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From: Tony Williams
Subject: Re: Voltage Controlled Resistor?
Date: Thu, 12 Dec 2002 15:20:36 +0000 (GMT)
References: <email@example.com> <firstname.lastname@example.org> <3DF79D95.32BB71EC@mfi.net> <email@example.com> <3DF7D257.799D2C80@mfi.net> <firstname.lastname@example.org> <email@example.com>
NNTP-Posting-Date: Thu, 12 Dec 2002 16:04:07 +0000 (UTC)
User-Agent: Pluto/1.14i (RISC-OS/3.60)
In article <firstname.lastname@example.org>,
Winfield Hill wrote:
> It's a small piece of an analog multiplier circuit.
> IN o--+-----------------,
> | ______ |
> | | | |
> R1 | Q1 | | Q2
> | | |/ \| Vc
> +----+----| |---o
> | |\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
I'd already got as far as ln(I2/I1) = Vc*(q/kT) and
therefore the actual incoming control voltage would
have to be logged, such that Vc = kT/q * ln(Vin).
This would give I2 = I1 * Vin.
Rout = Vout/Iout = Vout/(I1+I2) = Vout/I1(1 + Vin).
And the easiest way to get rid of the non-linearity
due to the (1 + Vin) term is to buffer R1, so that
Iout does not have to include the I1 current to R1.
> 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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