The Cyber-Spy.Com Usenet Archive Feeds Directly
From The Open And Publicly Available Newsgroup
This Group And Thousands Of Others Are Available
On Most IS NNTP News Servers On Port 119.
Cyber-Spy.Com Is NOT Responsible For Any Topic,
Opinions Or Content Posted To This Or Any Other
Newsgroup. This Web Archive Of The Newsgroup And
Posts Are For Informational Purposes Only.
From: Winfield Hill
Subject: Re: Voltage Controlled Resistor?
Date: 12 Dec 2002 19:13:25 -0800
Organization: Rowland Institute
References: <firstname.lastname@example.org> <email@example.com> <3DF79D95.32BB71EC@mfi.net> <firstname.lastname@example.org> <3DF7D257.799D2C80@mfi.net> <email@example.com> <firstname.lastname@example.org> <email@example.com>
X-Newsreader: Direct Read News 4.20
Tony Williams wrote...
> Winfield Hill wrote:
>> 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 ...
OK, i'll indulge myself. Here's the full drawing and formula
for R ~ control voltage. R_in = R1 (Vset/Vref)(R3/R2), if
I avoided mixing up a numerator and denominator someplace!
. Rin = 1.0k Vset/10
. IN o----------------, ___ R3 ___ Vref = 10V
. G =_+1 | | 10k
. __ R1 ___/ |___| __ | ___ R2___ Vset
. | 2.0k \_| | __/ -|----+ | 20k 0 to +10V
. |_________ | | \_+|--G | |_________
. | | | | | | |
. | Q1 |/ Q2 \| | |/ Q3 \| Q4 |
. +-------| |---+--------| |-------+
. | |\V V/| |\V V/| |
. | __ |_____| matched NPN |_____| __ |
. '--|- \______| transistors |______/- |--'
. G --|+_/ \+_|-- G
If Q1-Q2 and Q3-Q4 are each matched to 25uV, and with a beta
approaching 1000, the formula should be accurate to about 0.1%
Alternately a few trim-point adjustments can be added.
Go Back To The Cyber-Spy.Com
Usenet Web Archive Index Of
The sci.electronics.design Newsgroup