toggle quoted messageShow quoted text
thank you Albert
On Sun, Jan 19, 2020 at 12:28 PM, Albert Otten wrote:
Actually the math appears to be surprisingly simple and in theory the
instrument shows EXACT results with LINEAR pots..
Suppose all fixed (plus their trimpots) and variables have equal resistance R
(5 k in our case).
Each pot is in parallel with one fixed resistor.
The wiper of a pot splits the pot resistance in xR (top part) and (1 - x)R
(bottom part) where x varies linearly with angle from 0 (ccw) to 1 (cw).
The effective resistance, say z, of one pot circuit is seen between wiper and
bottom terminal. The path consists of parallel (1 -x)R and xR + R = (1 + x)R.
Hence z = (1 - x)R*(1 + x)R / [(1 - x)R + (1 + x)R] = (1 - x^2)R^2 / (2R) =
R(1 - x^2)/2.
There is the quadratic function we need!
To make it complete:
F branch including the extra R to the meter: R + zF = R + R(1 - xF^2)/2 =
(3/2)R - (R/2)*xF^2,
A+B+C branch: zA + zB + zC = R(1 - xA^2)/2 + R(1 - xB^2)/2 + R(1 - xC^2)/2
= (3/2)R - (R/2)*(xA^2 + xB^2 +xC^2)
Balance when xF^2 = xA^2 + xB^2 +xC^2.
On Sun, Jan 19, 2020 at 04:15 PM, Roger Evans wrote:
Thanks for pointing that out, the calculator is quite different to what I
imagined. I thought the resistance would increase in the clockwisedirection
and each pot would just be connected as a two terminal variable resistor. of
That does rely on having a good approximation to a square law for the taper
the pot. As you pointed out the resistance increases ccw and I can't see
immediately why the maths works for the combination of fixed resistor in
parallel with the three terminal connected pot. It might be a rather clever
piece of algebra or an approximation that is good enough for the intended