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Q&A

Is a BJT 3-transistor Wilson mirror faster than a simple mirror?

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My question is trying to understand if there is an effect on the switching time, on both turn-on and turn-off, between the following configurations.

A mirror set up to switch a current, and there is switching on the input side of the mirror input (for other reasons), meaning the simple mirror's base voltage must ramp up and down from near zero each time. In this configuration, is there a speed benefit to the 3-transistor Wilson mirror? Talking about discrete components, ~1 MHz switch freq.

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4 comments

We don't know what you think a "half-Wilson" mirror is. If you want to compare two circuits, show the schematics of both of them. Also the current would be a step, not "ramping" up and down. Do you mean the voltage ramps up and down to nearly both rails? Olin Lathrop‭ 3 months ago

@Olin, sorry for not being more clear. Edited to hopefully clarify what I'm asking. I didn't mean that it is an intentionally controlled ramp, rather that the output (i.e. collector current sunk from the node at the top of the cap) takes measurable time to go from 0 to the value that it will have. I sort-of think the 3rd transistor would bring in extra current from the output side, but maybe this is offset by having another transistor to turn on, hence thought I would post the question. Pete W‭ 3 months ago

Use a simulator. Really, use a simulator. If you are not able to use a simulator let me give you some very, very good and strong advice: Get a simulator and learn how to use it. Andy aka‭ 3 months ago

@Andy aka -- yes, you are right. It's something I have been avoiding. Thanks for the push to get me there. Pete W‭ 3 months ago

1 answer

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First Circuit

Let's start by looking at the conventional current mirror circuit:

The base of Q1 (Not sure which one that is? Use component designators next time!) is driven to whatever it takes to pass the current dumped onto the collector. Q2 is assumed to have identical properties, so its base is also driven to support the same collector current as Q1.

The current onto Q1 is switched to be either 0 or 2 mA. Q2 will therefore sink either 0 or 2 mA. Since a bias of 1 mA is always supplied to C1, this will therefore switch between +1 mA and -1 mA onto C1. You say the switching frequency is 1 MHz, so 500 ns per phase. The voltage change on Q1 per phase is:

    (1 mA)(500 ns)/(470 pF) = 1.1 V

Since the voltage will change linearly with time each phase, the waveform on C1 is a 1.1 Vpp sawtooth.

As for speed, 1 MHz feels fast for discrete parts. One advantage is that Q1 never saturates, but you are still stuck with nothing removing the charges in the bases other than by causing collector current when the current is switched off. Also, care should be taken to make sure the collector voltage of Q2 stays high enough to keep Q2 out of saturation.

A current mirror like this is much more useful in integrated circuits where the two transistors will be physically close to each other on the same die, and likely well matched. Two separate discrete parts out of a bin won't be as well matched. One trick to deal with that in discrete designs is to add small emitter resistors. That cuts into the compliance range a little, but makes the circuit more tolerant of part variations.

Second circuit

In this case, the original current mirror is the other way around. Q2 is being driven, and Q1 follows. Q3 is an emitter follower. The input current drives Q3 to whatever it takes to ultimately cause Q1 to sink the input current. That also causes Q3 to sink that same current, so the whole thing does work like a current mirror when looking only at the three external connections. The inaccuracies due to the finite gains of the transistors will be a little less, but that doesn't seem to matter in this case since you asked about speed, not accuracy.

In general, more parts make things slower, unless those parts are specifically for driving the other parts harder and faster. That doesn't seem to be the case here. Draining the charges in the bases of Q1 and Q2 when the current turns off is still the same problem as before. Now there is yet another part with a non-zero response time to add to the mix. I don't see this as being faster. In fact, I suspect it will be slower in real life, especially with discrete parts.

However, the best way to answer the question is to try out both circuits and see.

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1 comment

Thanks. Spent a couple hours playing with it, hard to tell the difference between 2- and 3-transistor circuits, with the parts and equipment here. On solder-proto-board with 2n3904's, a tightly packed 3-transistor setup switched on in <25ns and switched off in <20ns, which was better than I expected. That output timing is about as fast as I think I can measure the edge of the input signal with 200MHz scope anyway. Input was from HP33120a into 51ohm resistor. Pete W‭ 3 months ago

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