Re: Looking for a better in circuit ESR meter


Chuck Harris
 

Reg,

The "o)" symbol is a BNC connector. The ")" represents the grounded
shield on the BNC connector, the "o" represents the center pin.

One "o)" connector connects to the function generator, and the other
"o)" connector connects to the input on the oscilloscope.

The "@" signs are simply bends in the schematic wire, or nodes when the
wires cross or "T".

The 10K resistor, when fed with a 10Vpp 100KHz signal limits the current
to 1ma P-P.

The capacitor under test is connected where the two "=>" probes are on
the right side of the screen.

Ohms law is: V = I * R, and works equally well for AC, and impedance.

ESR is the series resistance a capacitor presents to the circuit.

Z = Xc + ESR

The idea behind I suppose all ESR meters is that at the frequency of the
measurement, the Xc of the capacitor is low enough that the ESR will
dominate the measurement.

As an example, let's assume that the capacitor we are testing is 1uf.

Xc = 1/(2pi * f * C) , Where f = 100KHz, and C = 1 E -6 Farads.

So, Xc = 1.59 ohms @ 100KHz

V = I * R, where I = 1ma, and R = 1.59 ohm.

So, as to the question of diode conduction, the expected voltage across
the capacitor is 1.59mv. Which is well below the conduction voltage of
the diode clamp.

A 10uf, or a 100uf would be even lower...

ESR + Xc values would have to be higher than 0.5V, or 500 ohms to
cause the diodes to conduct.

The 10K resistor makes your ESR readings come out in K ohms per volt,
so, 50 ohm ESR would be 0.05Vpp, and 5 ohm ESR would be 0.005Vpp...

If that is too far into the weeds, you can easily decrease the 10K
resistor to 1K, or even 100 ohms, to gain some more voltage in your
reading...

-Chuck Harris



Reginald Beardsley via groups.io wrote:

Chuck,

I'm afraid I don't understand the diagram or the explanation. If there is a resistor in parallel with the cap, how much of the resistance is the resistor and how much is the ESR of the cap? You don't appear to address that. It appears to me you assume that there is no resistance in parallel with the cap.

If there is a diode and resistor in parallel with the cap, how are you going to keep the diode from affecting the result with a 10 V input?

It certainly seems to me that it is critical to not to forward bias the diode so that it conducts.

I've got a few dozen caps to check. That is what makes the EDS-88A so attractive. The caps are packed in so tight that lifting one leg is not possible and getting them back in would be a huge ordeal.

Please draw out a circuit consisting of a diode, resistor, inductor and capacitor in parallel and explain how you would separate the effects of the cap from the other 3 devices with a connection across the cap. That's the general in circuit test problem.

Have Fun!
Reg



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