Input impedance darlington tor

SneaKSz

Member
Hello all,

I was reading an audio amplifier book (by bob cordell) and he said the β8=100 and β10=50, the collector of Q4 would see 100*50*8=40kΩ.

I understand that this would be the case if R11 wouldn't be included and R13 is low and left behind, but R11 infects the input impedance.

Why can the input impedance be calculated by using the formula above?
Doesnt Q9 and Q11 also being active, not infect the collector impedance of Q4?

And finally shouldn't Q5 also be taken into account when determining the collector impedance of Q4? These are some questions I've been banging my head over lately.

Hopefully you guys can point me out in the right direction!

kind regards !

Last edited:

Jony130

Active Member
Well I use a small-signal mode to find the input impedance for Q8 Q9 and R11, R13, RL

[LATEX]Rin = \frac{(\beta 1 +1)(\beta 2+1)((R13 RL+R11 (R13+RL))}{R11+ R13(\beta 2+1)} = 28.688K\Omega[/LATEX]

Where β1 = Q8 and β2 = Q10.

And we don't need to include Q9 and Q11 and Q5 in our model. Error introduced by Q9,Q11,Q5 is practically negligible.

Last edited:

SneaKSz

Member
Hello jony, thanks for the answer!
Are you able to scan calculations you made to retrieve Rin?
Could you tell me why Q5 is neglected?

Here are my calculations, also with the PI model.

View attachment 67290

Kind regards!

Last edited:

Jony130

Active Member
Now I need to go to sleep but tomorrow I try to show you how I got this result.
But you can try to solve it yourself. Simply find Input current for this circuit Rin = V1/Ib1

As for Q5. We can ignore it because the output resistance of a Q5 (VBE multiplier) is very small ro=20Ω
http://paginas.fe.up.pt/~fff/eBook/MDA/Mult_Vbe.html

β

Last edited:

Jony130

Active Member
Hi, today I use Mathematica software and Jacob Shekel circuit analysis method to find Zin.
This method is described by The Electrician in this thread

So the full equation for Zin of a Q8 and Q10 look like this

[LATEX]Zin = \frac{(1+\text{$\beta$1}) ((\text{R13}+\text{re2}) (\text{re1}+\text{RL}) (1+\text{$\beta$2})+\text{R11} (\text{R13}+\text{re1}+\text{re2}+\text{RL}+(\text{R13}+\text{re2}+\text{RL}) \text{$\beta$2}))}{\text{R11}+(\text{R13}+\text{re2}) (1+\text{$\beta$2})} = 23.660K\Omega [/LATEX]

I assume β1 = 100; β2 = 50 and Ic8 = 20mA ; Ic10 = 0.1A and re = 26mV/Ic

And if we assume re1 and re2 = 0 we get the same equation that I post in previous post.

So Zin seen by Q4 collector is equal

Zin = 23.6K ||ZoQ7 ≈ 23.6KΩ

[LATEX]ZoQ7 = \frac{\text{R10} (\text{re1} \text{ro}+\text{R9} (\text{re1}+\text{ro})) (1+\text{$\beta$1})+\text{R10} \text{R9} \text{ro} \text{$\beta$1} \text{$\beta$2}+(\text{re1} \text{re2} \text{ro} (1+\text{$\beta$1})+\text{R9} (\text{re2} \text{ro}+\text{re1} (\text{re2}+\text{ro}) (1+\text{$\beta$1}))) (1+\text{$\beta$2})}{\text{R10} (\text{R9}+\text{re1}) (1+\text{$\beta$1})+\text{re1} \text{re2} (1+\text{$\beta$1}) (1+\text{$\beta$2})+\text{R9} (\text{re1}+\text{re2}+\text{re1} \text{$\beta$1}+\text{re2} \text{$\beta$2})} [/LATEX]

But if we assume re1 = re2 = 0Ω then

[LATEX]ZoQ7 = \text{ro}+\frac{\text{ro} \text{$\beta$1} \text{$\beta$2}}{1+\text{$\beta$1}}[/LATEX]

So if ro = 10K , β1 = β2 = 100 then ZoQ7 = 1MΩ (700KΩ is we use a full equation)

Last edited:
• SneaKSz

SneaKSz

Member
Hello,
great method to find the impedances and gains! With the attachments by the electrician, I understand how you created the matrix and got the Zin.

I manually determined the Zin for the darlington ( without re') and finally got the same equation.

Are you able to send me the pdf for the Zout of the current source?

I have 1 thing I dont understand about the active matrix. If I would have a bjt with collector and emittor a resistance, I have a 3*3 matrix ( standard). But when the collector is grounded ( darlington example), the row for the collector is deleted. Should there be 2 collums of zero's?

And if the emittor is connected to gnd, than only (1,1) and (1,3) of the standard bjt matrix stay..

Try to search in the posts of 'The Electrician' but didnt really got it.

thanks for the great help!

Last edited:

Jony130

Active Member
Are you able to send me the pdf for the Zout of the current source?
Sure I can.

I have 1 thing I dont understand about the active matrix. If I would have a bjt with collector and emittor a resistance, I have a 3*3 matrix ( standard). But when the collector is grounded ( darlington example), the row for the collector is deleted. Should there be 2 collums of zero's?

And if the emittor is connected to gnd, than only (1,1) and (1,3) of the standard bjt matrix stay..
I'm not quite understand your question. But if we want to included ro then simply connect resistor between collector and emitter.

Last edited:

SneaKSz

Member
Thanks for the help Jony!

well I'll try to explain my question a little bit more precize.

Here, you tried to analyse the circuit with this special method, you have a different array for Q2 when compared to the standard ( 3*3) BJT array. I thinks this is because the emittor is connected to the ground. (http://forum.allaboutcircuits.com/attachment.php?attachmentid=11669&d=1250886068).

And also what should be done to the standard BJT matrix (3*3) if the collector was grounded? This was done in the darlington configuration, but I dont really get why the active array is formed like that.

e.g. in the current source example, the array for Q1 is formed in a different way than usual. View attachment 67362
Hopefully you understand my question(s) a little bit better!

Kind regards

Last edited:

Jony130

Active Member
If any node is connect to ground we need delete row and column that shorted node. SneaKSz

Member
Ok, thanks I got this now.

And what about the positioning in the matrix? This is from the current source, why is it array 1 that's correct and not array 2?

kind regards!

Last edited:

Jony130

Active Member
I hope that this imagine explains everything Last edited:

SneaKSz

Member
Thanks, now I've got how you guys got those arrays.

Thank you very much Jony, for learning me this method! Could this also be applied to fets?

kind regards

SneaKSz

Member
Tried the Vbe multiplier's output resistance

View attachment 67373View attachment 67374View attachment 67372

Does this sound allright?

I reworked the Zout than so it would be clear that the output resistance is really low. Instead of the tor the Calculated Zout can be connected to node 1 and 3, so it could be incorporated for the Zin of the darlington right?

Last edited:

Jony130

Active Member
Could this also be applied to fets?
I don't know, but probably you can but I I've never tried. I only use it for BJT and op amps.
Maybe The Electrician can show as some light on this.

Tried the Vbe multiplier's output resistance

View attachment 67373View attachment 67374View attachment 67372

Does this sound allright?

I reworked the Zout than so it would be clear that the output resistance is really low. Instead of the tor the Calculated Zout can be connected to node 1 and 3, so it could be incorporated for the Zin of the darlington right?
As for Vbe multiplier output resistance. Your matrix looks god but you forgot to short one of a nodes ( GND is missing). See the pdf file.

Also you shoudl kept in mind that we can ignore Q5 because his low impedance ZoutQ5 ≈ 20Ω.
As for Q9 and Q11. If we assume that the bias current is equal 0.1A, then for the low output signals. When output stage work in class A. The impedance seen by Q4 collector is almost the same as for the class B ( Q9 and Q11 are in Cutt-off or Q8 and Q10). The circuit work in class B if IL > 2*bias current = 0.2A. And this give Vload = 0.2 * 8Ω = 1.6V.

SneaKSz

Member
As for Vbe multiplier output resistance. Your matrix looks god but you forgot to short one of a nodes ( GND is missing). See the pdf file.
I didnt know if I had to short the collector yes or no. That because the source is a current source. So I should also ground it when the supply is a current source?

Jony130

Active Member
But if you are measure resistance you measure it between two terminals, you don't measure open circuit resistance. So you need a reference node (GND).

Last edited:

SneaKSz

Member
Yea youre right, my fault. But what should normally be grounded than? The collector or emittor of Q5. The collector of Q4 is creating an AC signal, so then its more like finding the input impedance of Q5 and that would mean that the collector of Q5 is grounded and the input signal is attached to the emittor of Q4. Now we have two terminals to measure the input impedance.

Does this sound right?

Jony130

Active Member
It doesn't matter which of the two terminals (emitter or collector) you chose as a reference point. The impedance will be the same.

The Electrician

Active Member
Thanks, now I've got how you guys got those arrays.

Thank you very much Jony, for learning me this method! Could this also be applied to fets?

kind regards
Yes, it can be used with FETs; see the attachment.

Attachments

• 16 KB Views: 108

anhnha

Member
The Electrician : for the FET, why don't we replace 1/infinity by zero directly instead of 1/Rg? 