29-11-2013, 07:37 PM
The Power Factor (PF) of a capacitor is equal to the equivalent series resistance (R ) divided by the impedance (Z).
For a perfect capacitor R = 0 and so PF = 0. For a resistor Z = R and so PF = 1 or 100%.
Also for a capacitor of reactance (X), Z = SQRT(R^2 +X^2) so a capacitor with a PF of 70% would have R = X which is a pretty poor capacitor.
The only type of capacitor to show anywhere near this sort of loss is the electrolytic which usually has a value well above 100nF, so only the top range requires a PF control. Paper, film, ceramic etc types should all have PFs much less than 1%
Looking at the circuit with PF set at 70%, R2 = 100ohms (ie at max) which is in series with C6 = 1uF.
So to get a PF of 70% (1/SQRT(2) to be exact) the reactance of C6 must be 100ohm this would require a frequency of 1.5918kHz (2*Pi*F = 10^4). Is it possible that the bridge should run at this frequency?
Incidentally it is more usual to measure Q = X/R or D (Dissipation Factor) = R/X = 1/Q. Certainly the TF2700 does this.
Jim
For a perfect capacitor R = 0 and so PF = 0. For a resistor Z = R and so PF = 1 or 100%.
Also for a capacitor of reactance (X), Z = SQRT(R^2 +X^2) so a capacitor with a PF of 70% would have R = X which is a pretty poor capacitor.
The only type of capacitor to show anywhere near this sort of loss is the electrolytic which usually has a value well above 100nF, so only the top range requires a PF control. Paper, film, ceramic etc types should all have PFs much less than 1%
Looking at the circuit with PF set at 70%, R2 = 100ohms (ie at max) which is in series with C6 = 1uF.
So to get a PF of 70% (1/SQRT(2) to be exact) the reactance of C6 must be 100ohm this would require a frequency of 1.5918kHz (2*Pi*F = 10^4). Is it possible that the bridge should run at this frequency?
Incidentally it is more usual to measure Q = X/R or D (Dissipation Factor) = R/X = 1/Q. Certainly the TF2700 does this.
Jim
