I'm reading up on how SMPS's work and all the videos say that after the high-voltage chopping stage, a smaller transformer can now be used, because of this high frequency. My question is, how does higher switching frequency result in a smaller transformer?

In other words, what makes switching 50-60Hz AC require a bigger transformer?

  • 3
    \$\begingroup\$ A related question you might ponder is, "Why do aircraft use 400 Hz for their AC systems rather than more common 50 or 60 Hz?" \$\endgroup\$
    – Transistor
    Jul 14, 2016 at 20:08
  • \$\begingroup\$ @Omkar Gawde: Your statement that "This implies that inductance and frequency have an inverse proportion" using the formula "XL = 2*pifL" is incorrect, because the inductive reactance XL is NOT a constant. By changing either f or L, or both, XL will change accordingly. The simple reason for the smaller size of transformer in SMPS is due to the fact that the much higher frequency f (thanks to PWM) will allow the use of a smaller transformer with a much smaller inductance L, since the inductive reactance XL (product of f * L * 2 * pi) is still high enough to keep the current low (aka less heat l \$\endgroup\$ Nov 2, 2020 at 17:08

6 Answers 6


Imagine the 50/60 Hz transformer is an inductor connected to AC - disregard the secondary coils and concentrate on the primary winding. The primary winding is connected across the AC and it can be regarded as a simple inductor. How much current does this inductor take from (say) a 220V AC supply?

Common sense says we don't want it to take much because the current is wasted doing nothing other than magnetizing the core. So, in an average sized (generalism alert!) AC transformer it might be wound to have an inductance of (say) 10 henries. This will have an impedance at 50Hz of: -

\$X_L = 2\pi\cdot f\cdot L\$ = 3142 ohms.

This will take a current of 220 V / 3142 ohms = 70mA and that's OK in my book. When the unloaded secondary winding is added it still takes 70mA and when loaded it takes the "referred-to-primary" load current + 70mA.

A switching transformer operating at (say) 100kHz doesn't need to have anywhere near the same inductance - this is because it's operating at 100kHz (or 1MHz or whatever arbitrarily high frequency). It might have an inductance that is proportionally lower by the ratio of the frequencies i.e. 50 divided by 100,000 - this means it can have an inductance of 5 milli henries and still perform as well (but at the higher speed).

Ask yourself, what transformer is bigger - one which has a primary inductance of 10 henries or one that has a primary inductance of 5 mH?

EDIT - section on fly-back transformers

It's better-news for fly-back switch mode designs (as used in most low to medium power AC-DC converters) - the primary inductance becomes a "feature" of the design - it is used to store energy during one half of the PWM cycle and then that energy is released into the secondary during the 2nd half cycle. If the primary inductance is (say) 1000 uH and let's say it is "charged" in 5 us and "released" in the next 5 us, the energy per transfer can be calculated by first estimating the peak current: -

\$\dfrac{220V\cdot\sqrt2\times 5\times 10^{-6}}{1000\times 10^{-6}}\$ = 1.556 A

  • The above formula is just V = \$L\dfrac{di}{dt}\$ re-hashed
  • 220V x sqrt(2) is the rectified and smoothed DC voltage obtained from the AC

Then this current converts to energy = \$\dfrac{L\cdot I^2}{2}\$ = 2.42 mJ

This can be turned into power by multiplying by 100,000 (the switching frequency) i.e. 242 watts. Using a fly-back topology allows you to utilize the primary inductance and lower it beyond what you reasonably could do in a linear power supply. Hope this makes sense.


For purposes of intuitively understanding the size issue, imagine that the transformer is a bucket. Imagine that at the given frequency you take water from a well and put it in a pool.

Say that you need to supply 60 liter per minute. If you are getting 1 bucket of water ever 10 seconds, then you need a 10 liter bucket. However, if you take 1 bucket of water every 2 seconds, then you only need a 2 liter bucket.

By increasing the speed, you reduce the size requirement, and since it is pretty easy nowadays to make very fast electronics, the size of transformers has gone down dramatically.

Note that this is not how SMPS actually work, the flyback described by "Andy aka" is closest to this, but this should give you an understanding of the impact of frequency.


Take the first answer. The resistance is the product of 2, pi, frequency and inductance. To get the same result you can reduce the inductance (size) when increasing the frequency.

  • 2
    \$\begingroup\$ Can you expand on this answer and give some example calculations? It's hard to tell how it answers the question. \$\endgroup\$
    – skrrgwasme
    Jul 14, 2016 at 20:10

Simple equation analysis would result as follows:

XL = 2*pifL

This implies that inductance and frequency have an inverse proportion.

In other words, for higher frequencies, the inductance can be reduced for the same impedance value.

So for an instance, I want my transformer to consume the minimum current, then I would select the highest possible impedance.

Let's take the same example as provided by Andy aka.

Current consumed by the transformer primary should be less than 70mA

Let's say that the voltage is 220V @ 50Hz frequency (Just an example)

Now, V = 220V and I = 70mA gives, R ~ 3142E.

In our case R = XL = 3142E.

When f = 50Hz, L ~ 10H

When f = 500Hz, L ~ 1H




When f = 100kHz, L ~ 5mH

Lets see the different parameters affecting size of the Inductor.

  1. No of Turns (More turns more inductance)

  2. Coil Area (Increase the area for more inductance)

  3. Coil Length (Inductance increases with increase in Length)

  4. Coil Material (Greater the magnetic permeability of the inductor the more the inductance)

The above parameters suggests that for the same material used the size of the Inductor will increase for an increase in Inductance.

Hence, from all the above analysis it can be said that,

To reduce the transformer size we need to increase the frequency.

Note: Moderation required. This is my analysis and hence should be considered only if enough experts support my say.

  • \$\begingroup\$ Good clear explanation \$\endgroup\$ Apr 1, 2019 at 11:32

You can imagine the transformer core being 'fully' magnetized by the input voltage. This represents a certain amount of energy. Next this energy is provided to the output. This cycle is repeated over and over.

Each cycle transforms a maximum amount of energy, limited by the amount of magnetism the core can 'hold'. Hence a higher frequency transfers more power.


If you supply the transformer at lower frequency the emf in one turn at primary is lower than higher frequency, it follows when the transformer is supplied at lower frequency the input current will increase , so the cross section of primary winding will increase same comparison can be done for number of turns at the secondary side which leads to the fact that the number of turns will increase for the transformer with lower frequency


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.