[mngeow]

As seen in the question. I have to find the value of IRef. Here's what I have so far :

For BJT Q2 , we want Ic=1.5mA. Since the gain is 100 , Ib would be 15μA.

For BJT Q1, it can be seen that the Collector is shorted to the base. Therefore Vc=Vb , and Ve (Emitter Voltage ) =0.
Assuming Vbe=0.7 , Vb = Vb-Ve = 0.7 -0 = 0.7. It can also be seen that IRef = IC1.

Here's where I get stuck. Since the collector is directly shorted to the base. It could be said the Ic=IB1 + IB2. IB1 can then be expressed in terms of Ic and the value of Ic can then be solved. However this value of Ic obtained is not correct. Is there something I'm doing wrong? I think for this question I'm having problems visualising the current flow for Q1 and Q2.

Thank you.

 

[Arouse1973]

Hello
Think about what voltage you have across the resistor Rref. You know the current you need in Rref to ensure the other transistor matches this. Can you work this out now?
Adam

 

[Ratch]

As seen in the question. I have to find the value of IRef. Here's what I have so far :

For BJT Q2 , we want Ic=1.5mA. Since the gain is 100 , Ib would be 15μA.

Who cares about that? Ref is in the collector circuit of Q1.

For BJT Q1, it can be seen that the Collector is shorted to the base. Therefore Vc=Vb , and Ve (Emitter Voltage ) =0.
Assuming Vbe=0.7 , Vb = Vb-Ve = 0.7 -0 = 0.7. It can also be seen that IRef = IC1.

No, IRef = Ib2+Ib2 , but who cares?

Here's where I get stuck. Since the collector is directly shorted to the base. It could be said the Ic=IB1 + IB2. IB1 can then be expressed in terms of Ic and the value of Ic can then be solved. However this value of Ic obtained is not correct. Is there something I'm doing wrong? I think for this question I'm having problems visualising the current flow for Q1 and Q2.

Why get hung up on current? You know what the current is supposed to be in Ref. That's a gimme. And you know the voltage at each end of the Ref. Why can you calculate the value of Ref from that information?

Ratch

 

[mngeow]

I know that the voltage across Rref is 5-0.7 = 4.3V. But would the current across Ref for Q1 have to match that of Q2?

 

[Ratch]

I know that the voltage across Rref is 5-0.7 = 4.3V. But would the current across Ref for Q1 have to match that of Q2?

The circuit you submitted is called a "current mirror". That means the current in Q1 is the same as the current in Q2. You know that is true because both transistors are identical and their Vbe's are the same. Since a BJT is a tranconductance device (voltage controls current) and the Vbe's are both identical, so too are the currents. If you set Q1 to 1.5 ma like the problem states, then Q2 will also pass 1.5 ma. Where are you getting 15 ua from? Doesn't the answer match (5 - 0.7)/1.5 ma ?

Ratch

 

[mngeow]

I got IB1 = 15μA due to the fact that Ic2=1.5ma and beta is 100. But I understand now, this helped to clear up a lot. Thanks!

 

[Ratch]

I got IB1 = 15μA due to the fact that Ic2=1.5ma and beta is 100. But I understand now, this helped to clear up a lot. Thanks!

Y'er welcome.

Ratch

 

[mngeow]

Would just like to confirm if what I understand is correct. Since the current Ic is dependant on Vbe and gain. If two transistors have the same gain and Vbe, then their current Ic must be identical?

 

[Ratch]

Would just like to confirm if what I understand is correct. Since the current Ic is dependant on Vbe and gain. If two transistors have the same gain and Vbe, then their current Ic will be identical?

Two identical BJTs with the same Vbe will pass identical currents when operated in the active region at the same temperature. That is because BJTs are transconductance devices. Identical BJTs will have the same transconductance gain. See image below.

Ratch

 

[LvW]

Would just like to confirm if what I understand is correct. Since the current Ic is dependant on Vbe and gain. If two transistors have the same gain and Vbe, then their current Ic must be identical?

Just to avoid misunderstandings: When writing "gain" - do you mean "current gain" (beta)? In this case, you are in error. The value of Ic depends on Vbe only. It is only the base current that differs in case of different beta values.

 

[Laplace]

When the problem statement mentions "identical" transistors, that is a code word for 'matched pair'. That means two identical transistors formed in close proximity on the same monolithic substrate and fabricated into one package so that both transistors will operate at the same temperature. So both transistors will always have the same Vbe vs Ibe characteristic during operation. In this circuit configuration both transistor bases are directly connected so each transistor has the same Vbe which causes the bias current to split equally going into each base. Since the current gain of each transistor is the same, the collector current in each transistor must also be the same.

 

[LvW]

So both transistors will always have the same Vbe vs Ibe characteristic during operation.
.

...and Vbe vs. Ic characteristic.
However, to be exact we should also mention that BOTH base currents are to be provided by the left part of the circuit. This causes a kind of unsymmetry within the current mirror. More than that, the voltages Vce differs considerably - thus causing another source of unsymmetry (different EARLY effects).