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I have read here that temperature stability is one of the performance issues in current mirrors. Can anyone what does temperature stability mean in current mirrors? Thanks.

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  • \$\begingroup\$ You do know Vbe is used for thermometers so thermal mismatch matters \$\endgroup\$ – Tony Stewart EE75 Dec 19 '19 at 7:36
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    \$\begingroup\$ A perfectly stable device would have current that is unaffected by temperature. If it is 1.000 mA at 25C, it will be 1.000 mA at 50C also. But real devices will experience some variation in current with changes in temperature. If those changes cause you problems in your circuit then you are having a performance issue due to temperature stability (stability over changes in temperature). \$\endgroup\$ – mkeith Dec 19 '19 at 7:44
  • \$\begingroup\$ Thanks @mkeith for the comment. It make sense \$\endgroup\$ – user15847 Dec 19 '19 at 7:50
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In an NPN current mirror:

enter image description here

\$I_{in}\$ is converted into a voltage by \$Q_1\$ so we get \$V_{BE,Q1}\$.

That \$V_{BE,Q1}\$ is applied to \$Q_2\$ which converts that voltage back into a current \$I_O\$.

This all works by the fact that the voltage <=> current relation:

(\$I_C(V_{BE})\$ and \$V_{BE}(I_C)\$)

of both transistors are absolutely identical.

Unfortunately this relation is quite temperature dependent.

That means that if \$Q_1\$ and \$Q_2\$ do not have the same temperature then their voltage <=> current relations aren't the same and that means that the currents \$I_{in}\$ and \$I_O\$ will not be identical in value.

Possible solutions are:

  • couple the transistors such that they will be at very similar temperatures.
  • Use a component where two identical transistors share the same package
  • use this on an IC where all transistors share the same silicon die
  • use emitter degeneration to make the voltage <=> current relation less dependent on the transistor:

enter image description here

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