#### Re: 7603 weirdo inverted Z axis

Albert Otten

Hi David,

It seems that you forgot about R1155?
4.8 V across R1155 gives 0.48 mA. That 0.48 mA reduces the zener current and reduces the current via R1159 by the same amount.
I think you took 130 V across R1120 while it is about 128.4 V, this gives 0.21 mA less via R1120 and 0.21 mA more via R1159.
Together this yields 0.27 mA less via R1159 or 4.7 V lower output voltage. Then our results are the same.

Albert

---In TekScopes@..., <davidwhess@...> wrote :

On 24 Jan 2016 11:46:17 -0800, you wrote:

>Hi David,
>>With no input at Q1107, I calculate a 21 volt output at R1157 which agrees with your measurement and should blank the beam. That points to a problem in the DC restorer.
>
>I calculate it at 16 V, never mind.

I did not find a way to get 16 volts but my work can be checked:

With Q1107 disconnected there are three currents into or out of the
base of Q1156:

1. R1141 is -1.7mA
2. R1121 + R1122 is -1.15mA
3. R1120 is +2.3mA

The later two connect at the top of VR1148 but affect the base of
Q1107 the same way except for an offset of 11 volts.

2.3mA - 1.15mA - 1.7mA = -0.55mA

The amplifier is inverting and note that it does not care whether the
currents are applied to the top or bottom of VR1148 as long as VR1148
is conducting which R1141 guarantees so 0.55mA * 17.4k = 9.6V

The output is offset by VR1148 and the Vbe of Q1156 so:

9.6V + 11V + 0.6V = 21.2V

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