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I've just analysed a Wilson current mirror (part (ii)) and was wondering if someone could check that I've done it correctly. The input current should be mirrored to the load current right? Isn't that the whole point of a current MIRROR? It's just I saw another example where the load current was 10 times more than the input current...

Would I instead need to calculate I(C) of Q2?

Also any ideas for part (i)? Not sure what a load plot line is and couldnt seem to find one anywhere.

Thanks![enter image description here]1

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  • \$\begingroup\$ The currents do not have to be equal. A current mirror that gives a 10x higher output current is still a current mirror ! Actually the name current mirror is a bit strange, current copier would be a better description. But they're called current mirrors so there. Most analog ICs contain loads of these. \$\endgroup\$ – Bimpelrekkie Aug 26 '15 at 14:50
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You got it partially right. The wilson mirror will reflect the reference by physical propriety of the transistor junction. The following relation show that: \begin{gather} \frac{I_{0}}{I_{ref}} = \frac{1}{1+\frac{2}{\beta ^{2}}} \end{gather}

In your particular case, you assume both transistor are identical physically therefore and because the beta = 100 you can assume: \begin{gather} \frac{I_{0}}{I_{ref}} = 1 \\ I_{0} = I_{ref} \end{gather}

To have that relation, you need that the beta of each transistor equal and both transistors to be matched.

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  • \$\begingroup\$ Thanks. So my calculations of I(ref =) and I(load) are correct? I believe the transistors will always be identical in my upcoming exam questions so I don't think I'd need the first equation you provided, although it could come in useful in future. \$\endgroup\$ – Physicsman Aug 26 '15 at 18:01
  • \$\begingroup\$ @Physicsman In most case, it will be true. I think for what I understand of your level, you should understand what are the advantage of the wilson mirror current compare to a regular mirror current. \$\endgroup\$ – MathieuL Aug 27 '15 at 15:44

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