How does a capacitor double the transformer output?

Question

I recently got a transformer with a ratio of 10/1 and fed 40 VAC into it. I got the ~350 VAC on the secondary which was expected.

When I connected a 1000 V, 1 μF AC capacitor to the secondary its output was 600+ VAC (the multi-meter doesn't read higher than that). What is going on? How is the capacitor doubling the AC voltage?

Answer 1

We can use a nonideal transformer model to explain the behavior:

^{simulate this circuit – Schematic created using CircuitLab}

Lm represents the magnetizing inductance (open circuit measurement). XFMR1 is 4:35 turns ratio, and ideal (zero leakage inductance, infinite magnetizing inductance), so the 600mH is reflected as \$\left(\frac{35}{4}\right)^2\$ times larger or 46H on the secondary side (close enough to your 40H measurement). LL represents the leakage inductance (short circuit measurement), 'p' for "primary referred" since it's modeling all of it on the primary side. We can use the same transformation if we like (3H, close enough to your measured 4H) and place it on the secondary side equivalently.

If we pull everything to the secondary side (source included), we get a simple LC circuit:

^{simulate this circuit}

L2 is large enough we can ignore it (it acts in parallel (Thevenin's theorem) with L1, reducing its, and V1's, value slightly). An LC circuit is resonant at \$F_0 = \frac{1}{2 \pi \sqrt{L C}}\$, or 92Hz. Evidently we're below resonance (oh, I forgot to ask what line frequency was, but that's fine, it's a small change to substitute through). Which doesn't really mean much, and response is infinite at resonance anyway (there is no resistance shown in this circuit). But we can solve for the voltage off-resonance quite simply by taking the impedance divider:

$$ V_o = V_i \frac{Z_\textrm{C1}}{Z_\textrm{L1} + Z_\textrm{C1}} \qquad Z_\textrm{C1} = \frac{1}{j \omega C_1}, \quad Z_\textrm{L1} = j \omega L_1 $$ where \$j = \sqrt{-1}\$.

This simplifies to,

$$ V_o = V_i \frac{1}{1 - \frac{\omega^2}{{\omega_0}^2}} \qquad \omega_0 = \frac{1}{\sqrt{L C}}$$

which interestingly is a real result (no imaginary component), again, because we've not included any loss (resistance) component. So, this will be off by some in practice. Anyway, we have \$F_0\$, and we only need the ratio to \$F\$ so we don't care about the \$2 \pi\$ of the \$\omega\$s, and it comes to \$V_o = 1.74 V_i\$ or about 610V. Which is close to the simulation.

Sources of errors in this analysis:

• The 'L' model (that is, one series and one shunt inductor, they form an 'L' -- well, a rotated one, okay) misses the fact that there is some magnetizing and leakage inductance associated with each winding by themselves; a 'T' or 'Π' network is more correct. The ratio is still large enough to get within, say, 10% of the correct result, which is fine given the other limitations.

• As mentioned, resistance isn't accounted for. No real resonant circuit has infinite Q factor (even superconductors get "only" into the Q ~ 10⁷ range), and both core and winding resistances will matter here. And the capacitor's, to a lesser extent (capacitors are usually much better Q than inductors, in practice). Including this in the model, will give the correct phase shift and magnitude, given the measurements that is.

• The measurements are probably a bit mismatched (with respect to primary/secondary referred values/measurements) due to nonlinearity of the core. Laminated iron has less permeability for small signals ("initial permeability") than large. Permeability eventually drops again at very large signal level (saturation); there is a sweet spot in the middle, say around half nominal voltage at [lowest] rated frequency. Your measurement probably uses a small voltage (~1V?) on either winding, so the HV winding should measure low, which seems consistent with the observation.

  In contrast, the leakage measurements specifically short out the magnetizing field, so the core largely isn't involved; the measurements are mismatched in the opposite direction apparently (LLs measures higher than LLp, LMs measures lower than LMp!), which might be due to how the windings are arranged, or something more nebulous (I'm not sure offhand, and would have to think about it a bit more, or see and test the part personally).