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Questions about IRFZ44N MOSFET

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DjinnG

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I have built the simplest possible linear power supply using the LM338 IC.
Since I was stepping 28V down to 16V the power dissipation was a problem.
I found what seemed to be an elegant (& simple!) solution in the following document: https://www.linear.com/pc/downloadDocument.do?navId=H0,C3,P1243,D4099 - figure 5. It is a freely oscillating switching circuit that is supposed to keep the voltage over the LM338 at a minimal level to lower power dissipation.

In my newbieness, I choped off half the components (the 10K, 2.2K resistors, protection diode & coil), replaced the 2N6667 with an IRFZ44N MOSFET, and used the opamp differently. This got the MOSFET hot enough to start smoking. It never occured to me that the FET was in danger since a quick glance at its datasheet showed that Ids Max is 49A, and my load should draw only 3-4A.

My questions are:
1. It seems the FET is still OK. How can I test it to make sure it is not damaged?
2. Has anyone tried somthing similar? How can the power dissipation be calculated for the FET (so I can properly heat sink it)?
3. Assuming the FET is on, according to its datasheet, it should not get hot by a 3A load current (rDS(on)=0.022). Does the switching cause it to heat up? Connecting it serially - in an on state - with a 3A current still made it heat up. Could it mean it was damaged previously and now has a much higher rDS(on)?
4. From the IRFZ44N's datasheet I learned that Max Vgs=20V, though Vgs=10V will completely saturate it (assuming Vds is high enough). Is that correct?

Thanks for you help.
 

flatulent

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to220 dissipation free air without a heatsink

The power dissipation is the current through the transistor times the voltage across it. As far as testing the device, if it still works in your circuit it must not have been too damaged.
 

VVV

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why mosfet fail

The MOSFET ratings confuse many people. The current rating of the MOSFET tells you what the maximum current can be with the MOSFET fully on and the case temp 25°C. Under these conditions, the power dissipation is Pd=Rdson^2*I. For this power dissipation, with the case at 25°C, the junction temperature reaches its maximum permissible, either 150 or 175°C.

This rating is good only for COMPARING PARTS. Keeping the case at 25°C with the junction at 150 or 175°C is really impossible, so the test is done with low duty-cycle pulses. So, as far as I am concerned this is just an aboslute maximum before the bonding wires inside melt.

In normal operation, at the maximum ambient temperature the junction should be at is 110°C maximum.
So, how do you do that? Look for the junction to ambient thermal resistance. For a TO-220 package, like your IRFZ44, it should be about 60-65°C/ W. That means that for each 1W the transistor must dissipate, its junction temperature increases by 60°C. So if your maximum ambient is 50°C then the maximum power the TO-220 MOSFET can dissipate in free air (no forced-air cooling) is Pmax=(110-50)/60=1W. That's it, without a heatsink, your MOSFET can only safely handle 1W at 50°C ambient, free air.

Adding a heatsink lowers the thermal resistance and allows more power to be dissipated. When you select a heatsink, follow the calculations above, but use the thermal resistance provided by the heatsing manufacturer. If there is an insulator between the MOSFET an the heatsink, add the thermal resistance of the insulator (typically 3-6°C/W for a TO-220 package).

Now, you cannot just remove parts without knowing what you are doing. The "protection diode" and the inductor, together with the MOSFET form a buck regulator, which has greater efficiency than a linear regulator (mainly because the transistor switches on and off, so either the current of the voltage across it is low, leading to low power dissipation).

Since you eliminated those parts it is likely your MOSFET somehow got into its linear mode and the power dissipation increased, because both the voltage and the current are relatively high.

I am not sure how you connected the MOSFET, since it is an N-channel and the original transistor was a PNP.

For calculations of power dissipation in the MOSFET of the switching regulator you can try to search for appnotes on the ON Semi website, or post and I will reply with some formulas.
 

    DjinnG

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DjinnG

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djinng

VVV said:
Since you eliminated those parts it is likely your MOSFET somehow got into its linear mode and the power dissipation increased, because both the voltage and the current are relatively high.

I was afraid of that, but I couldn't quite figure out how it got to its linear mode, since the opamp was driving the Gate with 28V (through a 47K resistor...)

VVV said:
For calculations of power dissipation in the MOSFET of the switching regulator you can try to search for appnotes on the ON Semi website, or post and I will reply with some formulas.

I would appreciate that. Thanks for your help :)
 

VVV

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+mosfet +heating +irfz44

First, I have a correction to make: when using a heatsink and insulator, use the thermal resistance JUNCTION TO PACKAGE for the transistor. The rest is correct.

The N-ch MOSFET needs a voltage on its gate positive with respect to its source. If you connected the MOSFET with its drain to the input voltage and the source to the input of the LM331 and if the output voltage was set to a fairly low value and without the diode/ inductor, then it would have been possible to turn on the MOSFET. In effect, you created a LINEAR regulator ahead of the LM331. For a linear regulator the power dissipation in the MOSFET is simply the voltage across it times the output current. So at 3A, with only a few volts across the MOSFET you easily exceeded the 1W I was talking about in the previous post.

The intention of the author of that circuit was to have a high-efficiency switching regulator that would keep the voltage across the LM331 constant at about 3V. That limits the dissipation of the LM331, allowing high output current (the same rule applies to the LM331: P=Iout*Voltage across it, so keeping the voltage at a minimum, for a given power dissipation you can get higher current).
The overall efficiency is improved, since the switching regulator has good efficiency, say 70%. However, the linear regulator has a much cleaner output with better transient response, so you get the best of both worlds.

The configuration is that of a buck regulator, working in hysteretic mode. You can get an idea about these circuits by consulting for example the datasheet for the MAX1744 from Maxim. You will find there some formulas for power dissipation, too.
If you still need help, please post the maximum output current. Please note that this part does a similar job, but not identical to what you are trying to do.

Another important thing is that the opamp cannot drive the N-channel MOSFET in the switching regulator configuration. That is because the MOSFET's gate has to be driven higher than the source by about 10V. When the MOSFET is fully on the voltage at the source will nearly equal to the input voltage. So the opamp would have to be powered from a voltage about 10V higher. You can use a P-channel MOSFET or a bipolar PNP as in the original schematic.
I will think about another solution.
 

    DjinnG

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DjinnG

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Re: MOSFET abuse

VVV,

What you said about the opamp not being able to drive the FET is quite interesting. I think I am beginning to understand why MOSFET drivers are needed. If the source is not at a 0V potential or worse yet, in an adjustable regulator is undeterminable, how can one properly drive the gate? on the one hand the 20V max Vgs cannot be exceeded AND on the other, the maximum voltage allowed should be applied in order to switch the FET efficiently.

Anyway,
I got an IRF9540 (P-CH) to replace the 2N6667 (in the original schematic) which is apparently obsolete, and since I couldn't find a 1MH 10A coil, I settled for a 200uH 10A coil.
Now that the buck coverter is in place (and with the "right" transistor), it works swell(!!) at about 16V 3A.
The voltage across the regulator remains at Vout+3V and it barely heats up.

Naturally, I have a new set questions... :)

- The coil buzzes like an angry bee when a load is applied :) What do I do about it?
- The MOSFET gets pretty hot after a minute or so. It may be normal but two things are not "proper" yet:
1. The coil in the original schematic is 1mH and I am using a 220uH one. could that cause excessive load on the FET?
2. I imagine more adjustments should be applied to the circuit when replacing the biploar with the FET. any ideas?
3. How can I protect the FET from excessive gate voltage and at the same time drive it efficiently?
4. You mentioned in your previous post that you can help with the power dissipation in a switching circuit: the max current I am aiming for is 10A.

Thanks.
 

VVV

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Re: MOSFET abuse

The coil buzzes because the switching frequency is very low. That is rather strange. I thought is would be around 3kHz (it should whine, not buzz). You can adjust the frequency by changing the 68 pF capacitor in series with the resistor in the opamp's feedback. Can you check it's really 68pF?

The lower inductor can definitely have something wih the MOSFET heating up. I will try to take a closer look at the circuit to see what the dissipation should be at 10A. Maybe it is normal for it to get hot even at 3A. Some heatsink will be needed. You recall the thermal resistance comments. "Hot" for your hand could mean 55-60°C (sensation of pain), which means it actually dissipates 0.5W or so.

You should not need to make any adjustment. You simply replaced one type of SWITCH with another. If you drive it correctly, there is no problem.

To protect the gate in this circuit, use a 12V Zener diode between the gate and source (K to S, A to G) and keep resistor in series with the MOSFET's gate and output of opamp. This will slow down the MOSFEt somewhat, so it only works if the switching frequency is low, which I expect it is. Do you have a scope to check a few things, such as frequency, etc?

Many MOSFET's have Vgs of 20V max. and 10V for saturation. That is why you want to drive them with 10-12V.
 

    DjinnG

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DjinnG

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Re: MOSFET abuse

Yeah, "buzz" may be a wrong term... It does whine, and with different frequency as the load varies :)
The hysteresis cap is something I dont understand. All the reading I've done show a resistor only. Anyway, it is definitely a 68 pF.

About the adjustments - that was my question actually: will the original circuit drive a FET correctly?

My aim was to try to increase switching efficiency to reduce power dissipation by:
1. Manipulating the switching frequency by modifying the hysteresis loop though I dont know whether I should increase or decrease it. (haven't tested it yet)
2. Allowing more current to drive the gate, i.e. reduce the 10K and 1K resistors. I've tried a few combinations that retain the 10 to 1 ratio (e.g. 1K & 100ohm) - they showed no improvement or were even more problematic. I also tried a few with a 3-2 ratio - e.g. 2K & 1K - which did not drive the FET at all. (I figured that this way I could protect the FET from excessive voltage since this is a voltage divider...)

Your point about hot to the touch is well taken. I should try to get me a thermometer or something...
Alas, I do not have a scope :( I have plans to build an ADC to interface my computer...

Two things that came to mind during this writing:
1. Maybe I should just scrap the switching idea and add parallel BPT's to the regulator.
2. Is a depletion type N-ch MOSFET interchangeable with an enhancement type P-ch MOSFET? I can't seem to find any theoretic differences...
 

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Re: MOSFET abuse

OK. I have news.

I guess my assumptions about heating were "a bit" off.
I decided to use the TO-220 heatsink I bought despite the technical difficulty of placing it on a solderless breadboard (where my circuit is currently housed). The results were amazing.
Without the sink the FET would burn my finger in about 30 seconds of use. With the sink I could barely feel a difference (compared to room temprature) after 2 and 3 minutes of pumping 4A. I even touched the drain tab to make sure the FET is not getting hot without transferring it to the sink - Nadda. This is so exciting! :)
The shamless LM338 regulator gets hot, but the FET seems to handle it with ease.

First, I'd like to thank you, VVV, for all your help - I have learned a lot.
I am still interested in the answers to the questions I posted, if you've got the time.

And I've 3 more:

1. The FET's source is connected to 28V, and the Gate receives pulses of 0V. By all accounts it should be dead, shouldn't it? (I have yet to place the Zener to protect it - I've read somewhere that doing so can amplify gate drive problems)

2. The original design was for 3A. Assuming that the FET and regulator can take 10A of load, what inductor-capacitor combination would be necessary to support that current?

3. The circuit doesn't start by itself. I have to "jumpstart it" manually with a touch of the negative lead to its Gate.
My suspicion: In the original circuit there is a 2.2K resistor connecting the circuit's input with the output which is missing from my circuit. Now the notes for the circuit do mention a 4.7K(???) resistor that allows it to start... Could it be the 2.2K resistor they were referring to? I did try a 10K in place of the 2.2K and it did nothing to help the circuit start to oscillate.
 

VVV

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Re: MOSFET abuse

Hi DjinnG,

You are welcom, first of all. I am glad you sorted out the MOSFET heating issue. Like I said, the hand isn't the most reliable thermometer.
OK, so the circuit appears to work. (What about the noise?)

1. Many components can withstand more than their ratings. The same applies to your MOSFET G-S voltage. Please add the zener and the 1k resistor. Hopefully nothing will change at this low frequency of operation.

2. As long as the inductor does not saturate at the peak current (Iout + ripple), you do not need to change anything in the schematic. Just be sure nothing gets hot and the inductor does not saturate.

3. You are right, the 2.2k resistor is supposed to jump start the regulator. This may fail to happen under heavy load, however. Imagine for instance that you have adjusted the voltage to minimum, 1.2V and you have a 3A load connected, that is you have a 0.4Ω resistor at the output. The 2.2k and the 0.4Ω resistor form a divider, dividing down the 28V to 0.4/(0.4+2200)*28=5mV. This input is then divided down again by the 10k 15k divider, so you only get a few mV at the input of the opamp. If its offset is higher than that in the opposite direction, it will never start. So, you can use a 2.2- 3.3MΩ resistor connected from +28 to the - input of the opamp, but this may increase a little the voltage across the linear regulator.
Or, use the circuit in the picture. You select R1, R2 so as to get about 0.7-0.8V across R2. This will apply through the diode a voltage to jumpstart the regulator. Then, even at minimum voltage, the 15k 10k divider will provide enough voltage (0.48V) at the cathode of the the diode to turn it off. It is best to have this divider powered from a regulated voltage, such as a Zener diode, instead of the 28V.

Hope this helps,
VVV
 

    DjinnG

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tkcmd

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Re: MOSFET abuse

Im also a noob at this, but have been running a few simulations here (am not 100% sure if it corresponds with the reality)
The 2K2 "jumpstart" resistor also has an infulence on the max and minimum output that he liniair regulator can give you
If you want to be at the lower range of the regulator you should take a resistor that has an higer value (something aroud 6k)
And if you want to be on the high end you should take lower but it for some reason if you try to push the ouput to high the voltage over the regulator drops to levels that would not make it work in real life
Anyone that could give some more info about this schem ?
 

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Re: MOSFET abuse

Yes, that is true, the jumpstart 2k2 can increase the output voltage at light loads/ no load. That is because it forms a divider with the load in parallel with 25k(10k+15k). If the load is not present, the only resistance left to GND is 25k, since the linear regulator cannot sink current. In other words, at light load the output voltage is not really regulated and with heavy loads there is a chance the regulator will not start up.
So, try simulating the circuit with the two resistors and the diode and see what happens.
 

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