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From: firstname.lastname@example.org (Winfield Hill)
Subject: Re: Voltage Controlled Resistor?
Date: 12 Dec 2002 06:18:08 -0800
References: <email@example.com> <firstname.lastname@example.org> <3DF79D95.32BB71EC@mfi.net> <email@example.com> <3DF7D257.799D2C80@mfi.net> <firstname.lastname@example.org>
NNTP-Posting-Date: 12 Dec 2002 14:18:08 GMT
Tony Williams wrote:
>> Received thank you. Still sitting here looking
>> at it. 00:20 in the morning is not my best time.
> And I still can't see it. Because I can't get past
> the interpretation that Q2's emitter is forced to
> sit at -Vbe(Q1) permanently, by the opamp. So the
> range of Vcontrol looks limited to the Vbe difference
> between two transistors with their emitters joined.
It's a small piece of an analog multiplier circuit.
| ______ |
| | | |
R1 | Q1 | | Q2
| | |/ \| Vc
| |\V V/|
| __ |_____| matched NPN
'---|- \ | transistors
The current in R1 and hence Q1 is I1 = Vin/R1, so Vbe1 from
Eber-Molls is Vbe1 = VT ln(I1/Is), where VT = kT/q and Is is
the reverse saturation current (which in a complete setup
would ultimately disappear from our equations). Seeing that
Vbe2 = Vc + Vbe1, we can determine I2 as a function of Vc,
I2 = Is e^ (Vc/Vt + ln(I1/Is)). This can be manipulated into
an I2 = I1 * K form, where K is an exponential function of Vc.
So we see Rin = R1 in parallel with a programmable resistance.
Isolating R1's current from the IN terminal with a follower,
and deriving Vc from a current into a logging transistor to
fix the exponential-function issue, can clean up the control
equations. Two more transistors and three more opamps later
we'll obtain a resistance proportional to Vc looking like,
R = (Vc/R2)/(Vref/R3). I'll post a new drawing and explain
the equations tonight or early tomorrow morning; right now I
have to walk the dog and go to work. :>)
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