simulate this circuit – Schematic created using CircuitLab

I have a few questions regarding variable power supply units powering boost converter circuits with transformers and the relationship between input power and output power.

I have a PSU supplying power to a boost converter (circuit and transformer - as the above simplified schematic) which is converting the input voltage to approx. 20kV, and at the open circuit (no load where the variable resistor is) the PSU is supplying 3.21W.

When I attach a 10M\$ \Omega \$ load across the secondary coil of the transformer (the output, where the variable resistor is) the PSU supply increases to 3.29W. If I then attach a 1M\$ \Omega \$ load the PSU supply increases to 3.41W. The PSU input power increases as I decrease the output load all the way down to about 50\$\Omega\$.

Q1. If I measure the power across the variable resistor can this be related to the PSU input supply? (i.e If at 10M\$ \Omega \$ the power dissipated across the resistor is 0.01W and the PSU is supplying 3.29W. Every time thereafter the PSU supply is 3.29W can I infer that there is 0.01W dissipating across the load attached to the secondary coil, or the transformer is producing 0.01W power when the input is 3.29W, the transformer and circuit are 0.3% efficient at that PSU input power?)

Q2. If I have a resistor that I don’t know what the resistance is and place it across the secondary coil (where the variable resistor is) and the PSU supply reads 3.26W, can I conclude that the value of that resistor is between Open Circuit and 10M\$ \Omega \$? If I was to measure the PSU supply at different resistances between Open Circuit and 10M\$ \Omega \$ (200M, 100M 50M etc) and find where the PSU reading is the same, could I say that is the resistance of that unknown resistor?

Q3. Is the open circuit power demand of the transformer the lowest it could possibly be or is it possible that the PSU can provide a power lower than the open circuit value (i.e lower than 3.21W open circuit value) if a specific load is attached (inductor, capacitor etc)?


1 Answer 1


Q3. An AC power transformer draws "magnetizing current", which creates magnetic flux in the core, even when there is no load on the secondary. If there is an appropriate amount of capacitance on the secondary, a resonant circuit is created, and the current in this resonant circuit may contribute to the magnetizing current, thus reducing what is drawn from the power supply. The magnetizing current may be thought of as current going through an inductor parallel to an ideal transformer. Although this current, adds to Volt-Amp requirements, because it is going through an inductor, it does not directly add to power consumption. (Refer to power factor for more info). However, there is ohmic loss of power from the current due to the resistance of the coil. The Volt-amps drawn from the power supply should be lowest (at least to an approximation) when the resonant circuit is tuned to the frequency of the power supply.

In radio frequency transformers, creating such resonant circuits is quite common. Adding capacitative loads is also practiced by utility companies where savings can be found. However, for mains frequency transformers for home devices, resonant circuits for transformers are generally not worth the trouble.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.