CURRENT SUPPLY - one more thing.
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Posts: 46,084
I should have mentioned that the higher value of R means better accuracy.
Think about this. Say you have a 1mA mirror with R=1K so Ve=1V. Then
mentally construct another mirror with R=10K and Ve=10V. We set Vb to be 1.6
and 10.6V respectively. Now, something happens that makes the BE drop go to
.62V instead of .6V (temperature or even some variation in the base voltage
could change Ve). So now Ve is 1.58 and 10.58. The output current changes to
1.58mA (1.58% change). The second mirror changes to 1.058mA (1.06% change).
Really just percentage change plus dividing by a big number instead of a
small number. Of course, you can't use too big a resistor, or you'll lose
compliance. Suppose you have a 12V supply. You can't use a 100K resistor
since that would require Ve=100V!
Regards,
Al Williams
AWC
*Stamp project of the month changing today (I hope):
http://www.al-williams.com/awce/som.htm
Think about this. Say you have a 1mA mirror with R=1K so Ve=1V. Then
mentally construct another mirror with R=10K and Ve=10V. We set Vb to be 1.6
and 10.6V respectively. Now, something happens that makes the BE drop go to
.62V instead of .6V (temperature or even some variation in the base voltage
could change Ve). So now Ve is 1.58 and 10.58. The output current changes to
1.58mA (1.58% change). The second mirror changes to 1.058mA (1.06% change).
Really just percentage change plus dividing by a big number instead of a
small number. Of course, you can't use too big a resistor, or you'll lose
compliance. Suppose you have a 12V supply. You can't use a 100K resistor
since that would require Ve=100V!
Regards,
Al Williams
AWC
*Stamp project of the month changing today (I hope):
http://www.al-williams.com/awce/som.htm