Hi,! I'm getting very confused when analyzing this amplifier.

The doubt is about the gain. There is no gain here?, because i(vin) equals I(r1),so gain=1...

How can I analyze the currents & voltages?, I'm lost and sick. Is there any method I miss?, Thevenin, Norton, superposition theorem, KVL,??

The transistors are in cut off region why are about 10 volts between emitter and collector of each one?

Is the ground ok? where I put the input ac?.
Any bibliography appreciated
thanks.

 

The output node is the emitter of both BJT`s.
Thus, both BJT`s must be seen as common collector stages.
Do you expect any voltage gain from these configurations?

 

Voltage gain?, no I mean current gain. gain=I(Input)/I(Output). it is 20ma/ma=1..

 

OK Nelson, do you realize that the pair of diodes at the input draw a lot of current?
To measure the current gain of the BJT only you must use the base current as an input.

 

In a power amplifier the (seems to me quite obvious) important parameter to analyze is Power Gain. Of course there is merit to looking at voltage gain and current gain but...
You can analyze anything you desire and it is helpful to analyze and understand a circuit from various perspectives.

So "keep on beating".

 

If you analyze the circuit you should get a current gain of 5.

Your simulation results don't make sense. If your input source is delivering 18mA to the bias network, where is it going?

Are you positive that the i(v1) waveform is using the left-hand axis?

 

Try read this
http://forum.allaboutcircuits.com/showthread.php?p=615183#post615183
Maybe it will help you understand what is going on the this type of a circuit.

 

I note the DC supply has no reference to ground.

 

I note the DC supply has no reference to ground.

That, my friend, is an excellent catch and completely explains the simulation results. The only two nodes connected to ground are the load resistor and the input source. Hence they are in series and HAVE to have the same current in them.

@OP. Replace the 20V source with two 10V sources in series and then ground the junction point between them.

 

Try read this
http://forum.allaboutcircuits.com/showthread.php?p=615183#post615183
Maybe it will help you understand what is going on the this type of a circuit.

Yep I saw this circuit last week, thank you.
But the problem was when working with only one battery, I got confused.
Cant understand why if the Q1 & Q2 transistor are both in cutoff region I have VCC/2 in each one. So they are working?, and what is The VBE vltage? 0.6 or more sorry, got sick again!!.

 

That, my friend, is an excellent catch and completely explains the simulation results. The only two nodes connected to ground are the load resistor and the input source. Hence they are in series and HAVE to have the same current in them.

@OP. Replace the 20V source with two 10V sources in series and then ground the junction point between them.

Thank you very much, like this one??. understood but not the first.Sorry. I made a virtual ground with a battery, hope this is the idea to bias with only one battery.


View attachment classAB2question.asc

 

That, my friend, is an excellent catch and completely explains the simulation results. The only two nodes connected to ground are the load resistor and the input source. Hence they are in series and HAVE to have the same current in them.

@OP. Replace the 20V source with two 10V sources in series and then ground the junction point between them.

Just like this?, I did it the first time, but when replacing by only one battery I got very confused. If the circuit is a serial, and the Q1 and Q2 transistors are in cut off region there no current?. if not --> they are working and what is the VBE in q1&q2,.... suppose Q1 have 4v, Q2 had to be 16v complementing KVL & KCL laws. ?
Thanks again

 

Just like this?, I did it the first time, but when replacing by only one battery I got very confused. If the circuit is a serial, and the Q1 and Q2 transistors are in cut off region there no current?. if not --> they are working and what is the VBE in q1&q2,.... suppose Q1 have 4v, Q2 had to be 16v complementing KVL & KCL laws. ?
Thanks again

Yes, just like that. Notice that the current gain is 5, just like would be expected.

Why do you say that Q1 and Q2 are in cutoff. If both are on, then the Vbe for each is going to be about 0.6V or 0.7V.

The key thing is to work out the voltages and currents starting with the constraints and don't impose constraints that don't exist. It's real easy to think that the transistor collectors are at ±10V, but they aren't. The difference between them is always 20V, but the actual voltage (relative to ground) is moving all over the place.

This is not a trivial circuit to understand and it has some subtleties in it. So let's walk through it a bit at a time and see how we might solve it.

In the attached picture, assume that all of the diodes and transistors are conducting and have 0.7V across them. What are the voltages at node {A,B,C,D}? What is the current flowing in the input supply (from supply to the diode junction) and in the load resistor (from node D to GND through the resistor)?

 

Thank you very much. !!, I'm working on it, first Analyzing in dc . Hope I can understand..REALLY very clever you answer & teaching Technic.