My aim is to get 12 volts output from a Stepdown transformer for a circuit I need to construct. Supply voltage at home is 220 volts AC. I can simply use a stepdown transformer available in market. But some times the supply voltage gets doubled (440 V), which can burn out the primary of transformer.

If I take two similar transformer with primary rated for 220V and secondary for 24Volts, and connect the primary of both in series, then naturally the output voltage will be dropped from 24V to may be nearly 12V. But here I may have benefit to protect my transformer from higher voltage fluctuations. I will take output from only one transformer. The second transformer will just act as a resistance in series of primary. Please comment ...

  • \$\begingroup\$ Now the concept is clear to me. I will design the circuit using option (b) or (c) \$\endgroup\$
    – Anwar
    Sep 15, 2017 at 15:26

1 Answer 1


Are you really saying that you have a 220 V supply that intermittently gives 440 V? How does anything electrical survive in your location?

You can't series connect two primaries on different cores like that unless you parallel the secondaries. Let's think about why:


simulate this circuit – Schematic created using CircuitLab

Figure 1. (a) Due to different secondary currents the primary impedances will not match. (b) and (c) will work because the secondary currents will be the same, therefore the primary impedances and currents will be the same.

  • A transformer's primary current will depend on it's secondary current (which will vary with the load). \$ \frac {N_P}{N_S} = \frac {I_S}{I_P} \$ where N is the number of turns, P is primary, S is secondary and I is current.
  • With no secondary current a transformer's impedance (AC resistance, if you like) is quite high - otherwise a high current would flow.
  • On the transformer with secondary current the primary impedance would be lower.
  • It should now be clear that you no longer have half the supply voltage on each primary. The one that is loaded will have a much lower voltage.

Connecting the secondaries in series or parallel ensures that both secondaries pass the same current and the primary impedances will match.

This only gives you enough theory to understand why your proposed solution is poor. You need to fix your incoming power supply.


I don't quite understand how impedance of transformer will have to change?

Given that \$ Z_P = N^2 Z_S \$ where \$ N \$ is the turns ration we can see that if a seconday is open circuit then Z is infinity on both sides of the transformer.

I think the connection shown in Fig. 1a will induced voltage tending to infinity across secondary, similar to case in current transformers.

It would if you could force current through the primary. You can't however. You only have your mains voltage and an infinite impedance. The primary of XFMR2 will appear to be an open-circuit to an AC supply. No current will flow. Full mains voltage will be across XFMR1 primary and XFMR1 secondary will have a voltage \$ \frac {1}{N} \$ times that.

Remember that we're dealing with ideal transformers for this discussion. Real ones will have some losses and leakage so the impedance will not be infinite.

  • \$\begingroup\$ Hmm. I don't quite understand how impedance of transformer will have to change? I think the connection shown in fig. A will induced voltage tending to infinity across secondary, similar to case in Current Transformers. \$\endgroup\$
    – Deep
    Aug 19, 2017 at 14:53
  • \$\begingroup\$ See the update. \$\endgroup\$
    – Transistor
    Aug 19, 2017 at 15:17
  • \$\begingroup\$ Oh yeah, I understood that now, I need to look closely at equivalent circuit ^^". \$\endgroup\$
    – Deep
    Aug 19, 2017 at 15:24
  • \$\begingroup\$ Good. Wait a day or two to give others a chance to respond. Then upvote and accept the best answer. You can upvote other good answers too. \$\endgroup\$
    – Transistor
    Aug 19, 2017 at 17:14

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