Re: Looking for a better in circuit ESR meter
Reg,toggle quoted messageShow quoted text
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
Reginald Beardsley via groups.io wrote: