I was reading about inverters in a textbook where the author says that

The size and cost of the circuit can be reduced to some extent if the operating frequency is increased but then inverter grade thyristors must be used which are costly.

How does an increase in frequency have an impact on the size of the inverter circuit(or does it affect the rest of the circuit too?).Is there some physics involved which causes this?


The largest single factor is usually inductor size. If you eg double frequency you can generally halve inductance (as the impedance of a pure inductor is proportional to frequency). In practice a number of factors apply so that it's not a directly linear relationship, but good enough.

If you need a peak current of say 1A then the time taken to ramp up from 0 to 1A is related mainly to inductance and applied voltage. If the inductor is say 10 x smaller the current ramps at ~ 10x the rate. The discharge time is also similarly speeded up and the overall cycle is faster so operating frequency is higher. You can look at this as the smaller inductor causing higher frequency operation or of the higher frequency allowing smaller inductors.

If the text mentioned thyristors in that context it is probably either an old one or is dealing with extremely high power levels. Nowadays, for most purposes inverters would usually use either MOSFETs or IGBTs. The very largest inverters may still use Thyratron valves - such as the many MegaWatt units used for DC to AC power conversion for DC submarine cables.

In typical portable modern applications an inverter which may have been operated at 100 kHz or less 10+ years ago is now liable to operate at 500 kHz to 2 MHz and a few operate at higher again. At 1 MHz+ and power levels of say a few Watts the inductor size may be 10%-20% of the size at 100 kHz and the inductor may still dominate the overall size.

Note that current carrying capacity ~ proportional to wire area but inductance is proportional to turns squared. This does not mean though that core size changes only with sqrt of frequency as you have issues of core cross section, core path length, winding window size and more to add to the fun.

  • \$\begingroup\$ Isn't \$Z_L = 2 \pi fL\$ ? How is impedance then inversely proprtional to frequency? \$\endgroup\$ – Vineet Kaushik Dec 1 '13 at 13:21
  • \$\begingroup\$ @VineetKaushik - Whoops - my brain and my mouth (hands) not in sync :-). -> proportional. ie I was trying to convey that as frequency rises physical size of inductor falls to achieve the same job. Also as Brian Drummond correctly suggested (deleted answer) power when energy is stored in an inductor as part of process is proportional to L x I^2 x f. As f increases you can use proportionally smaller inductor - eg water analogy: more buckets transferred faster along a bucket line deliver same flow. \$\endgroup\$ – Russell McMahon Dec 1 '13 at 23:36
  • \$\begingroup\$ @Brian Drummonds answer which he deleted was essentially correct and was useful. He said: Fundamentally an inverter transfers packets of energy from one circuit to another. As such, it needs to store or transfer a given amount of energy in a switching cycle. Keep the desired power constant and you reduce the energy transferred per cycle, allowing a smaller component to store or process it. \$\endgroup\$ – Russell McMahon Dec 1 '13 at 23:38

The use of higher frequency requires smaller capacitors, physically smaller inductors / transformers and their cores, and therefore reduces overall size of a design.

  • Capacitive reactance \$X_C = \dfrac{1}{2\pi fC}\$, so for any given reactance desired (filtering etc) a higher frequency f allows a lower capacitance C.
  • Inductive reactance \$X_L = 2 \pi fL\$ so again, for any given reactance, a higher frequency f allows for a smaller inductance L.

On the other hand, depending on the purpose intended, a high frequency inverter might not suit the purpose: For domestic power inverters, an output at least approximately close to the mains frequency is required for most equipment.

The way some sinewave inverters address this, is by operating at a far higher frequency, kilohertz to megahertz, and generating the sine waveform via PWM. Thus, the bulk of the power transmission occurs at the higher frequency, with a final stage low-pass filter to get rid of the higher harmonics from the PWM signal, and leave behind a smooth sine wave at the desired 50 / 60 Hz.

  • \$\begingroup\$ Sir, are you sure on this. Because many inverters does not have a sine wave. It does have a modified sine wave. \$\endgroup\$ – Standard Sandun Dec 1 '13 at 10:19
  • \$\begingroup\$ @sandundhammika Agreed that many inverters are not pure sine wave. The ones that are true sine wave, are the ones my answer refers to. \$\endgroup\$ – Anindo Ghosh Dec 1 '13 at 10:34

I was having the same problem and here’s what I found:

XL = 2πfL
Z= (R2 + XL )1/2
I= V/Z

When f increases XL increases.
When XL increases Z increases.
I is inversely proportional to Z therefore when Z increases I decreases.
Therefore increase in frequency results in decrease in current.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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