What makes the transformer in a SMPS smaller?

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.