Finding the maximum number of outputs

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mgr1397

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I'm wondering how iti s possible to find the maximum number of bjt current mirrors you can put together with an error that does not exceed 10% and β= 100.
 
interesting question...isn't it where the more current mirrors u put in parallel, the less the error is?

i would say 10% probably needs 1 or 2 max
 
The mathematical answer is 10.11 BJT's So I think that 10 is the answer

Iref/Io = 1/ ( 1 + (1 + n)/β )

n - number of outputs


Iref/Io = 1/ ( 1 + (1 + 10)/100 ) = 1/1.11 = 0.9009
 
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