Continue to Site

Welcome to

Welcome to our site! is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

Switching a transformer

Not open for further replies.


Apr 4, 2022
Reaction score
Trophy points
Activity points
Hi guys,

Based on the below image I want to switch the transformer( with high inductance and low resistance) the power supply is 220v ac 50Hz and the relay is ideal which have a 25Hz of switching frequency with 50% duty cycle.

As you see in the graph which shows the current passes through the transformer, there is no negative side. Is it normal?

Does it lead to transformer saturation and having inrush current?




a typical situation.
Switch ON close to zero cross of voltage. Magnetic field builds up. Then negative half wave ... magnetigc filed decreases to what it was before.
In real world this usually leads to saturation (about twice the normal magnetic field strength), high current. But with the use of serial resistances (source, transformer, wiring) the DC part decreases over the next few fullwaves to become close to zero..


there will be some ( possibly a small amount ) of remanance energy stored in the Tx at the end of the cycle resulting in some extra volt spikes when the relay opens ( and the terminal volts on both sides will ring towards zero volts across the terminals )

contrary to the above comments, starting at max neg and finishing at same completes a 0 to 0 cycle of the BH loop - so only the remanance will reamin.
--- Updated ---

assuming you have a true bipolar sine source and start at -ve pk and finish at -ve peak ....
Last edited:

contrary to the above comments, starting at max neg and finishing at same completes a 0 to 0 cycle of the BH loop - so only the remanance will reamin.
This is not contrary to the comments above.

Generally for ideal inductors:
If you assume to start with about zero B:
* then starting at any angle phi ... and ending at the same angle phi (full wave)
--> will always end at zero B

But not the start and end of B is the point of discussion, but the average of a full wave (DC), or in other words:the range of B.
In an ideaalistic view:
* B could be positive only when starting at positive zero cross
* B could be negative only when starting at negative zero cross
* B could by symmetric when starting at positive or negative peak of voltage
* or anything inbetween.

In a real circuit this all will relax over the next fullwaves and B will "move" symmetrically around zero.
Whether ... and how much this will lead to saturation and overly high current depends on the inductor.


The voltage waveform shown by the OP - since it is unipolar - will saturate the Tx before too long and effectively short circuit the driving source, also the volts across the Tx will include a large neg spike when the relay opens as the core tries to reset ....

Any DC current through a transformer can lead to core saturation. and high magnetizing current.
I agree ... and not.
Generally the statement is true.

If you have DC input voltage, then according the inductance the DC current will "integrate up" and surely will sooner or later lead to saturation.
But in the original case (idealistic pure AC input, no series resistance)..
There is no "input DC" thus it will not "integrate up". In ideal case it's just an initial "one shot DC" caused by the point when voltage is applied. (Phase angle). In an ideal circuit this DC will not increase ... but it will not decrease either.

In a real circuit the DC (magnetic field) will be attenuated due to series resistance.



text in post#1 says:
As you see in the graph which shows the current passes through the transformer,

if there was DC voltage I´d expect the DC current to increase. (integrate)
But it shows the current, thus I´d expect the input voltage to be free of DC ... with an initial DC component.

Still all this is only true for non realistic "ideal" parts. .. and maybe not worth to discuss... idk


The OP has posted incorrectly - you cannot get that current from the sine source and the circuit shown for a real world TX - only for an imaginary Tx with no saturation.


Thus my answers always differed between real circuit and ideal circuit.


Notice that the OP didn't further contribute to the thread. The original question has been clearly answered, the waveform is expectable with ideal components. We don't know if a related real circuit problem exists or it's just a theoretical question.
If you want to model a real circuit problem, you should use a saturable transformer and appropriate electronic switch models.

Not open for further replies.

Part and Inventory Search

Welcome to