偏执电流求解

薛定谔的太极拳

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图中电流镜偏置电路，不考虑体效应，问流过电阻R的电流是多少？大神稍微说明解释一下。

wfcawy

when the circuit achieve stable state, the current passing the resistor R is 20uA.
at the initial state, Pmos mirror sources40uA, while the Nmos mirror can only sink 20uA, so there is a net sourcing 20uA current charging to the parasitic capacitors of P/Nmos drain region. finally, the voltage on the upper node of the resistor raises, and push the PMOS to the linear region a bit, and make the sourcing ability weaker till 20uA, and then the balance is achieved.

kwankwaner

回复 2# wfcawy


   “at the initial state, Pmos mirror sources40uA, while the Nmos mirror can only sink 20uA”    有什么根据吗

薛定谔的太极拳

回复 2# wfcawy

Thanks。。。。。。。。

sodede

俺仿了一下，最右PMOS确实没出现40A的情况。

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红线为最右NMOS漏电流，黄线为最右PMOS源电流。

344567112

回复 1# 薛定谔的太极拳
没电源大小，没R大小，怎么求，最高20μA，也可能低于。

wfcawy

it is an assumption, and it shows the conflict of the design.
A transient simulation can catch everything with the consideration of all the parasitic.

But the final conclusion will not change.

semico_ljj

只可能20uA多一点 而且 PMOS处于线性区

jiang_shuguo

很老的一个笔试题目了。

薛定谔的太极拳

回复 6# 344567112


   这个是某资料中的一个笔试题，具体就那么多参数。。。