Op Amps: Active Loads

[@davidluther3955]

EXCELLENT LECTURE!WE DID NOT HAVE ANYTHING ON EARLY VOLTAGE IN OUR AS SOCIATE PROGRAM DEGREE,SO I BE LIEVE I HAVE LEARNED SOMTHING.

[@rudygomez6996]

super cool im learning so much from you 🙌

[@jimgibson9811]

Thanks

[@andymouse]

There are actual SOT23 packaged current mirrors available, never used one but that's got to be the best matching you can get ! so I will use them if the need arises....cheers.

[@laurensdegoey5108]

In theory the current into the load equals v_in/r'_e so Av should be R_load/r'_e = 10MEG/28 = 385000. The measured/simulated Av is just 1140 (Wilson current mirror). Why the big difference?

[@ElectronicswithProfessorFiore]

The internal impedance of the mirror is not infinity.

[@laurensdegoey5108]

@@ElectronicswithProfessorFiore Thank you very much.

[@SeanPorio]

Really great explanation! This would also increase the AC output impedance as well though, correct? I see those load 10M resistors snuck in there haha

[@SeanPorio]

That’s usually where current amplification/buffering would be added in an op-amp, right?

[@ElectronicswithProfessorFiore]

@@SeanPorio Right, this is the first stage of an op amp with gain stages following.

[@simonyoungglostog]

Is it possible to calculate the resistor equivalent for an active load NPN? I Can't see an obvious equation.

[@ElectronicswithProfessorFiore]

Yes, but it's probably easier to just measure it (by its effect) ;-)

[@simonyoungglostog]

@@ElectronicswithProfessorFiore Probably? That sounds ominous. I enjoy the maths, is it in any of your downloads?

[@ElectronicswithProfessorFiore]

@@simonyoungglostog By "probably" I mean you'd have to dig around for some device data. Measuring is very easy (including "in sim"). Start with an open load to determine the max Av. Then add a load resistor and compare how far the gain has dropped (there is a voltage divider between the output impedance set by the active load and the external load resistor, just use the voltage divider rule in reverse). For example, if the gain drops in half, then Zo = Rl. If it drops to 1/3rd, then Zo = 2 Rl, and so forth.

[@simonyoungglostog]

@@ElectronicswithProfessorFiore Thank you. I'll have a go (I'm wading through my power-amp design and slowly working out what all the sub circuits do.