Using IGBT or Mosfet

I found a weighted decision table would help me make a choice. Usually a decision
is not just based on efficiency, cost, availability, reliability, load, design support......all
enter into the decision. Even make a similar post on some of the heavy hitter high power
devices, like Analog Devices, IXSYS, Toshiba,........see what they say. But do so with
adequate details, energy source, load, environmentals......

Certainly a simulator is helpful, especially if model of device thorough. That
you can check by asking vendor product engineer or field engineer for an
"inside" opinion and guidance.

I worked with Plasma Cutter and Welding guys for several years, they did a lot
of characterization work on several vendors before committing. Just a thought.


Regards, Dana.

Thanks Danadakk for the reply, But as a thumb rule calculation:

IGBT loss: Vce*Ic= 1.65* 20A= 33W
Mosfet Loss I^2 Rdson= 2*(12)^2*0.016= 4.6W

Based on this , MOSFET beats IGBT. I know this is a simplistic calculation. But if costing is similar, then isn't Mosfet better option. What other parameters are critical.

Many more if you do the search "IGBTS versus MOSFET"


Regards, Dana.

sabu31 said:

IGBT loss: Vce*Ic= 1.65* 20A= 33W

Click to expand...

In post#1 you just mention 1.2kW. No current at all.

So how do you come to 20A? Did you do some calculations already? CCM, DCM? what inductance?

1.2kW / 20A = 60V (for DC only .. which you don´t have.)

Do you have any example design you refer to?

Klaus

We have ups within the load you mentioned of 1.2kw
They used 4 mosfets for the high and low signals
Every two are series together I think

When load more than this for example 3kw and 6kw and 10kw , etc I never seen mosfets for inverter circuit only IGBT’s

I think mosfets are good option in your case, and by the way the simulation is important, btw to vendors talk

Regards

Hi,

MOSFET vs IGBT:
* switching speed usually is faster with MOSFETs
* voltage rating is usually higher with IGBTs
* voltage drop (power dissipation). Which one is better depends on current, R_DS_ON of MOSFETs, V_CE_SET of the IGBTs

Klaus

KlausST said:

Hi,

MOSFET vs IGBT:
* switching speed usually is faster with MOSFETs
* voltage rating is usually higher with IGBTs
* voltage drop (power dissipation). Which one is better depends on current, R_DS_ON of MOSFETs, V_CE_SET of the IGBTs

Klaus

Click to expand...

Basically I need to operate at around 40 -50kHz.The voltage input is 48V and output is 192V. Vcesat from datasheet is 1.65V where as the Rdson is16mOhm. Based on this it seems, that the conduction losses is higher for the IGBT based converter.

KlausST said:

In post#1 you just mention 1.2kW. No current at all.

So how do you come to 20A? Did you do some calculations already? CCM, DCM? what inductance?

1.2kW / 20A = 60V (for DC only .. which you don´t have.)

Do you have any example design you refer to?

Klaus

Click to expand...

The operation is in CCM mode. The inductance used is 65uH.
However, I am not looking into inductor losses. But need to know for the same condition, which device (IGBT or MOSFET, assuming fixed frequency of 50K, CCM, and 80% duty ratio, 48V voltage input and output 192V, 1.2kW) will have more losses.

2 boosters in series with the same gate drive will be a much better plan - running a single stage @ D = 85% to get the required step up

is not a favoured approach in power electronic circles - the pulse currents are high

Easy peasy said:

2 boosters in series with the same gate drive will be a much better plan - running a single stage @ D = 85% to get the required step up

is not a favoured approach in power electronic circles - the pulse currents are high

Click to expand...

What is generally prefered for this requirement.

For such high conversion ratio ( 48 to 192), isolated dc-dc converters are a good option. You can get the gain from transformer turns ratio.

sabu31 said:

What is generally prefered for this requirement.

Click to expand...

Twin interleaved boost converters is another option. This allows a lower saturation rating for inductors. Allows reduced stress on all components.

sabu31 said:

But as a thumb rule calculation:

IGBT loss: Vce*Ic= 1.65* 20A= 33W
Mosfet Loss I^2 Rdson= 2*(12)^2*0.016= 4.6W

Click to expand...

I also did the same rough estimation, but with a little modification. It should be considered that the MOSFET current rises in a non linear shape, so we should calculate the RMS power, not the instantaneous one, and the average power would be rather somethink like this (If I'm not mistaken):

• P_MOSFET(avg) = r_ds(on).(√2/2). [ I_max² - I_min² ]
• P_IGBT(avg) = V_ce(on).(1/2). [ I_max - I_min ]

Ron = 0 is never ideal, unless your reactance parts are so lossy that they provide the necessary damping factor or Q at minimum load.

Thus, you need to define your max/min load ratio or discrete values with time and determine your loop stability margin in each state for a step load in either direction to avoid a massive overshoot in one of the conditions. Then you will realize that efficiency and stability or step load error or load regulation error are all tradeoffs. Unless you can predict disturbances you must rely on voltage AND current feedback with whatever phase or gain margin you have at a given Q. This means you need all these specs and more to decide which is best. such as OVP margin, OTP margin, thermal runaway margin. Certainly the switch be it diode, FET or IGBT with the lowest RC=Tau value and the right value of Ron for OVP stability AND temp rise is best. These are the SiC and similar types which have a much lower Ron*C(0v)=Tau than Si types. These are also indicators of fast reverse recovery time which affects dynamic power loss directly with time during transition..

FETs naturally have a much higher Cout than IGBT's with a BJT output. Depending on your low side PWM switching f and resulting slew rate can make dynamic losses greater than static losses. Your primary average impedance R = 48V^2/ 1200W= 1.92 Ohms. Naturally you might estimate for 1% loss use a switch< 1% of 2 ohms.

IGBT's have a quasi linear curve of Vout= 0.7+ Imax * Rce. You can also estimate Rce for any transistor or diode or IGBT by consider the heatsink, Pd required to dissipate all the heat with < 60'C junction temperature rise for adequate reliability. The case will be cooler. My rule of thumb is Rce< k/ Pd for k = 0.25 to 1. If > 1 don't use it. Lower is better quality.

Thus for any transistor , diode or IGBT, Rpn=Rce= (Vmax-0.7) / Imax which is then approx = k/ Pd for k = 1 max for worst case.

Not to dwell on this but there is high correlation between thermal resistance and electrical ON resistance and 0V capacitance.

Read more what the experts say about switches and also read about how to design the optimal RC-D values of the high-side voltage booster.

Hi,

andre_luis said:

so we should calculate the RMS power

Click to expand...

I don´t think there is something like "RMS power". For heating we use the "average of the power".
You may have RMS current and RMS voltage....

andre_luis said:

PMOSFET(avg) = rds(on).(√2/2). [ Imax2 - Imin2 ]

Click to expand...

is here I_Max the peak of the current ripple ... and I_Min the minimun of the current ripple?
if so, then if I have pure DC (no ripple) then I_Max = I_Min.
According your formula the P_Mosfet becomes ZERO. Is this correct?

To be honest, I don´t know the exact formula. But I´d rather expect something like "(I_Max^2 + I_Min^2)/2" ...
... and then somehow multiplied with the duty cycle. .. because when the MOSFET is oFF then there is no power dissipation at all.

(At least no conduction loss. Still there may be switching loss and body_diode_loss, depending on circuit)


Klaus

Use of Average versus RMS power :



Regards, Dana.

KlausST said:

is here I_Max the peak of the current ripple ... and I_Min the minimun of the current ripple?
if so, then if I have pure DC (no ripple) then I_Max = I_Min.
According your formula the P_Mosfet becomes ZERO. Is this correct?

Click to expand...

The terminology used above would be wrong if the analysis were made in a "time" domain P_(t), but we are considering the power varying on an ascendent slope as a function of the "current" domain P_(i) - linear for BJT [P=k.I] and quadratic for MOSFET[ P =K.I² ]; In any case, read min as initial and max as final. The purpose was just to provide a handy formula to make a rough estimation for a switching cycle in the case where current do not falls downto zero, e.g continuous mode.

19 m-ohm ( worst case for the fet ) x 1.8 for 100deg C in the die, divided by 2 mosfets = 17.1 m-ohm

IGBT = 1.4 V worst case @ 30 A - plus these have poorer turn off losses

48 V to 192 V at 1200W output give s a D == 0.85 to allow for losses

so at steady state, I in = 26.3 amps, I out ave = 6.25A, so the output caps sees 21A pk ripple

Assuming an L that gives +/- 15% current ripple = Ipk on input is 30.3 A, I start each cycle = 22.4 A

rms current in the mosfets is:

= 24.43 amps @ D = 0.85

Mosfet cond losses = 10.2 watts ( 5.1 watts each )

IGBT = 34.2 watts losses ( 17.1 watts each )

turn on losses are dictated by Rgate ( on ) and can be similar for both devices ( igbt's can turn on really fast if you allow it - ~ 20nS )

Turn off losses can be near zero for the mosfet due to its non linear Cds, for < 30nS Vgs 10V to zero the losses ( turn off ) are very small indeed

for the igbt however: best case is 0.22 mJ turn off losses @ 30A, 192V, which at 50kHz = 11 watts, 5.5 W ea.

@danadakk The fact that the senseless term "RMS power" is used in an ADI application note doesn't give it more sense. See https://en.m.wikipedia.org/wiki/Audio_power#Continuous_power_and_.22RMS_power.22

Thanks to Easy peasy for the detailed calculation. I would usually answer the design question with a short consideration.
1. IGBT isn't well suited for 48V input due to high Vce,sat
2.IGBT operation at 50 kHz is no good idea

Hi,

FvM said:

The fact that the senseless term "RMS power" is used in an ADI application note doesn't give it more sense

Click to expand...

at least they clarify:

You do not want to calculate the rms value of the ac power waveform. This produces a result that is not physically meaningful.

Click to expand...

and

The rms value of this power waveform is 1.225 W.
...
The power dissipated by a sinusoidal 1 V rms across a 1 Ω resistor is 1 W, not 1.225 W. Thus, it is the average power that produces the correct value, and thus it is average power that has physical significance.

Click to expand...

and

The rms power (as defined here) has no obvious useful meaning (no obvious physical/electrical significance), other than being a quantity that can be calculated as an exercise.

Click to expand...

...

So litterally they say you can do RMS_power calculation, just as an exercise without being physicall/electrically useful.

Klaus

added:

FvM said:

I would usually answer the design question with a short consideration.

Click to expand...

and in addition: One can say:
* paralleling two MOSFETS will result in half of total power dissipation
* paralleling two IGBTs here works in best case with just 25% reduction (@ 60A DC). But thermal runaway will realistically decrease it maybe only by 10%.

So paralleling IGBTs is less effective and not reliable, thus usually not worth the cost/effort.

Klaus