain ... that's (roughly) Rc/Re, innit?

Hmm; changing emitter R to 20Ω gives me almost 2V of output. Much better.

And changing Rc to 400Ω gives me 3.6V (with what looks like the beginnings of non-linearity).

Thanks for getting me off of dead center here.
 
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Blocking cap (on the input, just after signal source, assuming that's where you were suggesting putting it) did nothing. But changing the source impedance (Rser) of the voltage source got me some output (~800 mV).
The coupling cap between the signal source and the base/R1/R4 junction.
But now I need more gain ... that's (roughly) Rc/Re, innit?
Yes, gain is Rc/Re. Normally a capacitor is connected in parallel with Re to increase the AC gain.
 
Try setting the source Ton and Tperiod to non-zero values ;).

Edit: You need an input coupling cap of several uF. The output coupling cap could also be increased.

Edit2: Reduce R4 to, say, 1k to prevent the transistor saturating.
 
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How do we even know the source impedance? It's one of LTspice's voltage sources: aren't these essentially "perfect" voltage sources with near-infinite impedance? Is that a parameter I can change in the simulation?

And besides, wouldn't a low source impedance work in our favor? I thought the problem was when one has a high-impedance source with a low-impedance load, not the other way around.
You have it backwards.
"Perfect" voltage sources have zero internal impedance.
"Perfect" current sources have infinite internal impedance.

The problem with a zero source impedance in your circuit is that it shorts the DC bias voltage from R1 and R4 to ground so the transistor has no bias (zip, zero, nada). ;)
 
The transistor is not biased correctly. Without the signal generator shorting its base to ground, measure the DC collector voltage. It should be a little higher than half the supply voltage but instead you have it biased wrongly with its collector saturated very close to ground.
 
The green trace is the collector voltage.
The blue trace is the base voltage.
Note: C2 is added. The base resistors are changed to bias the amp so the Collector voltage will want to go to 1/2 supply.
upload_2014-12-9_7-15-45.png
 
OK, now I'm getting somewhere. Here's the latest iteration of this experiment, which is now a (working) part of my previous experiment, a triangle-wave generator:

Amp experiment 2.gif


(full simulation file attached below)

And here are the resulting output waveform plots:

Triangle wave output 3b.gif


Not bad if I don't say so myself. I added an emitter-follower output stage to give a low output impedance, and therefore a better ability to drive a load. If this was a working instrument I'd probably replace the emitter resistor with a potentiometer to adjust the output. (Well, and the whole thing still needs a bit of tweaking to improve linearity, but that's a separate issue.)

One question: what would I do to bring the bottom of the output down to 0 volts? (Level-shifting, right?) And what if I wanted to make the output signal swing symmetrically around 0 volts? I noticed in playing with earlier versions of the amp experiment that I got an output that did just that, swinging from ~-2 to +2V. Is this even possible with a single supply, or do I need + and - supplies to do this? Trying not to complicate this too much (but we do loves complexity, don't we?).

You have it backwards.
"Perfect" voltage sources have zero internal impedance.
"Perfect" current sources have infinite internal impedance.

Ah, yes. Thanks for the correction. Kind of counter-intuitive (for some reason I think of high impedance=high voltage), but I'll try to remember that.

Lastly, it appears that additional coupling capacitors are completely unnecessary here. They don't seem to affect things in the simulation one way or the other, in any case.
 

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  • Triangle-wave generator 3b.asc
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Add a resistor (e.g.10k) from the output to ground and your output voltage will swing above and below ground.
 
Your new transistor amplifier has a voltage gain of roughly R2/R3= 200/50= 4 times as you show.

The resistor values for R1 and R4 for the base current of the first transistor are too high for the high current caused by the low resistor values of the emitter and collector.
If the supply voltage is 9V then the collector to emitter current in the first transistor is about 20mA. Its hFE is about 200 so its base current is about 0.1mA. The current in R2 and R4 is only slightly more than 0.1mA but it should be 10 times the base current to allow for a transistor having an hFE lower than average.
R1 should be 4k but nobody makes 4k so use 3.9k. R4 should be lower than 2.2k because in your circuit the value of R1 was much too high then R2 did almost nothing.

The emitter-follower does nothing because the value of its emitter resistor is much too high. An emitter-follower is not needed anyway because the value of the collector resistor R2 is so low at only 200 ohms.
 
One question: what would I do to bring the bottom of the output down to 0 volts? (Level-shifting, right?) And what if I wanted to make the output signal swing symmetrically around 0 volts? I noticed in playing with earlier versions of the amp experiment that I got an output that did just that, swinging from ~-2 to +2V. Is this even possible with a single supply, or do I need + and - supplies to do this?
As Alec_t said; add a resistor to ground. When you added C1 things changed. If you looked at the OUTPUT with a scope or a meter (anything with some resistance) you will find the center of the waveform is at ground with peaks going + and -. You have lost the ability to pass a very low frequency signal or DC.
Lastly, it appears that additional coupling capacitors are completely unnecessary here. They don't seem to affect things in the simulation one way or the other, in any case.
This is not a real world situation. You are looking at OUTPUT with out a load. A 1meg ohm resistor in a scope probe will cause the output to do just what you want. Blue is the collector voltage while green is the OUTPUT voltage with a 10k to ground. This is normally not done because the amplifier (capacitor) will not pass low frequencies.
upload_2014-12-10_7-54-2.png
 
Add a resistor (e.g.10k) from the output to ground and your output voltage will swing above and below ground.

Ah, yes, a load, like you'd find in real life. Note to self: always simulate a load. Thanks for that!
(Although I'm still mystified by how a single-ended circuit like this, one with a single supply, can produce a negative output at all ...)

You have lost the ability to pass a very low frequency signal or DC.

Yes, I know, because of the small value of the coupling cap. I did that to shorten the simulation time. In Real Life I'd use a larger cap, since I see this being used to generate a range of frequencies.

I really must get myself a 'scope ...
 
I'm still mystified by how a single-ended circuit like this, one with a single supply, can produce a negative output at all
Consider a cap with its left end at 4V and its right end connected to ground (0V) via a load resistor R. The cap will initially charge slowly via R. Once fully charged its right end will be at 0V. Now shift the left end down abruptly by 1V. The right end will instantaneously also go down by 1V, from 0V to -1V (since the 4V voltage across the cap can't instantaneously change). Subsequent slow discharge of the cap will bring its right end back to 0V.
 
.........................
Ah, yes. Thanks for the correction. Kind of counter-intuitive (for some reason I think of high impedance=high voltage), but I'll try to remember that.
..............................
To help remember, here's an intuitive way to think about it:

You want a voltage source to deliver a constant voltage independent of the load current. To do that there can be no IR voltage drop in the voltage source due to the load current, and thus its internal impedance must be zero (which can be achieved very closely with regulated power supplies)

Similarly you want a current source to deliver the a constant current independent of the load resistance and output voltage. To do this its output impedance must be infinite (or practically, a value much higher than the load resistance).
 
So thanks to all so far for your indulgence. Wonder if I can pick your brains a little bit more.

Here's my output amp so far:

Amp experiment 3.gif


Works OK, sorta, but I want more. I'm only getting a gain of about 4 here.

Here's what I'd like: more gain (~10 or so), a relatively high input impedance and a low output impedance. I think I have the low output Z with the emitter follower. Seems like this should be doable with these little xistors, no? I'm not asking too much here, am I? (And a Vcc of 9 volts.) Ultimately I'd like to be able to get maybe 3.5-4V p-p out of this (again, not asking too much of a 9-volt supply, am I? or is that too close to the "rails"?).

I know I should be hitting my books and figuring this out for myself, and really, I plan on doing that, someday soon, but in the meantime I hope someone will forgive my laziness and at least give me a hint or two ...

(afterthought) Well, increasing the supply to 12V certainly seems to help. Can anyone tell me what the maximum (relatively undistorted) output (p-p) one can expect on a 9V supply? Maybe I just need to up the voltage.
 

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  • Triangle-wave generator 3b.asc
    6.2 KB · Views: 124
The gain is a little less than R8/R9= 220/47= 4.68 times only if the input signal has a low impedance because of the negative feedback provided by R7. Increase the gain by increasing the value of R8, reducing the value of R9 or both. It works well with a higher output level when R9 is 33 ohms.
The emitter follower has a high output resistance for the downward swing of its output so it should be eliminated or added to the input of the gain transistor.
 
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But when I change R9 to 33Ω, the output waveform degenerates:

Triangle wave output 3b2.gif

(this is with a 12V supply)

The emitter follower has a high output resistance for the downward swing of its output so it should be eliminated or added to the input of the gain transistor.

Hmm, not quite sure what you mean by that: are you saying to put the emitter follower before the gain stage?
 
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