Cascode current mirror of BJT

For MOS transistor, i have known that the cascode current mirror will put a higher output resistance and therefore it will dicrease the channel length modulation effect.


The formula is : Rd = 1/ λId

I have understood that , when we put the current mirror in cascode it will increase the output resistance (Rd) and therefore it will dicrease the channel length modulation effect(λ). And thererefore we can get a a constant current.


What about BJT,? I know by cascode current mirror , it will increase the output resistance. What is the main factor that increases the output resistance?

Is it the Ic or something else. Thank you.

I suggest u check on this book Analysis and design of analog IC by Gray Hurst Lewis Meyer

You can check the formulas but here is some intuitive explanation. Let's call the transistor on the bottom of the cascode transistor 1 and that on the top - transistor 2. If you have a simple current mirror (one transistor only) - both in cmos and bjt - the voltage at the output terminal of the mirror (drain or collector) can vary depending on the load. This as you know causes channel length modulation or base-width modulation (the Early effect). The modulation causes variation in the current - so, you have a variation in the voltage which causes variation in the current, which is in fact a caracteristic of a resistor - Ro of the transistor. If you can make the drain or the emitter voltage more constant , you'll have less modulation and hence the current generated by the transistor will stay constant. Now, in the cascode configuration transistor 2 acts like a source or emitter follower for the drain/collector of transistor 1 (transistor 1 is the load for this follower). If we connect the gate/base of transistor 2 to some relatively constant voltage we will also force its sourse/emitter to be constant as well. And transistor 1 sees just a small voltage variations at its drain/collector, so less modulation of the current i.e. the constant current generated by transistor 1 goes directly into transistor 2 and to the output. If the output voltage varies, those variations go as a variation in the Vds/Vce of transistor 2, but the current remains constant. Looking into the drain/collector of transistor 2 this is equivelent to a very big output impedance.