Requirement of emitter resistors in current mirror circuit

[brainbaby]

In a current mirror circuit ..presence of an emitter resistor (R1 and R2) increases the output resistance which minimises harms of early effect...how?
why its advantageous to have a high output resistance in current sources/current mirror circuits?

 

[meBigGuy]

Think about how the early effect manifests itself. As the output voltage changes, the output current changes. That indicates a finite output impedance. Since adding the emitter resistors increases the effective output impedance, it reduces the change in I caused by a change in V.

delta I = deltaE/deltaR. If R is infinite, no amount of E change will change I.

 

[LvW]

why its advantageous to have a high output resistance in current sources/current mirror circuits?

Because - by definition - an ideal current source is identical to a voltage source with an infinite source resistance.

 

[brainbaby]

Think about how the early effect manifests itself. As the output voltage changes, the output current changes. That indicates a finite output impedance. Since adding the emitter resistors increases the effective output impedance, it reduces the change in I caused by a change in V.

delta I = deltaE/deltaR. If R is infinite, no amount of E change will change I.

nice explanation ...thnks

 

[Baluncore]

R1 and R2 help linearise the mirror for higher currents. They reduce the effect of V_be negative temp coefficient and reduce the thermal imbalance.

By placing another transistor above the mirror you can equalise the collector voltages and so make a mirror with a better thermal power balance.

 

[brainbaby]

R1 and R2 help linearise the mirror for higher currents. They reduce the effect of V_be negative temp coefficient and reduce the thermal imbalance.

By placing another transistor above the mirror you can equalise the collector voltages and so make a mirror with a better thermal power balance.

Are you talking about this circuit(refer attachment)..
The transistor Q3 is involved just to cancel the effects of base current ...i mean base current is quite small in magnitude (that should be neglected ) then what is the requirement to minimise base current...and involve Q3?

 

[meBigGuy]

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

 

[Baluncore]

See attached. The accurate current balance here is done by Q1 and Q2. The rest of the circuit is there to help them.
For high currents R1 and R2 swamp the V_be of Q1 and Q2 so as to prevent thermal offsets or runaway.
Think cascode, Q2 and Q3 make the collector voltages of Q1 and Q2 very similar in magnitude, so heating is the same in both Q1 and Q2.
D1 keeps Q1 and Q2 on the edge of saturation, a small collector voltage makes less heat.
Q5 produces the base current for all other transistors. If Q5 is high beta then most base drive comes from Vcc, not I_in.
Only Q1 and Q2 need a common thermal environment. Temperature of other components is unimportant.

 

[The Electrician]

See attached. The accurate current balance here is done by Q1 and Q2. The rest of the circuit is there to help them.
For high currents R1 and R2 swamp the V_be of Q1 and Q2 so as to prevent thermal offsets or runaway.
Think cascode, Q3 and Q4 make the collector voltages of Q1 and Q2 very similar in magnitude, so heating is the same in both Q1 and Q2.
D1 keeps Q1 and Q2 on the edge of saturation, a small collector voltage makes less heat.
Q5 produces the base current for all other transistors. If Q5 is high beta then most base drive comes from Vcc, not I_in.
Only Q1 and Q2 need a common thermal environment. Temperature of other components is unimportant.

Shouldn't the transistor designators above in red be as I've shown?

There's more on the topic at: http://en.wikipedia.org/wiki/Wilson_current_mirror

 

[Baluncore]

The Electrician is quite correct, my typo.

Note that the diode in my diagram can be replaced with a transistor to give base current cancellation. See attached.
This is probably a buffered, improved, Wilson Current Mirror.

The problem with the Wilson mirror is that the V_ce voltages are different for the critical pair of transistors. Power dissipation, and therefore heating differs by a factor of two between the two matched transistors.