Calling all circuit analysts

Discussion in 'General Electronics Chat' started by The Electrician, May 9, 2009.

1. The Electrician Thread Starter AAC Fanatic!

Oct 9, 2007
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It seems circuit analysis problems on the Allaboutcircuits board are usually homework problems. I thought I'd bring attention to something that's not a homework problem (but could be involved in homework).

I'm sure T N K, steveb and hgmjr will want to look at this. I've noticed that if you look for information about models for a simple common emitter amplifier on the web you will find that various models use two different values for the base resistance, r∏ (sorry that looks a little funny; the ∏ is too big). Some use β*re and some use (β+1)*re. Which is correct?

Here:

are some Powerpoint slides with models for various configurations from a book by Boylestad. I'll attach some images from there for the simple common emitter amplifier with no emitter resistor. I'll ignore ro for my discussion.

If you derive an expression for the voltage gain using β*re for the base resistance, you get Av = -Rc/re. According to this, the gain is independent of β, and in fact remains constant even if β goes to zero; this seems wrong.

If the base resistance is taken to be (β+1)*re, then Av = (β/(β+1))*(-Rc/re). Now the gain goes to zero as β goes to zero, as one would expect.

The difference in the gain expressions is small, and for reasonable β they will give essentially the same result.

But, why not use the more exact expression? It's not that much more complicated. And, even though for this simple amplifier, there's not much difference, in some other topologies it's clear that (β+1)*re is the right value to use. I'll show those in subsequent posts.

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• Slide 6.jpg
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Last edited: May 9, 2009
2. The Electrician Thread Starter AAC Fanatic!

Oct 9, 2007
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Here's a topology where you can really see the difference between using β*re or (β+1)*re for the base resistance.

If you calculate the voltage gain using β*re, you get:

Code ( (Unknown Language)):
1.         (Rf-re)*Rc
2. Av = - -----------
3.         (Rf+Rc)*re
an expression which also doesn't vary with β, so it should be true for β=0.

If we use (β+1)*re for the base resistance and derive an expression for voltage gain, we get:

Code ( (Unknown Language)):
1.         (β*Rf - (β + 1)*re)*Rc
2. Av = - -----------------------
3.          (Rc + Rf)*(β + 1)*re
4.
It was this gain expression for this topology that led me to notice the difference in the r∏ expressions used by various authors.

Have a look at slide 32. Imagine that β goes to zero. Then the βIb current source at the output becomes non-functional and vanishes. The output voltage is simply determined by the voltage divider comprised of Rf and Rc, and is given by Av = Rc/(Rf+Rc). Interestingly, the 180° usual phase inversion vanishes too. Since we know that if β>0, there is a phase inversion, there must be a value of β for which the gain is exactly zero.

The first gain expression given above, where r∏ =β*re, doesn't equal Rc/(Rf+Rc) when β goes to zero, because it doesn't even involve β.

The second gain expression above, derived with r∏=(β + 1)*re, does approach Rc/(Rf+Rc) as β goes to zero.

I take this to be strong evidence that r∏=(β + 1)*re is the correct value to use.

I'll show another case later.

Notice that in slide 33, Boylestad has apparently made some approximations in deriving his expression for Av.

Has anyone else noticed all this?

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3. studiot AAC Fanatic!

Nov 9, 2007
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The model used by Boylestad and many others is called the hybrid-pie model or the simple hybrid-pie model.
As you observe it contains the simple βib current source.

The full hybrid pie model uses the proper β+1 relationship. Opportunity is also taken to account for other effects in the full model and β becomes a complex function of frequency.
The simple current gontrolled current source is replaced by a voltage controlled current source.

I have attached a couple of sketches, sorry about the quality. I can't find links to point you at. Look up the Early effect.

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4. AdrianN Active Member

Apr 27, 2009
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is re*(β+1), and here is why:

re is the resistance "seen" between emitter and base.

re = vbe/ie and then re = α*vbe/ic (that's an alpha).

re = α*vbe/(β*ib) , and because rπ = vbe/ib then

re = rπ*(α/β).

Now we know that α = β/(β+1), and replacing α in re,

rπ = re*(β+1).

The relation rπ = re*β is customary in the industry because people consider re = VT/IC instead of re = VT/IE, in effect making α≈1. Nothing wrong with it. I saw this in many books. It is still a good approximation. This model is also called the "re model". Slide 32 should have had (β+1)re in the input.

5. mauro.laurenti Active Member

May 8, 2009
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Hello

In practical problems the reason why they use just β instead of (β+1) is due to the fact that β has a big tolerance anyway. This means that the β from the datasheet could be 100 but but the max value could be 130 or more, so there are no big differences in considering β instead of (β+1).
An other reason for considering just β is when you have a feedback amplifier and you multiply everything for the gain loop. The gain loop also help to make the system more independent from β variations (or the amplifiers chain).

Anyway, a good design should consider the max and min values you can get from the components, and make sure that you are always within spec.

Ciao,

Mauro

6. steveb Senior Member

Jul 3, 2008
2,431
469
I've noticed this before. It seems to crop up in many situations where approximations are made. The assumptions and limits of any approximation need to be in the back of your mind.

In this case with beta and beta+1, there is a bit of art that should be employed. I agree with you. Why approximate something as beta when beta+1 is not very complicated. If you end up with a term with beta/(beta+1), so be it.

For example, I could say that ic=beta*ib and also say that ie=beta*ib, which is accurate when beta is large. However, then I end up with a strange result if I employ a current node equation.

ib=ie-ic=0

Ok, we know ib is small compared to ie and ic, but it is not zero. -Just another example of taking approximations too far, and getting nonsensical predictions.

Last edited: May 10, 2009
7. mauro.laurenti Active Member

May 8, 2009
68
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each approximation can be "exact" if what you get is what you need.

After all the model you are shoving is another approximation to represent the component.

Ciao,

Mauro

8. steveb Senior Member

Jul 3, 2008
2,431
469
True enough. However, if you are not careful, you may think you got what you needed, when if fact you made a huge mistake. This can be embarrassing in the best case, costly in the usual case and deadly in the worst case.

9. mauro.laurenti Active Member

May 8, 2009
68
0
...you are perfectly right.

We don't know each other, but I'm one that does not trust spice simulation...
...I have been playing with model spice and I have seen too many times that the results where strange...
...than the manufacturer said...this behaviour is not part of the model!

...I will never say that something is right if the formulas and the lab are not saying the same things and they are within the expected spec.

In this case the model gives what you need and it's the right one

Ciao,

Mauro

10. mishu.daniel Member

May 13, 2009
19
0
Hi all,
Can someone explain me how is calculated the input impedance Zi in slide 32 please?
Also, I don't understand why ro is shown only in slides 10/11? Why not in the other cases too?
And the last question, why in slide 23 Rb is not taken into account?

For the other cases i think is clear (Yes I am beginner).

Thanks.

Last edited: May 13, 2009
11. The Electrician Thread Starter AAC Fanatic!

Oct 9, 2007
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Assume that a current source of 1 amp is applied to the input, then calculate the voltage that appears there. To do this, you can set up a pair of nodal equations:

Vi/(β*re) + (Vi-Vo)/Rf = 1

(Vo-Vi)/(Rf) + Vo/Rc = -β*Ib

substituting Vi/(β*re) = Ib, the second equation becomes:

(Vo-Vi)/(Rf) + Vo/Rc = -Vi/re

Solving these two simultaneous equations, you will get:

Code ( (Unknown Language)):
1.        Vi            β*re*(Rc + Rf)
2. Zi = ------  =  ----------------------
3.      1 amp      (β + 1)*Rc + Rf + β*re
The expression the author gets in slide 32, re/(1/β+Rc/Rf) can be rearranged to:

Code ( (Unknown Language)):
1. β*re*Rf
2. ----------
3. β*Rc+Rf
Apparently, the author of slide 32 has made the assumption that Rf >> Rc, that (β + 1)*Rc + Rf >> β*re, and that (β + 1)*Rc ~ β*Rc. With these assumptions, the larger expression for Zi derived above from the two simultaneous equations can be reduced to the expression given in slide 32.

I think the reason may be that in the other slides, if you include ro, the various expressions become more complicated. He probably could have ignored ro in slides 10/11 too, because ro is usually much greater than Rc in a typical amplifier circuit.

Rb is not taken into account in the expression for Av in any of the slides where it is present in the circuit, because when you calculate Av, it is assumed that you are driving the input with an ideal voltage source which has zero output impedance. Rb therefore is in parallel with an impedance of zero ohms, and has no effect.

You will notice that Rb is taken into account in the expression for Ai.

Last edited: May 13, 2009
12. The Electrician Thread Starter AAC Fanatic!

Oct 9, 2007
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Notice that the expression for Zi derived from the two simultaneous equations in the previous post (and also the expression given in slide 32) will reduce to zero ohms if β goes to zero, a plainly bogus result.

This is a result of the Boylestad's taking the value of r∏ to be β*re.

If r∏ is taken to be (β+1)*re, then the expression for Zi becomes:

Code ( (Unknown Language)):
1.      (β+1)*re*(Rc+Rf)
2. Zi = ----------------
3.      (β+1)*(re+Rc)+Rf
which has a limit of:

Code ( (Unknown Language)):
1.       (Rc+Rf)*re
2. Zi = ------------
3.        Rc+Rf+re
when β goes to zero, a more reasonable result.

13. mishu.daniel Member

May 13, 2009
19
0
Vi/(β*re) + (Vi-Vo)/Rf = 1

(Vo-Vi)/(Rf) + Vo/Rc = -β*Ib

If we look at slide 32, the current arrows produce some sign changes in your formulas above:
Ii + I' = Ib <=> 1 + (Vo -Vi)/Rf = Vi/(β*re) (this is the same)
and
I' + Ic = Io <=> (Vo - Vi)/Rf + β*Ib = Vo/Rc (this is different)

Am I wrong?
Thanks.

14. The Electrician Thread Starter AAC Fanatic!

Oct 9, 2007
2,720
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A common convention in nodal analysis is to assume currents leaving a node to be positive, and currents entering a node to be negative.

The flow of current in a resistor is from the more positive end to the more negative end. Notice that Ib is downward, away from the Vi node. Boylestad's choice of direction for Io is contrary to this.

If the voltage Vo is taken to be positive (that's how he shows it), then the current flow in Rc is Io = -Vo/Rc.

In other words, you should have:

I' + Ic = Io <=> (Vo - Vi)/Rf + β*Ib = -Vo/Rc

Otherwise, I think the calculated input impedance may be negative for some values of Rc and Rf.

15. mishu.daniel Member

May 13, 2009
19
0
Yes, in this case you're right. But if we look at slide 30 Io is correctly pointed from the most positive end of resistor to the most negative one.
After translation, on slide 31 due to the ideal Vcc (I think, with internal r=0) the positive and negative ends of resistor are changing but Vo doesn't change the polarity. This is a little blurry for my logic ...
I'm afraid that I will end up with wrong results when I will design a schematic. Anyway, I have many things to learn so I will continue to dig.

Thanks for your suport 16. The Electrician Thread Starter AAC Fanatic!

Oct 9, 2007
2,720
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We are considering the small-signal (AC) currents here, so the output node is considered positive for the signal currents. In both slides the current is shown as flowing from ground into the node, so by convention it is taken as a negative current.

17. mishu.daniel Member

May 13, 2009
19
0
I didn't pay too much attention at Ai calculation, and following your explanation I was trying to get the result Io/Ii (in slide 6) using Rb also.
I get formulas using voltage values so I arrive to Ai formula using Av. But how about Ai = Io/Ii = (β*Rb*ro)/(ro + Rc)*(Rb + βre) ?

Thanks.

18. The Electrician Thread Starter AAC Fanatic!

Oct 9, 2007
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That's the expression I get, and it's also the expression in slide 6. What is your question? Don't you think it is correct?

19. mishu.daniel Member

May 13, 2009
19
0
Sorry I was not clear, it is about Av and Ai. (slide 6/7)
I understood why Ai = -Av (Zi/Rc)
But why Ai = BRbro / (ro + Rc)*(Rb + Bre) ?
Is the current flowing through ro from the most positive end to ground?
I don't manage to exclude Vo/Vi from formulas ...
I have to find a very good tutorial about small signal model, until now, all documents I have don't contain some useful explanations.

Last edited: May 14, 2009
20. The Electrician Thread Starter AAC Fanatic!

Oct 9, 2007
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If you understand why Ai = -Av (Zi/Rc), then take the expression for Zi from slide 5, which is Zi = β*re*Rb/(β*re+Rb), and the expression for Av from slide 6, which is Av = -(Ro*Rc)/(re*(Ro+rc)) and substitute into -Av (Zi/Rc). You will then get Ai = B*Rb*ro / (ro + Rc)*(Rb + B*re)

That is the assumed direction. When using the nodal method, any impedances connected from a node to ground are assumed to have a (signal) current flow away from the node toward ground.

Realize that "most positive end" means the end that is connected to the output node, the collector. We're not talking about most positive DC voltage here. Our analysis will work for PNP or NPN transistors; it's independent of the DC bias voltages.

Ordinarily, because of the 180° phase shift in a common emitter amplifier, the current gain would have a negative sign. But, because Boylestad assumed the current flow in Rc to be upward, into the node, and because he defined the current gain as the ratio of the current in Rc to the input current, the sign of the current gain is reversed.

The current gain may be calculated by using the two nodal equations I set up in post #11 to derive the input impedance, making some changes to account for the fact that we're now considering slide 5 (Rb and ro are now present and Rf is absent). I injected a 1 amp current into the input and solved for Vi. Since the solution of those two simultaneous equations also gives an expression for Vo, we can use that to get Io (Io = -Vo/Rc), and that will give us the current gain. The result is:

Code ( (Unknown Language)):
1. Io   -Vo/Rc             β*ro*Rb
2. -- = -------  =  ---------------------
3. Ii    1 amp      (Rb + β*re)*(Rc + Ro)