[SOLVED] Certain BJTs more efficient than certain MOSFETs?

Hey all
New member here, first post. Please excuse if it is in the wrong section.
Here's what I'm currently working on: an H-Bridge implementation to drive a 56V17A DC Series motor.
Now, I have a TON of TIP142T/147T Darlington pairs as well as IRFP250N Mosfets. The voltage and current rating can be somewhat ignored since the motor will be undervolted. I'd prefer to use these (High side switching is not a problem for me here) since I just have them lying around.
Now to my main concern- if my calculations are correct, since the voltage drop in the BJTs will be max 3V (high+low), the max power dissipated can only be V x I ie 3 x 17 = 51.
I am somewhat new to mosfets, but assuming I fudge the heat sinking and the the mosfets are running hot, the max Rds(on) will be 3 ohms @175 degrees. That's 6 ohms in series with the motor. Assuming I am able to push the required amount of current through it (my calculations indicate otherwise), the power dissipation will be I^2 x R ie 17^2 x 6= 289 x 6 = 1734 W.

The Rds will be a minimum of 1 ohm at room temp, so that's 289 x 2 = 578 W!

This leads me to believe that the MOSFETs will be less efficient than the Darlingtons. I do this as a hobby, and I am sure I have overlooked things, but is my estimation correct? I intend to parallel the Darlingtons to get the current I want.

RM2488 said:

the max Rds(on) will be 3 ohms @175 degrees.
The Rds will be a minimum of 1 ohm at room temp

Click to expand...

That would be hopeless for your application. There are other MOSFETs with much lower Rds(on), but if you want to use what you've got then the Darlingtons would be better.

When you parallel them, be sure to add emitter resistors to help with the current sharing.

Thanks for the quick reply! Yes, I'll be using separate resistors for the sharing, so there shouldn't be a problem.
The reason I asked the question is, I was pretty sure my calculations had to be wrong. At these specs, the IRFP250s appear to be rubbish for ANY high current application. Why on earth would they have that rating? 30A@25C, 21A@100C! And that's the continuous rating!

RM2488,


In order to perform a quick comparation, I use the following calculus :

P(BJT) = Vce(on)*I
P(MOSFET) = Rds(on)*I^2

Click to expand...

So, if Vce > Rds*I the best choice is the MOSFET.
Of course, current nonlinear behaviour at MOSFET must be considered, and thershold limit is not exactly the one presented at above formula.


+++

You -could- pick a BJT:NMOS matchup that favored either,
if you were only looking for an argument position.

BJTs can be better for high current, but they are really
bad (the ones optimized for high current) for switching
losses. To get good Vce(sat) you need to conductivity-
modulate the collector resistance way down, which
means mucho stored charge C-B when that diode is
driven forward, hard.

This is why you see IGBTs relegated to sub-100kHz
applications. IGBTs give you "base current for free"
more or less, at the cost of a higher Vce(on) - the
current carrying device is more like Baker-clamped
than driven to saturation.

Oops, hang on, rewind.

You didn't read the datasheet right. There's one here: http://www.datasheetcatalog.org/datasheet/SGSThomsonMicroelectronics/mXywzvy.pdf

On page 2, it says Rds(on) is 0.085Ω max at 25°C.

The graph on page 5 shows "Normalized On Resistance vs Temperature". That means "compared to the value at 25°C", so the value of 2 at 125°C means it's 2 * the value at 25°C i.e. 2 * 0.085Ω = 0.17Ω.

Sorry I didn't pick that up before.

andre- Thanks, I will try out that formula, currently I go through the entire above calculation.
dick_freebird- I tried to understand, and some of it makes sense, but I should read it when I am not as sleep deprived (it is 1 AM local time).
godfreyl- I may have read the datasheet wrong, but here's the funny thing- the (wrong) values matched! Ambient is 40 here, and with V(GS) at 9v, I measured 0.9 ohms resistance!

EDIT: I made a huge blunder whilst verifying the datasheet , er, data. As it turns out, the PROBE lead on my multimeter is adding the resistance, it appears to have developed a break. Of all things... I did not check the zero reading before I measured it (it was fine this morning!). So problem solved! Thanks a lot, I learnt something new from you today- reading a normalised graph!

I solved the high side problem by using two 9v batteries to provide the greater-than-source gate voltage- not ideal, but the application is only a demo, intended to run only for twenty minutes, but at up to 100% duty cycle. Therefore a charge-pump did not have the appeal.

Hmmm...
1) Are you sure you measured right?
2) Did you get the MOSFETs cheap on ebay?
(if so, they're probably fakes)

Ok, I edited just about the same time you posted. No, the transistors were not bought on ebay, I am lucky to have a high volume retail store nearby. So they are the real deal. Now all I have to do is calculate using the new figures and fix a heatsink!

The calculations were correct, thank you!

- - - Updated - - -

The calculations were correct, thank you!