Continue to Site

Welcome to our site!

Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

  • Welcome to our site! Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

Impedance for a BJT stage, with collector-base feedback resistor

Status
Not open for further replies.
audioguru,

What is the point and conclusions of your simulations?

Ratch
My simulations show severe distortion when there is no negative feedback.
They also show that the distortion is reduced very much when there is a high resistance ahead of the feedback resistor so that it allows much more AC negative feedback that you disbelieve.
Your emitter resistor reduces the maximum output level, my input resistor adds the same negative feedback as your emitter resistor but does not reduce the maximum output level.

I simulated your idea of separating the AC negative feedback from the DC negative feedback. It makes no difference.
 
audioguru,

My simulations show severe distortion when there is no negative feedback.

Especially if driving the transistor emitter base (e-b)with a voltage source and maybe at too high a level. Response is not linear because the emitter current responds exponentially to voltage on the e-b.
They also show that the distortion is reduced very much when there is a high resistance ahead of the feedback resistor so that it allows much more AC negative feedback that you disbelieve.

Putting lots of resistance in the base circuit with a base resistor or emitter resistor linearizes the Ib-Ic response. It turns the voltage source into a current source. Perhaps I was not clear. I believe R2 does add feedback, but I don't think that is the proper way to do it because it conflicts with the bias requirements. Just a matter of opinion.

Your emitter resistor reduces the maximum output level, my input resistor adds the same negative feedback as your emitter resistor but does not reduce the maximum output level.

The emitter resistor causes feedback, so it is expected that the gain will be reduced. R4 will attenuate the source voltage, but it is not in the feedback path like R2 is. Therefore, R4 does not provide feedback.

I simulated your idea of separating the AC negative feedback from the DC negative feedback. It makes no difference.

In this case, for this particular value of R2, perhaps not. But if the value of R2 were smaller and the feedback higher, then who knows? That idea is not mine. It was suggested my Millman and Halkias in their book Electronic Devices and Circuits 1967.

Ratch
 
My new simulation shows the output level is reduced when I remove the C2 filter capacitor so it proves that there is AC negative feedback through R4 and R2.
 
MrAl,

[R2 being there for DC and AC feedback]
That is the point I am trying to make. I believe R2 should only be used for bias and not AC feedback.
That's not the way you originally presented your arguments as you stated that R2 "was not there for negative feedback", hence some of my replies were targeting that.


If you want negative AC feedback, then use a emitter resistor.
Well, again, we already have negative AC feedback. True it could be done a different way, but that doesnt mean there is no AC negative feedback now. The resistor R2 does both with one resistor. I also wont argue that an emitter resistor might help, but that's a different design.

Yes, R4 will reduce the gain and protect the emitter-base junction, but not enter into feedback considerations.
Not sure why you are stating this. R4 affects the gain, so it works with the feedback. It doesnt have to be in the feedback path itself, who cares. Also, dividing R2 in half does not remove it completely from the AC gain considerations. With R2a left side and R2b right side to collector, sized cap in the middle to ground, R2b does not enter in the AC gain equation but R2a still does.


I don't have any problems with my analysis.
Ok that's great, but check out R2a above.
 
Last edited:
audioguru,

My new simulation shows the output level is reduced when I remove the C2 filter capacitor so it proves that there is AC negative feedback through R4 and R2.

No question about it. R2 does provide feedback. I said as much in my previous post, did I not? Furthermore, its feedback is what stabilizes the bias. But I don't agree that the feedback path is through R4. Even though R4 does affect circuit performance, it is not due to its feedback.

Ratch
 
MrAl,

That's not the way you originally presented your arguments as you stated that R2 "was not there for negative feedback", hence some of my replies were targeting that.

I have since clarified that statement in post #22.

Well, again, we already have negative AC feedback. True it could be done a different way, but that doesnt mean there is no AC negative feedback now. The resistor R2 does both with one resistor. I also wont argue that an emitter resistor might help, but that's a different design.

I have said previously that R2 provides negative feedback for both the bias and AC. I consider it more of a problem than a help. I would use R2 for the bias and an emitter resistor for the AC feedback. To each his own.

Not sure why you are stating this. R4 affects the gain, so it works with the feedback.

To answer audioguru's comment. Yes, but I wanted to emphasize that R4 is not part of the feedback network that many texts designate as "beta".

It doesnt have to be in the feedback path itself, who cares.

Neither does the collector resistor. So who cares about that?

With R2a left side and R2b right side to collector, sized cap in the middle to ground, R2b does not enter in the AC gain equation but R2a still does.

If there is so little AC after the capacitor, there won't be any AC feedback from R2a.

Ok that's great, but check out R2a above.
I have

Ratch
 
MrAl,



[R2a and R2b, and how R2a may or may not affect the gain]

If there is so little AC after the capacitor, there won't be any AC feedback from R2a.

Ratch

Well again if you look back you'll see that i never said there would be feedback from R2a, i just said that "R2a is still in the gain equation". That's why i said that we cant completely remove R2 from the gain considerations, even after dividing it into two section with a cap to filter out AC variations from the collector. It may be better to make modifications overall but the AC gain due to R2a still has to be considered. That's why i pointed your analysis back to R2a.
 
Last edited:
patroclus,

I'm getting trouble on getting (using PSPICE-like software) the input and output impedance of a common-emitter BJT amplifier with collector-base resistor, to try to see how it compares with a more common emitter degeneration feedback (or both). I attached schematic pictures of my set-up. I hope someone can help me see where am I missing something.

OK, let's find out the imput and output impedance of your circuit. Using a general purpose model of your circuit we equate R[g]=R[4],R[f]=R[2],R[l]=R[c], and Z=h[ie]+(beta+1)R[e], where the values in the brackets are subscripts.

Setting the equations for v and e[o] we get (v-e[g])/R[g]+v/Z+(v-e[o])/R[f] = 0 and (e[o]-v)/R[f]-h[fe]*v/Z+e[o]/R[L] . Solving for eo/eg we get the transfer equation -R[L]*(Z+h[fe]*R[f])/(-R[g]*R[f]-R[f]*Z-Z*R[L]-R[L]*R[g]+R[g]*h[fe]*R[L]-R[g]*Z)

Now using the general immitance theorem, we set the denominator of the transfer function to zero and find -R[g]=Z*(R[L]+R[F])/(Z+R[F]+R[L]*(h[fe]+1)) . The right side of the equation is the input impedance of the circuit looking into it from the right of R[4].

To find the output impedance, set the same denominator to zero and find -R[L]=(R[g]*R[f]+R[f]*Z+R[g]*Z)/(Z+R[g]-R[g]*h[fe]), which is the output impedance looking at the circuit from R[L].

Ratch
 
Last edited:
MrAl,

Well again if you look back you'll see that i never said there would be feedback from R2a, i just said that "R2a is still in the gain equation". That's why i said that we cant completely remove R2 from the gain considerations, even after dividing it into two section with a cap to filter out AC variations from the collector. It may be better to make modifications overall but the AC gain due to R2a still has to be considered. That's why i pointed your analysis back to R2a.

I assume you mean AC gain. R2a can be in the gain equation all it wants. If there is no AC present, it might as well not be there.

Ratch
 
MrAl,



I assume you mean AC gain. R2a can be in the gain equation all it wants. If there is no AC present, it might as well not be there.

Ratch

If we talk about AC gain and how R2a affects it, how exactly does the viewpoint of zero AC input help here?
 
MrAl,

If we talk about AC gain and how R2a affects it, how exactly does the viewpoint of zero AC input help here?

The capacitor between R2a and R2b filters out the AC so that no AC current exists past R2, thereby insuring that no AC feedback exists from R2 being present.

Ratch
 
Hello,


Here's an equation for the AC voltage gain of the original circuit [B=Beta]:
Av=-(B*R2*R3)/(B*R3*R4+R3*R4+R2*R4)

Note that the gain is negative because the output is 180 degrees out of phase with the input.


Also, it looks like a good approximation for the output impedance is:
Rout=(R3*R2)/(R2+R3*B+R3)
 
Last edited:
MrAl,

Here's an equation for the AC voltage gain of the original circuit [B=Beta]:
Av=-(B*R2*R3)/(B*R3*R4+R3*R4+R2*R4)

Note that the gain is negative because the output is 180 degrees out of phase with the input.

Yes, the full equation for the gain is e[o]/e[g] = -R[L]*(-Z+h[fe]*R[f])/(R[g]*R[f]+R[f]*Z+Z*R[L]+R[L]*R[g]+R[g]*h[fe]*R[L]+R[g]*Z) , where Z = h[ie]+(beta+1)*R[e] . Assuming R[e]=0, h[ie]=1000, and h[fe]=100, we get a gain -4.18 from the full equaton. Using Z=0, we get your equation for a gain of -4.23, which is quite close.

Also, it looks like a good approximation for the output impedance is:
Rout=(R3*R2)/(R2+R3*B+R3)

Could you explain why R3 is in the output impedance equation.? I thought that R3 was the load resistor R[L]. From the general immittance theorem, I get an output impedance of (R[g]*R[f]+R[f]*Z+R[g]*Z)/(Z+R[g]+R[g]*h[fe]) .

Ratch
 
Hi there Ratch,


Well, i would have found doubt with an equation that did not include R3. Here's why...

As you probably know, the transistor acts almost like an 'impedance decreaser' in that it makes the resistor R2 reflect to the output as a much lower impedance. In other words, the transistor ends up acting like a somewhat low value resistor relative to the other major circuit values.
Now if we wanted to show a single steady state equivalent circuit i guess we could replace the transistor with a single resistor from the output to ground. But that resistance would only reflect the influence from R2. To get the equivalent output impedance (which in this case is output resistance because we are only considering lower frequencies) we'd have to consider the resistance due to R2 and the transistor in parallel with the resistance from the collector to Vcc. There's no way around it, because two series resistors acting as a voltage divider have output resistance equal to the parallel combination of the two resistors.

That's a rough analysis taking the resistance due to R2 and the resistance due to R3 as separate entities, but of course if they are both in the equation to begin with as they should be in a more thorough analysis, then it makes sense that R3 wont cancel out.

For a shorter example, if R3 was 5000 ohms and R3 was not in the equation and we calculate 450 ohms for Rout, then what happens when we reduce R3 to just 450 ohms (more than 10 times lower), we must end up with a lower output resistance somehow.
Given a smaller change, 5000 ohms to 4900 ohms, we must still see at least some decrease in the output resistance.

I'd be happy to follow your analysis through but it's a little hard to follow with all those brackets. Nonetheless, if you'd defined all your input variables (explaining what each one is and why they are there) i'd be happy to give in another shot. For example, if everything was in terms of the actual resistors used like R1, R2, R3, etc., it would be a lot easier to follow.
In many equations like these you'll find abbreviations like Rg, Rf, etc., rather than R[g], R[f], etc., and it makes it easier to follow. It's easy to note that the 'g' is a subscript in 'Rg' for example because it is lower case.

You might also note that the equation i gave for the output impedance is quite accurate given the assumption of a constant base voltage, or to put it another way, a tiny test current in the output.

We could go over these in more detail if you like. I'd like to see how you developed your equation too. We could use a circuit simulator to check both formulas and see how they compare side by side.
 
Last edited:
MrAl,

Could you explain why R3 is in the output impedance equation.? I thought that R3 was the load resistor R[L].
Ratch

Have a look at this circuit:
http://www.ecircuitcenter.com/Circuits/trce/trce.htm

The resistor which provides DC bias to the collector, RC, is separate from the load resistor, RL, which only receives an AC output signal.

Apparently MrAL assumed that R3 is the bias resistor, and the load resistor is not shown in the OP's first post.
 
Have a look at this circuit:
http://www.ecircuitcenter.com/Circuits/trce/trce.htm

The resistor which provides DC bias to the collector, RC, is separate from the load resistor, RL, which only receives an AC output signal.

Apparently MrAL assumed that R3 is the bias resistor, and the load resistor is not shown in the OP's first post.


Hi there Electrician,

Yes that's perfectly correct, i did in fact 'assume' that R3 is the 'bias' resistor which is not part of the load, and with very good reason. If you look back at the OP's original post, you'll see in his second picture attachment a current source connected directly to the 'output' of the amplifier, and along with his brief introduction he mentions that this current source (AC) is to be used to test the output impedance, to find out what it is for this circuit.

Using a current source like that is indicative of how we would test a transistor amplifier that already has a collector resistor (which you called a 'bias' resistor and appears as 'RC' in your informative link), where that resistor is not part of the load but intrinsically part of the amplifier itself. This is usually done so we know what class of load we can connect to the output. Thus, this design was done with the intent of connecting an external load.

This current source testing procedure is very typical and shouldnt come as any surprise to anyone. Would be interesting though if Ratch assumed otherwise.
 
Last edited:
MrAl,

This current source testing procedure is very typical and shouldnt come as any surprise to anyone. Would be interesting though if Ratch assumed otherwise.

What I don't understand is why the B-E terminals are shorted out. The output impedance depends on what is connected to the input terminals, so why short them out?

Ratch
 
What I don't understand is why the B-E terminals are shorted out. The output impedance depends on what is connected to the input terminals, so why short them out?

Ratch

The OP deals with this in post #5. Crutschow points out that the short is in the wrong place. The OP says "V1 is an AC source, so it isn't zero AC impedance to ground.", but if in fact it's a voltage source rather than a current source, it will have zero AC impedance to ground. But nonetheless the OP put an explicit short to ground and got about 508 ohms output impedance.

Since normally an amplifier like this would be driven from a voltage source, the output impedance would be calculated with a short on the input. For this circuit the short should be to the left of R4. The presence of a resistor such as R4 can account for any non-zero impedance of the source.
 
Last edited:
Hi again Electrician and Ratch,


Yes i agree that the base to emitter should not be shorted out and yes i thought we were beyond that issue in this thread.

What i proposed as an approximation device was to call the base emitter a constant voltage of some reasonable value like 0.6 volts or near that, for the equations that is. Of course a voltage source with a small resistance in series would have been better.

Im still not sure Ratch what you are calling the 'load' resistor and what you are calling the collector resistor, or if you are considering them to be the same resistor or what. In your equations you included "RL" to start with, but then later calculated a value for this same variable "RL", so perhaps you can clear that up. Also, a quick numerical run through would be great too and what value you intend to use for "Re" in your equations.
Also, when you declare "Zb" do you mean the resistance looking across the base emitter at the zero AC input voltage operating point (ie DC operating point)?
 
Last edited:
MrAl,

I'd be happy to follow your analysis through but it's a little hard to follow with all those brackets. Nonetheless, if you'd defined all your input variables (explaining what each one is and why they are there) i'd be happy to give in another shot. For example, if everything was in terms of the actual resistors used like R1, R2, R3, etc., it would be a lot easier to follow.
In many equations like these you'll find abbreviations like Rg, Rf, etc., rather than R[g], R[f], etc., and it makes it easier to follow. It's easy to note that the 'g' is a subscript in 'Rg' for example because it is lower case.

You might also note that the equation i gave for the output impedance is quite accurate given the assumption of a constant base voltage, or to put it another way, a tiny test current in the output.

We could go over these in more detail if you like. I'd like to see how you developed your equation too. We could use a circuit simulator to check both formulas and see how they compare side by side.

Yes, using subscripts in text messages makes it hard to follow. The attached file should make it clearer. Ask and I will answer.

Ratch
 
Status
Not open for further replies.

Latest threads

New Articles From Microcontroller Tips

Back
Top