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? **broken link removed**
These are some questions I've been banging my head over lately.
Hopefully you guys can point me out in the right direction!
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
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.
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..
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!
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?
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.
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?
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.