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How Does This Circuit 'Zero?'

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jamesinnewcastle

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Hi

My first post - for a very specific reason. I am an electronics engieer of 40 years standing but the vast majority has been in management and writing reports and documents, using no more than ohms law if I'm lucky. So I have got a lab together at home and I am starting electronics as a hobby. I am repairing an electronic voltmeter and I am simply trying to analyse the circuit for fun and to exercise the grey cells! Sadly I am not surrounded by competent engineers at work so I have to ask on the web.


I need to see if anyone can see and explain how this electronic AC Voltmeter circuit can produce a 'zero' on the meter when there is no input voltage. I have stripped out all the capacitors so you are just looking for the DC quiescent conditions. I have simulated this circuit using SPICE and that simulation backs up what I think and that is that the meter will have a substantial deflection even with no input.

To give you a 'head start' in the analysis, the diode in the collector load of Q6 should be roughly at half the supply voltage so that niether of the diodes D1 and D2 are conducting - that would be a nice zero as you would have the diode forward voltages as a buffer against any offset from the other components in the circuit.

Applying an AC voltage to the input should cause D7 to swing above and below 'zero' volts, the currents flowing in D1 and D2 meaning that the meter only sees current in one direction and hence sees the input as a full wave rectified signal.

The problem is that the bias current flowing in R14, while very small, generates a relatively large voltage across R14 which is a mad 470K. This voltage is amplified and appears at the output making the meter move away from zero!

The question is - have I missed something? This is the correct circuit and values as I have traced it all out on the meter itself! How can this circuit produce a 'zero' on the meter - can anyone work it out?

Cheers

James
 

FvM

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Did you use a correct model for the OA90 (a germanium diode)?.

Anyway, it's meaningfull to measure some voltages in the original circuit and compare with the simulation to understand the cause of differences.

As another point, the meter idle current most not necessarily be zero, the mechanical zero can be adjusted.
 

jamesinnewcastle

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Hi FvM

I haven't used correct models for any of the semiconductor components as my SPICE is a little primative (read cheap!). But poor SPICE modelling should not preclude an accurate paper analysis. I don't understand why the 470k resistor isn't say 4k7 for instance.

Sadly the meters compact, multi-layer construction makes in-circuit measurement impossible. I can only measure resistor values and check transistor 'diodes'.

The Meter is a Multi-Meter so that offsetting the needle physically would just mess up the zero on the DC, ohms and amps ranges and I know that that does not happen.

James
 

dick_freebird

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R19 and R20 should be first-order cancelling the base
current induced voltage in R14 (470K vs 480K sum).
The op amp (discrete, but still) only wants to make
Q6 collector equal GND (any offsets aside). Then you
have set up a "sort of" fixed current in R18, D2, R16,
D1 and the voltmeter reads Iset*R16.

What I do not see, is any indication of which element
is the adjustable for calibration. If it's an AC meter
then the capacitors you have deleted, may hold the
key (enforcing a zero balance or closing the loop
with a rectified-AC value perhaps).
 

FvM

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You must a least consider the different forward voltage of OA90 and 1N4148, otherwise you won't get meaningful results.
 

jamesinnewcastle

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Hi Dick_

You are right - daft of me not to spot the balanced input resistances which, as you say, should cancel out the transistor bias currents - I'll have to study it a little closer to see where the my SPICE model is going wrong.

When the 'op-amp' includes the diodes in its feedback loop then their forward voltage drop should be negated by the feedback mechanism. You get the 'perfect rectifier' op-amp configuration. (Perhaps my transistors haven't enough gain and the 'op-amp' isn't really acting correctly?)

Just to salve your curiosity here are the capacitors, as you can see the feed back is actually further on in the circuit than for the DC conditions (necessary I guess to remove the forward diode drop). Also for info, the circuit description I have (very sparse) states that the DC feedback path is negated for AC voltages by C10 and C11.

Hi FvM

Yes the SPICE model was for a germanium device, the part numbers on the circuit were direct from the SPICE directory.

As a general note this circuit is from a meter designed by a very famous Multimeter company so it ought to be good.


Thanks to all

James
 

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