I just realized I screwed up the math on this series-parallel resistor network - and this is one of the most basic there is.
Trying to break it down into a written explanation got my mind inverted.
Using the schematic values:
1.5k + 1.5k = 3.0k for the series combination of R211 and R212.
To solve a parallel resistor network, we add the reciprocals of the individual resistor values:
(1/3) + (1/1.2) = .333 + .833 = 1.167
And then find the reciprocal of the above result:
1/1.167 = 0.857 kOhms
Using the actual color code values of 1.2 kOhms:
1.2k + 1.2k = 2.4k
In parallel with 1.2k:
(1/2.4) + (1/1.2) = .417 + .833 =1.25
And the reciprocal of the above:
1/1.25 = .8 kOhms.
Still, the measurement of 1925 Ohms is over twice what is expected. With this much change, the shunt peaking effect of the capacitors in the network will be excessive. Short of unsoldering one end of each resistor to measure individually, it is possible to tack solder some extra resistance across this network. By temporarily reducing the total resistance to 800 Ohms, we can see what effect it has on the overshoot spike. This will be much easier than removing the pc board to unsolder resistor leads.
By rearranging the parallel resistor equation:
(1/0.80) - (1/1.925) = .703
1/.703 = 1.369 kOhms
Using two 2.7 kOhm resistors (red-violet-red-gold) in parallel will get pretty close at 1.35 kOhms. Let's see what this
does.
RF+ Tech