Lets consider we have a HV DC source, implemented with voltage multipliers (first image), which is powering up a load resistor. If we want to look at the whole scheme as a DC power source with voltage Udc.s and output impedance Rdc.s, how can Rdc.s be estimated?
If Rin in the first scheme is high, then we could probably simply measure the short circuit current of the source. But if Rin has low resistance what is the proper way to estimate the output impedance of the HV DC source? Analytically or experimentally or both.
Boris, I can suggest an universal method to measure the internal resistance of any real voltage source. To implement it, you need a load resistor R and, of course, a digital (perfect, high resistive) multimeter. Here is the procedure.
First, measure the output voltage E without load (open circuit) by connecting the voltmeter; this will be the electromotive voltage (emf). Then, connect the resistor and measure the voltage drop V across it. Finally, calculate the internal resistance from the voltage divider equation: V/E = R/(R + Rin).
Now as i think again, the HV source works in a constant charge-discharge transient process and Rin only affects the charging speed. In such case measuring the short circuit current is not an option at all. Does this mean that the output impedance of the HV source is almost zero?
Boris, I can suggest an universal method to measure the internal resistance of any real voltage source. To implement it, you need a load resistor R and, of course, a digital (perfect, high resistive) multimeter. Here is the procedure.
First, measure the output voltage E without load (open circuit) by connecting the voltmeter; this will be the electromotive voltage (emf). Then, connect the resistor and measure the voltage drop V across it. Finally, calculate the internal resistance from the voltage divider equation: V/E = R/(R + Rin).
But here, as we can see, wee have three cascaded devices - an AC source with internal resistance Rin, a storying element (capacitor) and a load. As you have noted above, Rin is between the first two (voltage source and capacitive load) and it is inside the voltage multiplier. If we consider it as a black box, we will actually measure the internal resistance between the capacitive voltage source and the load... and it will be, as you have noted, very small...
Thank you for the clarification Cyril. So i can expect low output impedance which could eventually be estimated experimentally as you suggested. Although i believe the estimation would be inaccurate because of the high impedance of the load. The voltmeter simply won't be enough to measure the difference because of its accuracy. That's why i was wondering if it's possible to estimate the output impedance of the source analytically.
I think that, for the purposes of the measurement, the load can be low-resistive enough...
But I wonder what an impedance would be the measured impedance... It should be the impedance of the two lower capacitors in series, i.e., a fully capacitive impedance...
- Badges
- Science topic
- Similar topics
- Engineering
- Electrical Engineering