Hi all,

I am studying the current mirror circuit.
I am trying to understand its operation but stuck in the first place. Why do we need a diode connected BJT?
What is the purpose of using a diode connected BJT? Won't a normal Diode be sufficient in place of BJT?



I understand it is a very simple circuit but I am not able to understand how does this circuit establish current in Q1 and Q2 to be same?

Can someone explain me the basic concepts to help me understand this? What is the necessity to establish a constant current using this method? Why can't we use a single transistor and control the Vbe and Ib of it to produce a constant current? Why do we need two transistors to complicate this? Please help me understand this.

While considering Q2, proper biasing is giving by Q1. Vbe of Q1 and Q2 are same. But should it mean that collector current of both the transistors are same? I understand that both are identical transistors in their geometry. But how does the current be same?

I also understand that Q1 must be in saturation and for Q2 to establish a constant current, Q2 should be in active region,right? So that, if the voltage Vout is increased,Q2 will vary its Vce to provide constant Iout. Is this correct?

Sorry for the long query. Please correct my concepts if I am wrong. Thank you all.

 

Why do we need a diode connected BJT?
What is the purpose of using a diode connected BJT? Won't a normal Diode be sufficient in place of BJT?

For the mirror to work properly the two junctions must be matched in configuration and current.
A diode has all the current going through its junction while the BJT only has the small base current going through the Vbe junction.
That would give a huge mismatch between the diode current and the transistor current.

Can someone explain me the basic concepts to help me understand this?

With BJT transistors the value of Vbe determines the collector current.
So if you establish a collector current through Q1 with the collector connected to the base, then a certain Vbe will be generated which is characterized by the gain value of that transistor..
If connected to the base of a second transistor then, since the transistors are matched, the collector current of Q2 will be equal to the collector current of Q1, since they both have the same Vbe voltage.
In practice the current of Q2 will vary some with the value of Vout due to the finite collector resistance of Q2.

To reduce this change of current with voltage, you can go to more complex circuits, such as the Wilson current-mirror.
It uses negative feedback from a third transistor to greatly reduce the change in current with voltage.
This LTspice simulation below compares the the standard two-transistor current-mirror with the Wilson three-transistor mirror.
The standard mirror collector current (red trace) varies from about 875μA to 962μA (about 19%) for a collector voltage going from 2V to 12V.
Over that same range the Wilson mirror current (green trace) is essentially constant (about a 0.1% change).

Why can't we use a single transistor and control the Vbe and Ib of it to produce a constant current? Why do we need two transistors to complicate this?

Because controlling the Vbe and Ib to the desired value requires more complexity than one additional transistor.

 

See

 

This might be of interest -

https://wiki.analog.com/university/courses/electronics/text/chapter-11

https://en.wikipedia.org/wiki/Widlar_current_source


Regards, Dana.

 

To answer your question the current mirror
The biggest problem with discrete current mirrors are matching the transistors. Maybe they have current mirrors in IC form. The Wilson current mirror greatly reduces the affects of Beta on the circuit so it's will be more stable. in real life.