IGBT on resistance - conduction loss

[BlinkingLeds]

Hi, with MOSFETs the power losses are calculated P.loss=R(ds-on) * I^2 (I think)
how can i calculate the power losses of an IGBT if i can't find Rds (or Rce ?) in the datasheet?

Suppose we have a MOSFET 600v 20A and an IGBT also 600v 20A which will be better (less losses?) ?

Thanks

 

[(*steve*)]

IGBT's don't have an Rds. They have a saturation Vce.

Here is an IGBT datasheet (I just googled for IGBT datasheet and took the first hit)

On the third page there's a graph of the saturation characteristics. There are several lines, each showing the characteristics for a different gate voltage.

This is a 1A device, so we go across at 1A and we find that for pretty much ang gate voltage higher than 5V, the voltage drop is 2.5V.

So this device will dissipate about 2.5W at 1A.

You don't say what your possible IGBT's are, but you'll find that some voltage drop between 2V and 4V is not unusual.

This might sound a lot, but the dissipation is linear with current. In a mosfet is it related to the square of the current. That's why you do see IGBT's in very high current applications.

Imagine a current of 200A. Across 4V, this is 800W. If you wanted to use a mosfet, you would require one with a channel resistance of 20 milliohms or less or it would dissipate more power.

If you think that's a bit fanciful, here is a datasheet for a 1700V 2400A IGBT. It drops 2.6V at 2400A (that's 6240W dissipated). Imagine trying to find a 1700V mosfet with a channel resistance of 1 milliohm or less?

At any current though, you will need to do the math for both and compare. (I^2R for the mosfet and IV for the IGBT).

Note that switching losses may need to be calculated for both as well since a difference in switching speed could easily swing the equation the other way.

 

[BlinkingLeds]

IGBT's don't have an Rds. They have a saturation Vce.

Here is an IGBT datasheet (I just googled for IGBT datasheet and took the first hit)

On the third page there's a graph of the saturation characteristics. There are several lines, each showing the characteristics for a different gate voltage.

This is a 1A device, so we go across at 1A and we find that for pretty much ang gate voltage higher than 5V, the voltage drop is 2.5V.

So this device will dissipate about 2.5W at 1A.

You don't say what your possible IGBT's are, but you'll find that some voltage drop between 2V and 4V is not unusual.

This might sound a lot, but the dissipation is linear with current. In a mosfet is it related to the square of the current. That's why you do see IGBT's in very high current applications.

Imagine a current of 200A. Across 4V, this is 800W. If you wanted to use a mosfet, you would require one with a channel resistance of 20 milliohms or less or it would dissipate more power.

If you think that's a bit fanciful, here is a datasheet for a 1700V 2400A IGBT. It drops 2.6V at 2400A (that's 6240W dissipated). Imagine trying to find a 1700V mosfet with a channel resistance of 1 milliohm or less?

At any current though, you will need to do the math for both and compare. (I^2R for the mosfet and IV for the IGBT).

Note that switching losses may need to be calculated for both as well since a difference in switching speed could easily swing the equation the other way.

Ok thanks now i know what Vce sat is for.



So about the switching part... what if we have a 300vdc 20A heater switching at 5khz and duty cycle from 1% to 99%.

MOSET:
IPW60R075CP
Rds-on = 0.075ohm
max amps = 39A

and

IGBT:
IXYP15N65C3
Vce sat = 2.5v
max amps = 38A

So, which one would you choose without even calculating. Is there an easy answer?

Thanks

 

[duke37]

What is wrong with calculating?

Mosfet
P = 20 8* 20 * 0.075 = 30
IGBT
P = I * Vce = 20 * 2.5 = 50

 

[BlinkingLeds]

What is wrong with calculating?

Mosfet
P = 20 8* 20 * 0.075 = 30
IGBT
P = I * Vce = 20 * 2.5 = 50

Nothing wrong with calculating conduction losses but i meant switching losses . I know it's difficult to calculate.

 

[BobK]

Why are you using a 5KHz PWM for a heater? Lower switching frequency = less loss, and a heater is not going to be responding at 5 KHz.

Bob

 

[BlinkingLeds]

Well it doesn't has to be 5khz i am just trying to "dim" it make it consume less power when i want.

So let's say 500hz. Another question i have is will the switching losses be lower at 1% duty cycle or at 99%?
Thanks

 

[(*steve*)]

The switching losses are identical at 1% and 99%.

The switching losses are proportional to the frequency and the time taken to switch the load.

In general, the mosfet will be able to switch faster than the IGBT

Look at the switching speeds for both devices. (typically a rise time and a fall time)

Say they're 17ns and 7 ns.

Add these together (24nS of switching time each cycle).

Calculate the power being switched (V * A) = 300 * 20 = 6000W. That's 6000J/s

Calculate how many joules are dissipated during each switching event. the rough approximation is 1/2 * switching time * power switched

1/2 * 24x10-9 * 6000 = 0.072 mJ

Now we multiply that by the number of times the load is switched per second (let's take 2 cases)

at 500Hz -- 500 * 0.000072 = .036J
at 5000Hz -- 5000 * 0.000072 = 0.36J

Since this energy is dissipated over a second, the power is 0.036W for 500Hz, or 0.36W for 5 kHz.

This is very small compared to the total dissipation.

Those figures just happened to be for your mosfet (assuming the ideal gate drive).

Repeating them for the IGBT, you get something like

63ns spent switching
189mJ per cycle
0.0945W at 500Hz, 0.945W at 5000Hz

These losses are small, and represent the minimum obtainable losses -- you may not do so well if you do not have a good gate driver.

Note that there are other losses as well. In this case I believe these are the most significant. If you were driving a highly capacitive or inductive load, you would have to consider them.

 

[BlinkingLeds]

The switching losses are identical at 1% and 99%.

The switching losses are proportional to the frequency and the time taken to switch the load.

In general, the mosfet will be able to switch faster than the IGBT

Look at the switching speeds for both devices. (typically a rise time and a fall time)

Say they're 17ns and 7 ns.

Add these together (24nS of switching time each cycle).

Calculate the power being switched (V * A) = 300 * 20 = 6000W. That's 6000J/s

Calculate how many joules are dissipated during each switching event. the rough approximation is 1/2 * switching time * power switched

1/2 * 24x10-9 * 6000 = 0.072 mJ

Now we multiply that by the number of times the load is switched per second (let's take 2 cases)

at 500Hz -- 500 * 0.000072 = .036J
at 5000Hz -- 5000 * 0.000072 = 0.36J

Since this energy is dissipated over a second, the power is 0.036W for 500Hz, or 0.36W for 5 kHz.

This is very small compared to the total dissipation.

Those figures just happened to be for your mosfet (assuming the ideal gate drive).

Repeating them for the IGBT, you get something like

63ns spent switching
189mJ per cycle
0.0945W at 500Hz, 0.945W at 5000Hz

These losses are small, and represent the minimum obtainable losses -- you may not do so well if you do not have a good gate driver.

Note that there are other losses as well. In this case I believe these are the most significant. If you were driving a highly capacitive or inductive load, you would have to consider them.

So what's the ideal gate drive?

so switching losses aren't that bad compared to conduction losses at least with resistive loads. I was trying to calculate the switching losses using those huge equations but couldn't yours simplifies it A LOT.

 

[duke37]

PWM seems to be well over the top for a heater.
You could use phase angle control or a solid state relay switched by a timer circuit.

 

[(*steve*)]

The ideal drive is generally applied using a gate driver IC.

In your case, it's probably not that important and I question (along the lines of what Duke37 is saying) why you need to switch it so fast (for light, 100/120 Hz is fine. For heaters 0.1Hz may be fine.

 

[BlinkingLeds]

Well... I don't deed to switch it that fast i was just using it as an example .