If I want to make a sine-wave inverter where the output AC voltage is lower than the input DC input voltage, how can I see that using pulse-width modulation will be more efficient that just using a transistor as an amplifier if I have a sinusoidal gate driving signal at the desired frequency?

I've heard that transistors generally operate efficiently when "on" or "off", but not in an intermediate regime. My impression is that this is true for both BJTs and FETs, but I'm not sure. This rule of thumb is consistent with other things I've learned, like the fact that CMOS integrated circuits tended to be lower power than equivalent TTL chips, and also the fact that switched mode power supplies are efficient and popular. I've never really challenged this idea before (that switching a MOSFET with a given duty cycle is more efficient than using it as an amplifier). I spent about an hour trying to fact check it earlier and didn't manage to get through it.

What I've done so far: Firstly, I decided to focus on MOSFETs. Secondly, as an example I decided to look at a specific n-channel MOSFET, a Toshiba K3767, because I have a LD7550-based switched mode power supply that uses a K3767 as its power transistor. Thirdly, a bit of internet reading tells me the two main losses will be switching losses and conduction losses. So I guess I could do two calculations, one where I switch with a square wave at a high frequency, like 65 kHz as described in the LD7550 datasheet, and another scenario where I drive the K3767 with a sine wave at a low frequency like 60 Hz.

Am I on the right track here? Is there some really obvious answer, like I2R losses will be huge if I use the MOSFET as an amplifier with a sine wave on the gate?

How can I show that rapidly switching a transistor between on and off at a given duty cycle is more efficient than operating it as an amplifier to achieve the same average output?

  • 4
    \$\begingroup\$ Simply measure the voltage drop across the output transistors when "on" and multiply by output current in each case. That's wasted power. Which has more wasted power? \$\endgroup\$
    – user16324
    Commented Nov 29, 2016 at 23:24
  • 1
    \$\begingroup\$ Thank you. I was thinking about what you said for the last 20 minutes and see it now. In a trivial amplifier circuit whatever voltage is not dropped across the load is dropped across the transistor. This is fine if either the whole voltage is dropped across the load (none across the transistor) or zero voltage is across the load (in which case zero current will flow). In both cases the power dissipated in the transistor is zero. \$\endgroup\$ Commented Nov 29, 2016 at 23:57
  • 2
    \$\begingroup\$ Your remark about the \$I^2R\$ losses is already nearly the whole idea of the story. If the transistor is off, \$I\$ is zero, so no losses. If the transistor is fully on, \$R\$ is (nearly) zero, so again no losses. Somewhere in between: Both \$I\$ and \$R\$ have values notably different from zero, so there are losses. \$\endgroup\$ Commented Nov 30, 2016 at 0:00
  • \$\begingroup\$ Yes, it took me a while, but I have it now. Actually I think I can explain it to my 7-year-old nephew now: Imagine plugging in and unplugging a lamp. When it's unplugged, no power will be used. When it's plugged in, power will be dissipated in the light bulb, but very little in the wires. So quickly "plugging in and unplugging" a lamp will dissipate 100% of the power in the lamp. This is the world's most trivial switched-mode power supply. \$\endgroup\$ Commented Nov 30, 2016 at 0:14

2 Answers 2


The power dissipated in the transistor is P=IV where I is the current through the transistor and V is the voltage across the transistor.

When the transistor is fully on V is small and therefore P is small.

When the transistor is fully off I is very small and therefore P is very small

When the transistor is partially on both V and I are significant and so P is much larger.

Note though that to make a PWM based system efficient you need the right kind of filter, you don't want resistors in your filter because they waste power, so you would normally use a LC filter. You also need to make sure that when the transistor is off the inductor has a path to discharge, either by a diode or a second transistor.

  • \$\begingroup\$ Thanks. This is basically what I got out of Brian's comment after 20 minutes of wall-staring. \$\endgroup\$ Commented Nov 29, 2016 at 23:59

A transistor is just a variable resistor. (That is in fact where it name comes from.) What you do when you switch it, you have it either fully on, or fully off, hence very low resistance or very high resistance.

If you load your transistor with a load resistor, say R1 = 1k Ohm, you will essentially form two series resistors (one of them being the transistor, which can be seen as a variable resistance). You can try to calculate the power through it for both cases (very low resistance and very high resistance) and will then realise that in both cases, the power will be very low.

Some examples: (Rtrans = the transistor resistance, Vin = 5 V = input voltage)

  1. Rtrans = 1 Ohm

    The same current will go though both resistances and hence about 1000 times more power will be dissipated in R1 compared to in Rtrans.

  2. Rtrans = 1 MOhm

    Because of the high total resistance in both resistors almost no current will go through any of them and hence they will not dissipate much power.

However, if the resistance is something intermediate, like in an amplifier, say Rtrans = 1 kOhm. Then both R1 and Rtrans will be of equal magnitude and hence dissipate equal amounts of power. Only 2 kOhm will be seen from the voltage source so 6.25 mW will be dissipated in the transistor as compared to almost no nothing in the other two cases.


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.