Transformer Capacitive Coupling

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The discussion centers on unexpected voltage measurements between the primary and secondary terminals of a 4:1 single-phase step-down transformer, raising questions about capacitive coupling. Despite the assumption that low-voltage transformers exhibit minimal coupling, significant voltages were recorded, suggesting otherwise. Participants recommend using a capacitance meter to measure winding-to-winding capacitance, which can clarify the coupling effects. Suggestions for circuit analysis include considering the impedance of the measuring device and employing a 'poor man's Thevenin' equivalent to derive two equations for the unknowns. Safety precautions are emphasized, particularly regarding potential shorts and the handling of energized wires.
QwertyXP
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I had a 4:1 single phase step-down transformer. Primary winding was provided the mains supply (225V) and output (secondary) was 57V. I was under the impression that capacitive coupling b/w primary and secondary was noticeable only in high-voltage transformers, and thus in this case (low voltage) the two windings would practically be electrically isolated.
However, when I measured the voltage b/w each of the primary (P1, P2) and secondary (S1, S2) terminals, I got significantly large values:
S1->P1: 204V
S2->P1: 154V
S1->P2: 6V
S2->P2: 44V

Are these values because of capacitive coupling?

I tried modelling this situation (Please have a look at the attached image. The arrows above primary and below secondary show assumed polarity). Then I wanted to find out the values of x and y through simultaneous linear equations, using the above values. But the first problem was that the meter showed RMS values which couldn't be negative, and secondly, this method didn't turn out to be correct either because all the different equations actually boiled down to a single equation. Is my model correct? How can I find the values of x and y?
 

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QwertyXP said:
I had a 4:1 single phase step-down transformer. Primary winding was provided the mains supply (225V) and output (secondary) was 57V. I was under the impression that capacitive coupling b/w primary and secondary was noticeable only in high-voltage transformers, and thus in this case (low voltage) the two windings would practically be electrically isolated.
However, when I measured the voltage b/w each of the primary (P1, P2) and secondary (S1, S2) terminals, I got significantly large values:
S1->P1: 204V
S2->P1: 154V
S1->P2: 6V
S2->P2: 44V

Are these values because of capacitive coupling?

I tried modelling this situation (Please have a look at the attached image. The arrows above primary and below secondary show assumed polarity). Then I wanted to find out the values of x and y through simultaneous linear equations, using the above values. But the first problem was that the meter showed RMS values which couldn't be negative, and secondly, this method didn't turn out to be correct either because all the different equations actually boiled down to a single equation. Is my model correct? How can I find the values of x and y?

Doi you have a capacitance meter? That's the easiest way to measure the winding-to-winding capacitances. BTW, the Cww values give you common-mode coupling between windings, not differential-mode coupling.
 
Capacitance meter would be great.
If your DMM is so equipped...

In your circuit analysis, what did you assume for impedance of your meter ?

Try ten megohms, which if digital it probably is.

Then try a 'poor man's thevenin' equivalent by placing maybe 10k ohms across your meter and reading voltage again.

Now you should have two equations to work with for your two unknowns.

Reason for 10k ohms instead of zero ohms as in real thevenin work is: there could be a short between the windings and you don't want to wreck your meter. If there is a winding to winding short the 10k resistor might quickly get very hot, so be careful.

SO: Check for shorts with ohmmeter first.

AND don't leave energized wires where tiny fingers can reach them. You know how curious kids are.
So hide your power cord.
old jim
 
I am trying to understand how transferring electric from the powerplant to my house is more effective using high voltage. The suggested explanation that the current is equal to the power supply divided by the voltage, and hence higher voltage leads to lower current and as a result to a lower power loss on the conductives is very confusing me. I know that the current is determined by the voltage and the resistance, and not by a power capability - which defines a limit to the allowable...

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