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question about circuit operation

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peter g

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hi, i've just completed construction of an esr meter. can anyone explain why the meter goes to full scale when you short the test probes? wouln't the signal from the oscillator go to ground since one of the test probes is grounded. thanks. pg
 
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Oh dear, this is likely to end in tears.

However, trying to be a bit positive, lets just think about this a minute.

theres nothing wrong with the esr meter i built
OK, fair enough. But you say it appears to exhibit odd behaviour, what would you expect it to do when the probes are shorted together?

the oscillator section of all esr meters is the same
Really!, every design of ESR meter in the whole world uses the same type of oscillator?

Unfortunately without knowing the circuit of your particular ESR meter, it is not possible to predict what will happen when the probes are shorted together.

Show the schematic or give a link to where it may be seen and then we can help you.
Until then...
...I for one have no idea.

JimB
 
hi again, the circuit i'am referring to can be found at MZEntertainment.com (ludens) basically, the circuit consists of a square wave osc, that signal is amplified, detected and rectified then goes to a meter. the actual ac signal(100kc) is picked off and fed to the cap under test the resulting signal then is amplified and fed to the meter. the lower the esr the greater the meter deflection. you have to short the test leads to start the test, you adjust for max deflection of the meter, then test the cap. my question was why does the meter deflect full scale when you short the leads? one of the leads is tied directly to ground. doesn" that kill the signal?
 
my question was why does the meter deflect full scale when you short the leads?
Meter full scale is the zero ESR indication.
A capacitor with zero ESR will effectively short the two probes together.
In the original article
https://www.ludens.cl/Electron/esr/esr.html
shorting the probes together was the way to establish the zero ESR point.

one of the leads is tied directly to ground. doesn" that kill the signal?
Why do you think that one of the probes is tied to ground?
Neither of the probes is "tied to ground".
Why would this kill the signal? this is the condition of zero ESR in the capacitor under test.

JimB
 
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