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Old 4th June 2004, 09:20 AM   #1
Bricolo is offline Bricolo  France
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Default I've got difficulties to solve this electronic exercise

Hi all,


This is a part of an exercise I try to solve,
I have to draw the "equivalent small signal schematic" for this amp, and calculate the voltage gain from this small signal schematic

I'm totally lost. For a single transistor it's ok, but for a cascode I can't find how to calculate the gain
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Old 4th June 2004, 09:29 AM   #2
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It's easier than you think.

Just forget the cascode and pretend that you only has RC and a plain transistor stage.

Ic*40*Rc = gain... approx.=> 1 mA, 1 kohms = 40 in gain

Put the schematic into LTSpice and their you have the answer at least.
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Old 4th June 2004, 09:36 AM   #3
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And here's the equivalent circuit I find

finding the ic for T1 is straightforward:

ic=B*ib=B*Ve/h11


but now, what about T2? How can I calculate the voltage across h11? (h11 from T2)
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Old 4th June 2004, 09:43 AM   #4
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Well as a pointer :

The purpose of a cascode transistor T2 is to prevent the voltage
swing at T1's collector that would normally occur, thus preventing
the miller effect reducing the bandwith of T1.

T2's base is clamped by the voltage source and the voltage swing
appears on T2's collector. As the voltage source should have very
low impedance the miller effect is less consequential here.
As T2's base is clamped so is its emitter effectively.

I gave up on equivalent circuits, finding them more confusing
than helpful beyond a certain point but they are important.

As stated the gain is near enough identical to a single transistor stage.

sreten.
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Old 4th June 2004, 09:53 AM   #5
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The purpose of a cascode transistor T2 is to prevent the voltage
swing at T1's collector that would normally occur, thus preventing
the miller effect reducing the bandwith of T1.


agreed

T2's base is clamped by the voltage source and the votage swing
appears on T2's collector. As the voltage source should have very
low impedance the miller effect is less consequential here.
As T2's base is clamped so is its emitter effectively.


What voltage source? Remember, my answears have to be based on the equivalent circuit.
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Old 4th June 2004, 10:10 AM   #6
sreten is offline sreten  United Kingdom
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Quote:
Originally posted by Bricolo
[B
What voltage source? Remember, my answears have to be based on the equivalent circuit. [/B]
R1, R2 and Cd, at AC this is a low impedance voltage
source, a divided [R2/(R1+R2)] version of rail voltage E.

sreten.
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Old 4th June 2004, 10:13 AM   #7
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I just realized that I forgot to post the picture


That's what I get for the equivalent circuit. It is correct?
I explained how to fond ic and ib for T1, but now how to calculate it for T2???
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Old 4th June 2004, 10:49 AM   #8
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T2 divides T1's current by the ratio of its Beta.

So ib2 + B2.ib2 = B1.ib1

ib2(B2+1) = B1.ib1.

ib2 = B1.ib1/(B2+1)

B2.ib2= B2.B1.ib1/(B2+1)

Output current is very near a single transistor for high Beta.

Note also this is one of those very poor circuits only found
theory textbooks, as voltage gain is dependent on B1, this
sort of thing should only be used inside a feedback loop.

DC output voltage is also very poorly defined.

sreten.
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Old 4th June 2004, 11:29 AM   #9
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Thank you, that makes sense.


So, for a high beta (180 in this case) ic is nearly the same as with a single transistor

ib1 is also very close to ib2, so V' is equal to Ve am I right?
So T1 isn't really running at constant Vce, but if the voltage gain is important, the Vbe variations can be considered null.


BTW, what would be a better circuit, with no (or lower) beta dependance, and DC stability? (one that wouldn't require feedback)
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Old 4th June 2004, 11:34 AM   #10
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Quote:
BTW, what would be a better circuit, with no (or lower) beta dependance, and DC stability? (one that wouldn't require feedback)
Simply add an emitter resistor to T1. And use a proper voltage divider for its base.

Regards

Charles
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