[SI-LIST] Re: question about caculation the emitter follower output quiescent power
- From: "Dr. Howard Johnson" <howie03@xxxxxxxxxx>
- To: "Si-List@xxxxxxxxxxxxx" <si-list@xxxxxxxxxxxxx>
- Date: Fri, 5 Nov 2004 14:15:16 -0800
Dear sduphz,
In the high state, the voltage (collector-to-emitter) across
the driving transitor is (Vcc-VH), while the current flowing
out through the load to VT is (VH-VT)/R.
In the low state, the voltage (collector-to-emitter) across
the driving transitor is (Vcc-VL), while the current flowing
out through the load to VT is (VL-VT)/R.
The formula expresses the average of these two values
(that's the "equal probability of ones or zeros"
assumption).
Best regards,
Dr. Howard Johnson, Signal Consulting Inc.,
tel +1 509-997-0505, howie03@xxxxxxxxxx
http:\\sigcon.com -- High-Speed Digital Design seminars,
books, and articles
-----Original Message-----
From: si-list-bounce@xxxxxxxxxxxxx
[mailto:si-list-bounce@xxxxxxxxxxxxx]On Behalf Of ³¯»Ô Åí
Sent: Tuesday, October 26, 2004 3:01 AM
To: si-list@xxxxxxxxxxxxx
Subject: [SI-LIST] question about caculation the emitter
follower output
quiescent power
hello everyone,
i am learning Howard's <the high speed digital design>.
there is a
question,in the chapter 2 and page 51, why the emitter
follower output
quiescent power is
P={(Vcc-Vh)(Vh-Vt)+(Vcc-Vl)(Vl-Vt)}/(2R).
thanks.
sdupzh
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