simple BJT audio amp experiment (need help)

alphacat

Joined Jun 6, 2009
186
Hey,

The output resistance of a CE amplifier is Rc || ro ≈ Rc.
If no RC is used, then the output resistance of the amplifier would be very small, which results in a large attenuation since the amplifier's gain consists of the term: Rout / ( Rout + RL), where RL is the load's resistance.

Just reminding that the output signal is current, therefore we would want Rout >> RL.
 

Wendy

Joined Mar 24, 2008
23,415
In this case there is no other resistance other than the speaker, so there is no parallel case to be discussed. Just the speaker.

In a CE amplifier, the gain is generally Rc/Re. Since there is no Re, the gain will be at max, the ß of the transistor. Since the gain goes down as frequency goes up (a given with any amplifier) the frequency response of this amp isn't exactly flat, but then we knew fidelity wasn't the strong point of this circuit.
 

Thread Starter

count_volta

Joined Feb 4, 2009
435
OK, first thing I see is the capacitor, which is a short to audio, is shorting the mic out. The schematic should look like this instead...



The mic is equivalent to a small AC generator, the cap allows the loop to be closed for it's current flow, which also happens to be through the BE.

Meanwhile, the cap blocks the DC, which is biasing the transistor into a linear amplification state.

See?

BTW, I didn't break it down, once I realized where the cap was I stopped looking, but your breadboard matches the original schematic from what I saw.

The previous graphic image I uploaded was my way of drawing protoboard layouts, for my various articles.
Cool, I got the connections right. :D

By the way Bill what do you use to draw the breadboard like that? It looks easier than the 20 minutes I spent drawing holes yesterday.

So maybe you are right about the connection of the capacitor. I will try it and see if I get acoustic feedback this time.

P.S. for acoustic feedback do I need to speak or yell into the microphone as its facing the speaker at a distance of like 1-2 inches apart? Thats what I been doing.
 
P.S. for acoustic feedback do I need to speak or yell into the microphone as its facing the speaker at a distance of like 1-2 inches apart? Thats what I been doing.
Allow the "mic" to face the speaker as close together as you can get them. You don't need to speak or yell into the "mic". If there's enough gain, the squeal will start on its own.

With my 2" speakers I got from Goodwill, they would feedback with a spacing of about 1.5", but with the little 1" Radioshack speakers, they needed to be almost touching.

Don't forget to try reversing the wires to one of the speakers because one polarity may not give the right feedback phase to cause squeal.
 

Wendy

Joined Mar 24, 2008
23,415
My brother used a tin coffee can to do something similar in our teens. Basically a coil mounted in back of the bottom lid. It was tinny (no pun intended), but it worked.

That protoboard drawing is part of my Paint as CAD (PaintCAD) package, you can find it in my "Introduction and Paint as CAD", which can be found here.

Bill's Index

Basically you can copy and paste the parts shown below to do something similar. I've found the neater you make a circuit, the more likely it is to work first time you apply power.

Sorry I haven't been paying more attention here, I'm just spread pretty thin. I bought some cheap 3" speakers I plan on using for this, if you read my articles (check out the 555 projects) I'm always keeping an eye out for project ideas.

Allow the "mic" to face the speaker as close together as you can get them.

With my 2" speakers I got from Goodwill, they would feedback with a spacing of about 1.5", but with the little 1" Radioshack speakers, they needed to be almost touching.

Don't forget to try reversing the wires to one of the speakers because one polarity may not give the right feedback phase to cause squeal.
I'm not saying it is, but what are the chances there is some magnetic coupling going on? Something that occured to me on my way to work, I do a lot of thinking on that short (20 minute) commute. Part of the reason I bought the bigger speakers is to increase the efficiency, which sounds like could be part of the problem.
 
Last edited:
In this case there is no other resistance other than the speaker, so there is no parallel case to be discussed. Just the speaker.

In a CE amplifier, the gain is generally Rc/Re. Since there is no Re, the gain will be at max, the ß of the transistor. Since the gain goes down as frequency goes up (a given with any amplifier) the frequency response of this amp isn't exactly flat, but then we knew fidelity wasn't the strong point of this circuit.
Bill, I think it would help avoid confusion if you would further qualify the word "gain" in these discussions. As you know, there's voltage gain, current gain and power gain.

Since you said "...the gain is generally Rc/Re.", I can tell that you're talking about voltage gain, but the novice reader might not realize that. Since some of the participants in this thread are still not fully up to speed on the relative roles of voltage and current gain, I think it's good to be very specific.

In a CE amplifier where there's no Re, the voltage gain isn't equal to the β of the transistor. When there's no Re, then the voltage gain is Rc/re, where re is the intrinsic emitter resistance, whose value is .026/Ic.

For the circuit I wired up, Ic was 50 mA, so re was .026/.05 = .52Ω, and the theoretical voltage gain should have been 8Ω/.52Ω ~ 15 (see post #134).

So even though the transistor's β was somewhere around 100 or more, the voltage gain was only 15.

If re is .52Ω and β is 100, then the input impedance at the base in this circuit would be about 100*.52 = 52Ω. Not a great match, but it's within an order of magnitude.

Notice that if you double Ic to 100 mA then re becomes .26Ω and the voltage gain doubles. Also, the input impedance is halved to 26Ω, a better match to the 8Ω "mic" source impedance. Doubling Ic is well worthwhile.
 

alphacat

Joined Jun 6, 2009
186
Bill, I think it would help avoid confusion if you would further qualify the word "gain" in these discussions. As you know, there's voltage gain, current gain and power gain.

Since you said "...the gain is generally Rc/Re.", I can tell that you're talking about voltage gain, but the novice reader might not realize that. Since some of the participants in this thread are still not fully up to speed on the relative roles of voltage and current gain, I think it's good to be very specific.

In a CE amplifier where there's no Re, the voltage gain isn't equal to the β of the transistor. When there's no Re, then the voltage gain is Rc/re, where re is the intrinsic emitter resistance, whose value is .026/Ic.

For the circuit I wired up, Ic was 50 mA, so re was .026/.05 = .52Ω, and the theoretical voltage gain should have been 8Ω/.52Ω ~ 15 (see post #134).

So even though the transistor's β was somewhere around 100 or more, the voltage gain was only 15.

If re is .52Ω and β is 100, then the input impedance at the base in this circuit would be about 100*.52 = 52Ω. Not a great match, but it's within an order of magnitude.

Notice that if you double Ic to 100 mA then re becomes .26Ω and the voltage gain doubles. Also, the input impedance is halved to 26Ω, a better match to the 8Ω "mic" source impedance. Doubling Ic is well worthwhile.
Hey Electrician.

I wanted to ask you please about a couple things you mentioned.

First, this is a transconductance amplifier, right?
that is since the input is a voltage signal (which is produced by the mic), and the output is a current signal (which drives the speaker).

So the gain of this amplifier is a transconductance gain, and it equals to:
Gm = [Rin / ( Rin + Rsource)] * gm * [Rout / (Rout / RL)]

If i'm correct about that, then i dont understand why would you say that low Rin (26ohm) is better than larger Rin (52ohm)?

The lower Rin, the larger the attenuation in the input stage, becaues of the term:
[Rin / ( Rin + Rsource)]

I'd be happy to hear what you got to say about this.

Thank you very much.
:)
 

alphacat

Joined Jun 6, 2009
186
In this case there is no other resistance other than the speaker, so there is no parallel case to be discussed. Just the speaker.

In a CE amplifier, the gain is generally Rc/Re. Since there is no Re, the gain will be at max, the ß of the transistor. Since the gain goes down as frequency goes up (a given with any amplifier) the frequency response of this amp isn't exactly flat, but then we knew fidelity wasn't the strong point of this circuit.
Thanks a lot Bill.
I got you on this. :)
 
Any device which can supply electrical energy will have an intrinsic impedance. For theoretical purposes, we can model a source as though it had an internal ideal voltage source (zero output impedance) in series with an impedance representing its actual impedance, or we can model it as though it had an internal ideal current source (infinite output impedance) in parallel with an impedance representing its actual impedance. Thevenin's theorem says (with certain restrictions that don't apply here) that either of these two models should give the same result when used as an element in a larger circuit.

A source with something other than infinite or zero output impedance can often be treated as though it did have one of those two output impedances, depending on what impedance it is driving. The determining factor is the ratio of the source impedance to the load impedance.

If the source impedance is much higher than the load impedance, then we will find that the result we get by assuming the source has infinite output impedance (is a current source) is the same as the result we get if we analyze using the actual output impedance to a very good approximation.

On the other hand, if the source impedance is much lower than the load impedance, then we can similarly assume that the source has zero output impedance (is a voltage source), and we will get a result that is a good approximation to the exact result derived from analysis with the actual output impedance of the source.

To give a concrete example, assume we have a source with an output impedance of 1000Ω. If we are driving an amplifier with an input impedance of 10Ω, we can proceed as if the source were a current source, and we will get almost the same result as if the source had 1,000,000,000Ω output impedance.

But if the amplifier has an input impedance of 1,000,000Ω (such as a vacuum tube or FET), then we can assume the source is a voltage source, and we will get very nearly the same result as if the source had zero ohms output impedance.

So, whether we approximate the source as a current source or voltage source depends on the actual impedance of the source relative to the actual impedance of the load.

In the case of the single stage CE amplifier we're experimenting with in this thread, the source "mic" has an impedance of about 8Ω, considerably lower than the input impedance of the transistor stage. We're justified in treating it as a voltage source.

Any amplifier can be treated as a "transconductance" amplifier where the input source is taken to be a voltage source, or as a "transresistance" amplifier where the source is taken to be a current source, if the actual source impedance is neither zero nor infinite, but something in between.

If the source impedance is in fact substantially lower than the input impedance of the amplifier, then treating the stage as a transconductance amplifier may seem more intuitive.

In that case, we pretend that the source impedance is zero and the input impedance of the amplifier is infinite. Then the voltage gain of the stage is -gm*RL. The transconductance of a BJT is given by gm = Ic/.026, so the larger Ic, the larger gm and the larger the voltage gain.

If you wanted to take into account the source impedance and the input impedance of the amplifier stage, you can derive a somewhat more complicated formula for the voltage gain (Rout is completely negligible in this circuit because our load is only 8Ω):

\(Av=\frac{-gm*Rin*RL}{Rsource+Rin} \)

Notice that this formula reduces to the simpler -gm*RL if either Rsource becomes zero or Rin becomes infinite.

If i'm correct about that, then i dont understand why would you say that low Rin (26ohm) is better than larger Rin (52ohm)?

The lower Rin, the larger the attenuation in the input stage, becaues of the term:
[Rin / ( Rin + Rsource)]
Let's calculate the input attenuation due to the loading effect of Rin on the source impedance. Rsource = 8 and Rin = 52 if Ic = .05A, so Rin/(Rin + Rsource) = 52/60 = .867. But if we increase Ic to .1A, then Rin = 26 and Rin/(Rin + Rsource) = 26/34 = .765.

The input attenuation did indeed worsen when we increased Ic, but, and this is important, gm doubled when we increased Ic from .05A to .1A. So low Rin is better because it is accompanied by larger gm, which more than compensates for the increased input attenuation.

The CE stage without emitter resistor can also be treated as a transresistance amplifier and if we analyze it as such, we get a result:

\(Av=\frac{-\beta*RL}{Rsource+(\beta+1)re} \)

From which you can see that if (β+1)re >> Rsource, then Av is almost independent of β, but is strongly dependent on re, which in turn is inversely proportional to Ic, so larger Ic means smaller re, which means larger Av.
 

Thread Starter

count_volta

Joined Feb 4, 2009
435
Hey, we have been building a single stage BJT amplifier in my university for the past week. I must say. Wow....

Some of the stuff I asked in this thread was so naive. :D

First of all, the ac small signal gain Av is very small, around like 5-10 V/V. Second of all the gain inverts the signal by 180 degrees.

Umm no wonder I couldn't hear anything, not only did tha transistor hardly amplify anything, the sound was out of phase by 180 degrees, i.e. silence.

And of course the fact that I wasn't really building a voltage amplifier. I was trying to build a current amplifier in this thread, but didn't realize that the low impedance of the speaker means very low power.

As they say, live and learn. Still this thread is full of very valuable information, and now that I understand these things I can go back and re-read everything here with a fresh outlook.

We are actually going to have a course where we learn to design amplifiers ourselves. Real multi stage amplifiers. I was just so anxious to jump ahead in the summer, that I was impatient and wanted to get this amp to work no matter what.
 
Last edited:
First of all, the ac small signal gain Av is very small, around like 5-10 V/V.
If you'll go back and have a look at post #134 in this thread, you'll see that I got a measured voltage gain of 13.

Umm no wonder I couldn't hear anything, not only did tha transistor hardly amplify anything, the sound was out of phase by 180 degrees, i.e. silence.
How do you figure that the sound being 180° out of phase leads to silence? Are you still driving a little speaker with the amplifier output?
 

Wendy

Joined Mar 24, 2008
23,415
I suspect he's thinking two waveforms 180° cancel. They can, but not too likely. The speakers aren't occupying the same space, and any distance between them can cause all sorts of interesting effects, but nothing major.
 

Thread Starter

count_volta

Joined Feb 4, 2009
435
I didn't really mean silence, I was thinking noise canceling headphones lol. There the output is Vo(t) = Vi(-t). Got phase mixed up with inversion.

Actually the output being 180 degrees out of phase with the input wouldn't create any noticeable audible effect would it?
 
Last edited:
Top