help reading a datasheet

Each of those specs measures something different. The spec that is most important to you, depends on the application.

Rds is the resistance of the fully turned on channel at some specified gate to source voltage (which can be different for different devices). To a first approximation, the lower the Rds, the lower the DC resistive heating the device will have for a given drain current.

The power rating tells you how much heat can be pulled out of the die into a heat sink that holds the cast at a given temperature. This is mostly about die area and package thermal resistance. If your application is a highly efficient one (that produces little heat in the die) this spec is not very important to you. If you will use the device as a load to test power supplies (intentionally dump lots of power into the die) this this may be the most important spec. The data sheet writers try to cover any possible application.

The maximum amps and volts spec also apply to high power applications, not high efficiency ones, since low loss applications will not apply such large voltages across the die while it also passes large current.

For more specifics, please post links to the data sheets, so we are looking at the same data. The devil is often in the details.

--
Regards,

John Popelish

Think basic physics. E = IR, P = IE.

Higher Rds(on) means that for a given current the device will burn up more power (I'll leave it up to you to figure out the relationship between current (I), resistance (R) and power (P) -- you can derive it from the above two relations). I suspect that those power figures are just the amount of dissipation you can expect at 31 amps, not any recommended operating range.

Each device has a certain thermal resistance to its case, and a certain maximum junction temperature that it can stand. It's your job to figure out how much power you can dump from the device (which may require that you design in heat sinks to conduct heat away from the device case; such design isn't trivial), then use that to figure out what Rds(on) you need to look for to meet your other requirements.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Do you need to implement control loops in software?
"Applied Control Theory for Embedded Systems" gives you just what it says.
See details at http://www.wescottdesign.com/actfes/actfes.html

IR is notorious for spec'ing absurd power dissipations and currents on their fets. 150 watts in a TO-220 is borderline crazy, and would need heroic heat sinking. Derate their power specs by at least 2:1.

But 31 amps into a 28 mOhm switch is only 27 watts, OK with a good heat sink. 12 mOhms is only 12 watts, even better.

John

wouldn't the least the resistance value be the better option in any cases? The way I see it, the more resistance Rds, the more power it will loose through heat sinking. E = IR, P = IE. so by doubling R, (P=I * E ) the power onto the heat sink will be 4x as much. Therefore I should be using a small Rds value. as I read above, the power column displays how well the MOsfter can dissipate heat. if I am wrong tell me. I intend to use a lot of current, therefore huge heatsinking. I never bothered with these values, but now they are critical and it affect design.

K

If you're switching some given load, I is probably pretty constant, so power dissipation in the transistor is linear on Rds-on, not squared.

P = I^2 * Rds, where I is probably constant.

But yes, the lower Rds-on, the lower the heat lost in the fet. Make sure you apply enough gate voltage to turn it on good.

OK, but it's better to buy more silicon (lower resistance fets, even several in parallel) than to use one and a huge heat sink. Heat sinks are big and expensive, transistors are small and cheap.

And it's creepy to run currents like 30 amps through one of those skinny TO-220 leads. More fets is better.

What's your actual load current?

John

All else being equal, yes (with due deference to John Larkin's "low aggregate Rds(on) through heatsinking").

But all else is rarely equal. Lower Rds(on) means higher gate capacitance which is a pain when you're switching fast, or a more expensive part which is a pain when you're trying to shave pennies out of a product, or a harder to second-source part, which is a pain when your vendor decides they want to discontinue the part or sells 100000 of them to Ford without bothering to tell the factory that they did it, or, or, or etc.

So I wouldn't run out and buy the World's Lowest Rds(on) without thinking hard about the other compromises.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Do you need to implement control loops in software?
"Applied Control Theory for Embedded Systems" gives you just what it says.
See details at http://www.wescottdesign.com/actfes/actfes.html

You are wrong, the power column tells you what to expect the transistor to generate in wasted power as heat. A lower number here is desirable unless other aspects are more important to the overall design.

HI,

well I am running about 1.7 amps (eventually at lot more) I am using a IRFZ44 and its heating up like crazy.. and melts. The gate runs at 22v while the drain and source are at 10v. This last info is taken from Pspice. I have measured 1.7 in my circuit. I would think there is something wrong with my circuit, its only 1.7 amps (no heatsink).

K

HI,

well I am running about 1.7 amps (eventually at lot more) I am using a IRFZ44 and its heating up like crazy.. and melts. The gate runs at 22v while the drain and source are at 10v. This last info is taken from Pspice. I have measured 1.7 in my circuit. I would think there is something wrong with my circuit, its only 1.7 amps (no heatsink).

K

** Absolute nonsense.

The Pd figure tells you the LIMIT on how much heat can be dissipated by the device under ideal heatsinking conditions and at a stated case temp.

** The reverse is true.

A higher Pd figure indicates that in a given application and with other parameters the same, the chip will remain cooler.

In the OP's example, the "R" device has a much lower thermal resistance from chip to case then the "VZ" device - ie 1.0 degrees C per watt as against 1.6.

Turns a 90 watt rating into 150.

...... Phil

Something is wrong. Under those conditions (zero volts drop drain to source (since both are at about 10V) there should be no heat at all. If you mean that the drain and source are close to 10 volts, then the heat should be low. For instance, if the source is at 9.9 and the drain is at 10 volts, then the device should be producing 0.1V*1.7A=0.17 watts of heat. you might notice that as a slight rise above stone cold, but not much more. No heat sink should be needed for this condition.

What voltage drop do you measure across the actual MOSFET?

--
Regards,

John Popelish

t d

Hi,

I posted a snapshot of my circuit (wellpart of it and the outcome) I am putting two batteries in series:

8290 you may need to download the jpg to see it better From the simulation , it gives me the same voltage in drain and source. K

Yes, that simulation shows almost zero volts drop across the device, so should produce almost zero heat.

Do you really have a separate floating battery in your physical circuit just to drive the gate, like the simulation shows?

--
Regards,

John Popelish

I have two optocoupler driving from a a 6v battery, the other batteries are 12v.

..

I will do this tonight when I get back at home thanks

k

So... not the schematic shown in the simulator. Please post the actual schematic that produces the hot transistor.

--
Regards,

John Popelish

..

John, I implemented the simulator schematic, do you want to see the full version of that? I have two circuits, using 3 batteries (A, B and C). from these circuits I am using mosfets to create a 24v (battery A, B) section charging into the 12v section(battery C) , or battery, once full I switch the circuit around 24v is now battery A and C. I added three screen shots, one odd thing, when i simulate the separate circuit I get 1.2 amps, but when I add the two circuits, I get 2.2 amps. step 2 circuit (step1 is just the same but reverse.:

9826

Full circuit:

8482

I decided to parallel my mosfet last night, I was missing one mosfet to finish all the pairing. when I turned on the switch that single mosfet blew up like I never seen before, half my breadboard is now black...

K

(snip)

On the step 2 circuit I see 12 volt batteries called V1, V10 and V14. On the full schematic I see V1 labeled A that looks like V1 on step2, V5 labeled B that looks like V14 on step2, but V7 labeled C is 6 volts, not like the 12 volt V10 on step 2.

Why is C a 6 volt battery?

Battery C is turned the wrong way round to be in series with battery A (both positive ends connected).

You have no current limiting resistors in the LED (input) side of your opto couplers. If you hook them up this way they will be destroyed very quickly. If you will drive these from something that can hold almost 5 volts while delivering say, 10 mA, then the resistor in series should be about 330 ohms.

I'll study these a bit, but they are an awful tangle. I would not be surprised if you have trouble following these as you build the actual circuit. I think the time you took to better organize these drawings would pay off in fewer blasted parts caused by mis-connections.

Pretty obviously, your mosfets formed a short circuit across a battery. The weak link acted as the fuse (which you should add to the circuit).

--
Regards,

John Popelish

I can post the same circuit without the voltage and current showing will that help?

K