Why does a current source on the collector of a BJT bring kaboons of gain ?

[Ehsan]

I read this article to learn how to make a current source with a BJT and two diodes:

http://www.ecircuitcenter.com/Circuits_Audio_Amp/BJT Current_Source/BJT_Current_Source.htm

I understand the concept fully, but then the author says at the end of the article that a current source on the collector of a transistor gives a lot of gain. Why is it so ? Can someone show me the mathematics behind it ?

Thank you.

 

[LvW]

I understand the concept fully, but then the author says at the end of the article that a current source on the collector of a transistor gives a lot of gain. Why is it so ? Can someone show me the mathematics behind it ?
.

The most simple answer is: Such an (active) current source presents a very large internal ac resistance (and at the same time a moderate and finite dc resistance). Thus, it is very advantageous to replace the ohmic collector resistance by such a current source. This allows large gain values.
Increasing the ohmc Rc has limits because each increase requires a corresponding increase of the supply voltage (100 kohm and 1mA give 100 Vdc).

Another explanation (based on currents): When the collector current "wants" to change (because of the input signal voltage) - however, it "cannot" chnage because of the constant current source in the collector path, the resulting difference appears as an input current to the next stage (as a measure for the signal voltage at the input).

 

[Ehsan]

Another explanation (based on currents): When the collector current "wants" to change (because of the input signal voltage) - however, it "cannot" chnage because of the constant current source in the collector path, the resulting difference appears as an input current to the next stage (as a measure for the signal voltage at the input).

Brilliant explanation. Everything makes sense now.

How much is that input current to next stage ? I mean how much will it be amplified ?

Current gain = h_FE = (I_c / I_b) and we know I_c can not change, then when I_b changes (because of change in input signal voltage) how much will be that current handed to the next stage ? Will h_FE vary during this process (assuming constant temperature)?

 

[Arouse1973]

Hello Ehsan.
What do you mean by the next stage? I don't quite understand.

 

[Ehsan]

What do you mean by the next stage? I don't quite understand.

Q1, D1 and D2 forms a current source and it is placed on the collector of Q2.

Q2's base is the input and Q2's collector is the output that goes for the next stage, you can connect the base of another transistor to this point or whatever your design demands.

 

[Arouse1973]

Don't forget RB. Without RB the circuit won't work.
Adam

 

[Arouse1973]

Brilliant explanation. Everything makes sense now.

How much is that input current to next stage ? I mean how much will it be amplified ?

Current gain = h_FE = (I_c / I_b) and we know I_c can not change, then when I_b changes (because of change in input signal voltage) how much will be that current handed to the next stage ? Will h_FE vary during this process (assuming constant temperature)?

It will only be Ie and Ib that changes and not Ic. Otherwise the circuit wouldn't be constant current. With Vce fixed and Ic constant then it can only be Ib and Ie that changes. So for the current to remain constant if Ib and Ie changes then the gain must change to compensate.
Adam

 

[LvW]

Yes - Ehsan is referring to typical amplifiers that consists of several stages - for example opamps.
When only moderate gains are required and no further stages are necessary and - in most cases - ohmic resistors in the collector path are sufficient.
However, for large gain values (opamps) with several stages the first stage is eqipped with a current mirror in the collector path of the first differential stage.
In this case, the principle as described above is applied.

 

[LvW]

It will only be Ie and Ib that changes and not Ic.

Ic will change as usual as a function of Vbe. However, because of the constant current source in the collector path the change is forwarded to the next stage. That is the core of the working principle. Remember - Ic is split into two other currents: One current into (or out of) the current source (current mirror) and the other current forms the base current for the next amplifying stage.

 

[Arouse1973]

Ah yeah sorry LVW I didn't see his last drawing. I thought it was just a copy of the first one.
Adam