Help Adding a Current Mirror to a Diff Pair Res Calc

Zero Cool

Member
Paid Member
2004-09-20 6:10 am
MN
I am experimenting with adding a Current Mirror to a Diff pair circuit. I want to understand how to calculate the load resistors. So for example I have an existing circuit as shown. If I add a CM, how do I calc the new resistor values for R22 & R23??
 

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The current mirror is the load.
You tap off the collector below the CM to feed the VAS.

The CM can use zero ohms for the two upper resistors. but it does not work well. I think IC opamps use near zero ohms because resistors add to the IC cost.
The emitter resistors can be any value from 1r0 to 10k, but most adopt something in the range 100r to 1k
Read S.Groner.
He shows that the CM works very well with R>470r
 

mlloyd1

Member
Paid Member
2001-02-25 7:10 pm
Northern Iliinois
ohms law is your friend.
:)
you know the current in each transistor of the diff pair - when the pair is balanced, it would be half the tail current you set with the current sink.
then you choose a corresponding resistor that has 50mV across it when said current flows through it.

and, by the way, you take the output (current) from the collector of the mirror transistor, not the emitter resistor as you have shown ...

mlloyd1
 

Mark Johnson

Member
Paid Member
2011-05-27 3:27 pm
Silicon Valley
{50mV across the emitter degeneration resistors} is one guy's opinion of how much insensitivity-to-PNP-mismatch, is typically needed. Another guy's opinion, mine, is: you need much more. I aim for at least 200mV and if the circuit will tolerate more (some circuits will!), I drop even more voltage across the emitter resistors.

Those able to run SPICE simulations are invited to simulate a PNP current mirror with 2.0mA flowing in the input leg, variable emitter resistors, and a 5 millivolt voltage source in series with one PNP transistor base but not the other. (This voltage source simulates transistor mismatch.) What's the current in the output leg, as you simultaneously vary both emitter resistors from 2 ohms (2mV drop) to 500 ohms (1000mV drop)? What does it mean?
 
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forr

Member
2004-12-01 6:46 pm
Next door
Voltage across D7 compensates for the Vbe of T6
The voltage drop across R16, 390 Ohm is about 0.6 V, the same as the voltage across D9.
This defines a current of 0.6/0.390 = 1.54 mA in the long tail pair.
If the currents in transistors T8-T7 of the input pair are well balanced, the voltage across R22 or R23 should be around 1.54/2 * 2.2 = 1.7 V. It is a quite unusual value here.
No help is possible without knowing the right side of the circuit.
 

Mark Johnson

Member
Paid Member
2011-05-27 3:27 pm
Silicon Valley
@Mark, isn't the purpose of the degeneration (Re) to swamp the effect of differing (re) in the mirror devices?

Not in discrete transistors with huge (compared to IC transistors) emitter areas and tiny (compared to IC transistors) series resistances. In discrete designs the purpose of emitter degeneration is to swamp the effect of different VBEs in the mirror devices.

Have a look at the LTSPICE model parameter values for extrinsic emitter resistance of discrete transistors, they are tiny. For the 2N3906: 0.1 ohm. For the ZTX951: 0.02 ohms.
 

Zero Cool

Member
Paid Member
2004-09-20 6:10 am
MN
Voltage across D7 compensates for the Vbe of T6
The voltage drop across R16, 390 Ohm is about 0.6 V, the same as the voltage across D9.
This defines a current of 0.6/0.390 = 1.54 mA in the long tail pair.
If the currents in transistors T8-T7 of the input pair are well balanced, the voltage across R22 or R23 should be around 1.54/2 * 2.2 = 1.7 V. It is a quite unusual value here.
No help is possible without knowing the right side of the circuit.

Here is the rest of the circuit minus the output devices.
 

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