Hi All,

I'm making a battery tester that will test batteries with small loads (25 Amp max) over a long period of time. I will be able to adjust how many amps load I want to put on the battery based on the manufacturers specification and than see how long it will take to drain the battery to 10.5 volts and then compare it with again the battery specs. I'm using PID loop in a PLC that has a 0-10V analog output to control an IRF1405 N channel enhancement MOSFET. http://www.redrok.com/MOSFET_IRF1405_55V_169A_5.3mO_Vth4.0_TO-220.pdf. I connected the GATE straight to the 0-10 volt output of the plc, the DRAIN to battery pos through a fuse and the SOURCE to battery neg. There is no resistive load to limit the current, however i put a fuse between battery pos and DRAIN to protect components and wiring if the MOSFET becomes fully saturated. What i don't understand, when the MOSFET opens more than i want it to and the fuse blows, the MOSFET gets damaged and it's like it stays fully saturated, open, it just breaks down. Why do you think that happens and what is the solution?

 

You want the MOSFET to be a switch, not the load.
Put a resistive load between battery positive and MOSFET drain.

 

Hello,

As said, it is not a good idea to take a mosfet as load.
When you want to regulate the power taken from the battery, use a fixed load and use PWM to regulate the power using the mosfet as a switch.
Now you are using the mosfet in linear mode and the mosfet must dissipate all power.
This will heatup the mosfet and give you the failure.

Bertus

 

You might want to study this application note at Linear Tech: A Closed-Loop, Wideband, 100A Active Load

I can only guess that your MOSFET is blowing either because it is overheating, shorting out and blowing the fuse, or possibly because the fuse is blowing simply because of excessive current, and in the act of blowing the abrupt cessation of current flow causes a catastrophic inductive voltage spike in the wiring between the MOSFET, fuse and battery, and this spike is what is destroying the MOSFET.

 

@OBW0549 Your signature at the bottom of your posts says (in part: "eat your vegetables !" That reminds me of a bible story - the one where Cane kills Able. You see, the brothers offered sacrifice to God. Able was a sheep herder and offered a sheep, whereas Cane was a farmer. His offer - yes, "Vegetables". The bible goes on to say that God looked upon Able's sacrifice with favor. This angered Cane, and so he killed Able.

Here's what I learn from that story: God don't eat no vegetables neither!

Not to stray from the original post: I have to agree with all the advice you've received so far (@Sebestyén Kárpáti). Depending on a MOSFET to act as a resistive load - - - um, lets consider this: Your device can handle so many amps. Like a resistor, we also need to calculate the wattage. So using a 12 volt battery as an example, lets put a 10 ohm resistor across the posts. 12 ÷ 10 = 1.2 amps. If you put a quarter watt resistor in that circuit you're going to see it glow brightly for a brief instant, then it will fail. You're pushing 14.4 watts through it. That's a lot of heat. Look at the data sheet and - um, here's where I'm not good with the numbers, calculate how many degrees (C or F) your part will rise at that kind of wattage. Even if your device is capable of handling 30 amps, and suppose you are using a 15 amp fuse, 1.2 amps is NOT going to blow a 15 amp fuse. But the MOSFET is going to heat up significantly. Likely it will blow. Quite possibly go dead shorted and then draw 30 or more amps through your 15 amp fuse - and yes, your fuse will blow.

 

Here's what I learn from that story: God don't eat no vegetables neither!

I don't give a fat rat's derriere what God eats; I eat all my vegetables and you should, too!

...Look at the data sheet and - um, here's where I'm not good with the numbers, calculate how many degrees (C or F) your part will rise...

Taking the TS's numbers, a MOSFET with 12 volts drain-to-source and passing 25 amps will be dissipating 300 watts. Looking at the data sheet, the maximum rated power dissipation of the IRF1405 is 330 watts, and that's with the case temperature held at 25 °C-- that is, with an infinite heat sink, which he almost certainly doesn't have. We're awful close to the max.

Another way of looking at it is through the thermal resistance characteristics of the MOSFET: the data sheet gives a junction-to-case thermal resistance of 0.45 °C/watt, so with 300 watts dissipation the junction temperature will be:

0.45 °C/W * 300 W = 135 °C above ambient (nominally 25 °C), or 160 °C.

The maximum rated junction temperature is 175 °C, so once again we're right on the hairy edge with, as before, an infinite heat sink.

So no matter how you figure it,

But the MOSFET is going to heat up significantly. Likely it will blow. Quite possibly go dead shorted and then draw 30 or more amps through your 15 amp fuse - and yes, your fuse will blow.

Yup. She's a-gonna blow.

 

You might want to study this application note at Linear Tech: A Closed-Loop, Wideband, 100A Active Load

Thanks for posting that. I particularly found the discussion at the end in "circuit testing" quite enlightening. Favorite quote: "Note 5: Sorry, no photo available. Uncontrolled 100 Ampere amplitude loop oscillation is too thrilling for documenting."

 

Hi: I assume you have a battery in the circuit when you are doing the setup and testing. That being said, how does the controller monitor the actual charging current. I suspect the controller may be overshooting and driving the current way past the safe level. The transistor you are using is a brute and should be able to handle the requirement providing you provide adequate heat sinking. As someone above pointed out, the transistor is de-rated linearly so that the dissipation allowed at 175 deg C (Die Temperature) is 0. Given that all is working correctly, the transistor is in a straight pass mode and should be operating in a linear region and not fully on. The dissipation in the transistor will be the charging current times the supply voltage minus the terminal voltage of the battery being charged. You can calculate the die temperature by using the specified Theta JC which is .5 deg C/watt mounted on a heat sink.

 

Hello,

As said, it is not a good idea to take a mosfet as load.
When you want to regulate the power taken from the battery, use a fixed load and use PWM to regulate the power using the mosfet as a switch.
Now you are using the mosfet in linear mode and the mosfet must dissipate all power.
This will heatup the mosfet and give you the failure.

Bertus

Hi Bertus, thanks for your reply! So if I replace my fuse with a fixed load (let's say 0.5 Ohm) that will essentially limit the current to 24A. But If I want to load a battery with only 10A the FET will still have to work hard. You reckon to gate the FET with PWM rather than a constant V? Is PWM the principle of SMPS?

Hi: I assume you have a battery in the circuit when you are doing the setup and testing. That being said, how does the controller monitor the actual charging current. I suspect the controller may be overshooting and driving the current way past the safe level. The transistor you are using is a brute and should be able to handle the requirement providing you provide adequate heat sinking. As someone above pointed out, the transistor is de-rated linearly so that the dissipation allowed at 175 deg C (Die Temperature) is 0. Given that all is working correctly, the transistor is in a straight pass mode and should be operating in a linear region and not fully on. The dissipation in the transistor will be the charging current times the supply voltage minus the terminal voltage of the battery being charged. You can calculate the die temperature by using the specified Theta JC which is .5 deg C/watt mounted on a heat sink.

Hey, thanks for taking time replying! I'm using a PID loop, so there is a feedback of the actual current, however I haven't fine tuned the PID loop yet and for sure it is overshooting. That's why i used the fuse so if it overshoots I blow the fuse not the FET. But it doesn't work like that apparently. The FET is mounted on a cpu heatsink with the fan and I'm monitoring the FET temperature directly and it didn't get hot.

You might want to study this application note at Linear Tech: A Closed-Loop, Wideband, 100A Active Load

I can only guess that your MOSFET is blowing either because it is overheating, shorting out and blowing the fuse, or possibly because the fuse is blowing simply because of excessive current, and in the act of blowing the abrupt cessation of current flow causes a catastrophic inductive voltage spike in the wiring between the MOSFET, fuse and battery, and this spike is what is destroying the MOSFET.

Hi, thanks for your time! I have no inductive load in the whole circuit. Why would a MOSFET that is rated to 169 Amps get damaged if a 25 amp fuse is protecting it? It's well under it's capabilities. Sorry I'm not questioning just asking. Want to learn.

 

Why would a MOSFET that is rated to 169 Amps get damaged if a 25 amp fuse is protecting it?

The fusing time of the fuse (what type?) must have been greater than that of the FET junction. It is often said that a transistor is there to protect a fuse .
If you read the small print of the FET datasheet you will see that the 169A rating is for very specific circumstances (unlikely to be encountered in practical circuits).

 

The fusing time of the fuse (what type?) must have been greater than that of the FET junction. It is often said that a transistor is there to protect a fuse .
If you read the small print of the FET datasheet you will see that the 169A rating is for very specific circumstances (unlikely to be encountered in practical circuits).

I'm using standard automotive blade fuse. So far this seems to be the most likely reason as far as i can see.

You want the MOSFET to be a switch, not the load.I didn't want
Put a resistive load between battery positive and MOSFET drain.

I know what you mean but what can i use as a resistive load? Thinking automotive, what draws 25 amps in a car that is designed to be on permanently. Whatever it is it will get hot and will need cooling. I could use 4 headlight globes in parallel but not so keen.

 

I know what you mean but what can i use as a resistive load? Thinking automotive, what draws 25 amps in a car that is designed to be on permanently. Whatever it is it will get hot and will need cooling. I could use 4 headlight globes in parallel but not so keen.

I use car headlamps.

 

Typical halogen headlamp runs about 55 watt. At 13.6 automotive volts (sometimes a little higher) comes to about 4 amps per bulb. Four headlamps would run 16.2 amps (@ 13.6 volts @ 55 watts {times} 4).

I wonder how much current a 120 volt 100 watt bulb would draw at 14 volts ? ? ?

 

I wonder how much current a 120 volt 100 watt bulb would draw at 14 volts ? ? ?

About 280 milliamps.

Keep in mind that in an incandescent lamp, the resistance varies drastically with filament temperature: at the lamp's rated voltage, it's resistance will be 10-15 times higher than the resistance when the filament is at room temperature.

I'm not too keen on using incandescent bulbs as dummy loads because of this variation, at least not without constantly measuring and logging the actual current through them.

 

the resistance varies drastically with filament temperature.

Yes, I know. I have a small appliance bulb who's rating can't be read, but it's generally in the 10 to 20 watt range. It's cold resistance is about 17 Ω. At 120 VAC that would translate to 7 amps. At 7 amps (times 120 VAC) you'd have a bulb of 847 watts. Um, I DON'T THINK SO! So it's startup amperage may be 7 amps, but once illuminated, it should be drawing 83 ma to 167 ma. Meaning the resistance at temperature should be around (taking the average of the two milliamperes of 125) 960 Ω. Yes, I know that a bulb at temperature has a much higher resistance.

I'm convinced that a 120 volt bulb won't light up at 13.8 volts. I'm just wondering how much thermal change would affect the bulb and its resistance at that voltage. I suppose I could set up a test and measure the current. I DO have a variable power supply that is capable of varying the AC and the DC voltages between 0 and more than 120 volts. (AC = 125, DC = 135) But right now I'm developing a Marble Machine for my grandchildren for Christmas. If I think of it I may do this experiment some other day.

 

I'm convinced that a 120 volt bulb won't light up at 13.8 volts.

It glows dull red, with enough brightness that you can see it in a lighted room.

I'm just wondering how much thermal change would affect the bulb and its resistance at that voltage.

Pretty easy to predict; the TCR of tungsten is 0.45%/°C, and you can go from there.

 

the TCR of tungsten is 0.45%/°C

There you go again, math. Don't let the balding head fool you. Grass doesn't grow on a busy street, but it don't grow on concrete neither.

 

There you go again, math.

Yup. Hold my beer and watch this:

R(t) = R(25°C) + (TCR * (t - 25°C))​

One addition, one subtraction, and one multiplication-- not exactly advanced calculus.

I don't see how you can do much at all in electronics without at least that much math; without it, all you can rely on is guesswork and superstition.

 

plug in 2550C for the terminal temp of the tungsten filament.

 

plug in 2550C for the terminal temp of the tungsten filament.

Oops. Good catch; that should be

R(t) = R(25°C) * (1 + (TCR * (t - 25°C))),​

not

R(t) = R(25°C) + (TCR * (t - 25°C))​