IGBT Power Dissipation

Discussion in 'General Electronics Chat' started by resilient, May 15, 2009.

  1. resilient

    Thread Starter Member

    Feb 2, 2009
    23
    0
    Hi all,

    I was reading up on generating sinusoidal signals from an AC inverter that uses IGBTs for switches. Basically, the site said IGBTs dissipate the most power when operating in the linear region and so its not smart to input a sine wave into the gate. They said that it is better to use sinusoidal pwm to drive the switches.

    Is this true? If so, why do IGBTs lose so much power in the linear region?
     
  2. Darren Holdstock

    Active Member

    Feb 10, 2009
    262
    11
    All types of transistor lose power in the linear region, it's just I x V. Switching between hard on and fully off uses much less power as the "on" voltage drop is very small when the current is high, and the "off" current is very small when the voltage is high. The transistor will pass through the linear region on the way from off to on and back again, and there will be power dissipated there, so the trick is not to linger for too long in this region, and balance the clawback in power from faster switching against increased dissipation from switching losses.

    A sinusoidal PWM is a sine wave represented in PWM form, rather than an analogue sine wave. The switched PWM waveform will be filtered at the output to reconstitute the original sine wave.
     
  3. mik3

    Senior Member

    Feb 4, 2008
    4,846
    63
    An IGBT (and every transistor) can be:

    1) Turned fully ON
    2) Turned fully OFF
    3) Something in the middle (linear)

    In case 1, the current can be high but the voltage across the IGBT will be small, thus I*V=small.

    In case 1, the voltage can be high but the current through the IGBT will be small, thus I*V=small.

    In case 3, both the voltage and current are significant and thus I*V=significant and the losses in the IGBT increase and it gets hot.
     
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