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Question: amplifying signal by current mirror

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airboss

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Hi,

i have a question regarding using current mirrors to process signals. this is something razavi mentioned in his textbook. given a reference mos ckt with ideal current source Iref and (W/L)ref. the other mos ckt mirroring current from reference ckt is Ix and (W/L)x. so razavi says that if Iref increases by delta_I, then Ix increases by delta_I*((W/L)x)/((W/L)ref).

my question is, how do i generate the "small signal"? how do i generate the "delta_I"? it doesn't make sense to me that we generate a small current signal while the drain is connected to an ideal current source.

thanks for your patience and help.
 

seanyang

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You can use an ac current source(like, DC=10u, Amplitude=10mV, Frequency=10KHz) as your Iref.
Then you will see the change in another site.
 

    airboss

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deepa

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Well i think wat the author meant to say is it,the circuit exactlu mirrors the change in current in one device to another,,If ideally u want I to flow,due to process variations and body effect the current might not be the intented one,and this change will be reflected in the other device..
 

antonio_eda

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A simple current amplifier consists of a current mirror and two DC current sources. The input is a small-signal current flowing to the input node. This way we have two current sources connected to the drain of the input MOSFET. The output is a small-signal current flowing out of the output node. If for example IinDC=10u, IinAC may have an amplitude less than 10u. Current sources in current mode amplifiers just set operating points of transistors. If the output MOSFET has greater width, so the current gain of the circuit is greater than unity. A.
 

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