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startup ciruit in BetaMultipler.... (Read 2224 times)

somisetty Community Member Offline Posts: 64	startup ciruit in BetaMultipler.... Sep 14^th, 2010, 12:20am Hi All... In the attachment,the startup circuit is used to avoid a condition "where the gate voltages of transistor M1,M2 become zero and M3,M4 become VDD and the circuit doesnt work". Can anyone explain,why the gate voltages of M1,M2 go to zero and M3,M4 go to VDD? Thanks a lot....
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Garrett.Neaves

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Analog Design
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United States

Re: startup ciruit in BetaMultipler....
Reply #1 - Sep 14^th, 2010, 4:38am

The drain of M1 is connected to the gate of M1. �The impedance looking into the drain/gate of this connection is reduced compared to the impedance looking into the drain of a common source connection. �M3 is a common source connection. �The result is that there is a (much) lower impedance through M1 to ground than through M3 to Vdd thereby pulling the gate of M1,M2 to ground.

Use similar reasoning to understand why the gate of M4,M3 is pulled to Vdd.

Garrett Neaves

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RobG

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Bozeman, MT

Re: startup ciruit in BetaMultipler....
Reply #2 - Sep 14^th, 2010, 12:24pm

somisetty wrote on Sep 14^th, 2010, 12:20am:

Hi All...

In the attachment,the startup circuit is used to avoid a condition "where the gate voltages of transistor M1,M2 become zero and M3,M4 become VDD and the circuit doesnt work".
Can anyone explain,why the gate voltages of M1,M2 go to zero and M3,M4 go to VDD?

Thanks a lot....

That is just another way of saying that all transistors are off - which is a stable state that you don't want, hence the startup transistors.

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