# Why transformer is heating without load [closed]

I'm working as technician in computer store and I have a lot of scrap transformers . Most of them small and have 6-8 pins .I wonder, can I use them for my projects ? If I remember correctly these transformers only work with high frequency and if you feed them low frequency they burn or explode But I don't know why they heat up without load on secondary coil . Also I think this the same reason you can't get 5000v with 9V transformer by connect it reverse

• Are you testing the transformers with 60Hz 120 volts? Commented Jul 27, 2019 at 3:17
• We use 220V 50hz but no I don't test them with 220V . I use 14v 50hz transformer for tests .Even 14V caused so much heating . Commented Jul 27, 2019 at 3:24
• you must realize that transformers have inductance which has impedance which includes resistance and if know OHm's Law Use Z(f)= X(f) +Rs = 2pif*L + Rs instead of just R. and losses that result from mismatch Commented Jul 27, 2019 at 4:48
• If problem caused by inductance , can I add capacitor to reduce reactive power ? Like a reactive power compensation ? Commented Jul 27, 2019 at 5:01
• Transformers you would find inside remotely modern computer gear would be part of higher frequency switching power supplies. Linear supplies used early in the early (pre-IBM) PC era were large and quite heavy when powering a comparable load. Keep in mind a transformer is ultimately a piece of wire across the inputs; only the response to AC of the designed frequency prevents it from being a short, run far below the designed frequency it's just a wire wound low value resistor / heater coil. Without specs this is unanswerable, but you are basically misusing the parts. Commented Jul 27, 2019 at 14:25

The transformer is relying on the magnetic field in the transformer core to limit the flow of current.

If you took away the core, and connected the winding directly to your 220V supply, then it would rapidly burn out, as the inductance is not enough to limit the current flow.

50Hz (and 60Hz) transformers normally use a laminated iron core. These are made of sheets of steel, with an insulating lacquer between them to stop eddy currents from flowing. Adding the iron core creates a strong magnetic field, which opposes the current flow through the transformer primary.

High frequency transformers use a different core material, often ferrite (a ceramic that contains a lot of iron). These cannot be magnetized as strongly as iron. So when connected to a 50Hz supply, the core "saturates" - it becomes as strongly magnetized as it can be. The magnetic field isn't enough to limit the current through the primary, and the transformer overheats.

• One can visualize it that way the core has to "eat" the applied flux. It's the integral of volts over time on one half-wave during magnetization before de- and countermagnetization happens in the second half-wave. As soon that area U*t sums up over the saturation flux of the core, the core cannot "eat" any more flux and becomes invisible magnetically. Its inductance drops to near-zero, so does the impedance and thus, the current rises extremely. For a small high-frequency core, this happens quite soon during a 50Hz half-wave. Commented Jul 27, 2019 at 14:58

Remember the transformer equation that relates Volts ,Frequency ,Turns ,Core Area and magnetic Flux density .You are dishing up 50Hz at 14 V .The surplus switchmode transformers Could be designed for 50KHz so if you must test at 50Hz you would want 250mV or less to avoid saturation of the ferrite core .Why not use a signal generater set at say 100KHz ?

Even without a load on the secondary of s transformer, even with the secondary open-circuited, there’s current in the primary. Why? Because the primary coil is an inductor.

That current is determined by the drive voltage and frequency and the inductance: $$I = V /{\omega L}$$

A transformer built for high frequencies, like yours, keeps the current reasonable with a pretty small inductance because the frequency is high.

If you run it at lower frequency, like 50Hz, this inductive current goes way up. Then even the small resistance of the coil combined with that large current generates heat.

You wouldn’t expect a transformer to work right at DC, right? By driving it at much lower than the designed frequency, you’re getting too close to that point.

You might be concerned that this inductive current violates to usual $$\I_s/I_p=n_p/n_s\$$ rule. That rule is associated with in-phase current, the kind that carries power. The inductive current in a proper transformer is out-of-phase, and doesn’t convey net power (as you’ve seen, that’s just an approximation, and the windings’ resistance can vary that phase enough to heat the windings)