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I had a circuit on PCB. I used solder with acid-core to solder it. This combined with my mess soldering caused various shorts between components.

One of those components was MOSFET transistor BUZ11 which due to shorts heated up to the point of slight discoloring of radiator and giving a little smoke before I shut off the power.

I wonder what could cause this heat up. Current that was fed through circuit was about 2A. Well within range of this component. Voltage was 15V. There were no oscilators, no inductors, just two capacitors.

enter image description here

So the question is: How to heat up MOSFET transistor without passing through it (drain-source) more current than it can handle or applying higher voltage than it can handle.

Kamil Szot
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  • Not switching it on very hard, allowing significant voltage drop across it, could heat it up. Is T1 turning on hard (close to saturation)? –  May 18 '15 at 21:42
  • Not sure what you have in mind by "not switching it on very hard". T1 was completely open at the time, but there were various shorts in the circuit so don't look at it too closely. My question is about MOSFET transistors in general. What can cause them to heat up? – Kamil Szot May 18 '15 at 21:45
  • What Brian is referring to is that dissipated power is Voltage *times* Current, not just current. So even a small current can create considerable heat if the voltage is large. In the case of MOSFETs, this drain-source voltage depends on how 'hard' it has been turned on (and this depends on the Gate-Source voltage). – apalopohapa May 18 '15 at 21:49
  • If you look at Figure 4 in the [BUZ11 datasheet](https://www.fairchildsemi.com/datasheets/BU/BUZ11.pdf), your worst case of 2A @ 15V is outside the safe operating area. – apalopohapa May 18 '15 at 22:00

2 Answers2

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Mirroring what Brian Drummond said above, it may not be switching on 'very hard'. What this means is that the MOSFET is not entering the 'active' region of operation. In the 'active' region, you can generally expect the MOSFET to have an \$Rds\$ that matches the specified \$Rds_{on}\$ (in the case of the BUZ11, 0.040 ohms).

What is instead happening if the MOSFET is not in the active region is that it is in the 'linear' region of operation. The conditions to be in either of these two regions is:

  • linear region: \$V_{GS} > V_{th}\$ and \$V_{DS} < ( V_{GS} – V_{th} )\$
  • active mode: \$V_{GS} > V_{th}\$ and \$V_{DS} ≥ ( V_{GS} – V_{th} )\$

(The above from This SO answer)

If you are operating in the linear region instead of the active region, your resistance between drain and source is going to be greater than the specified \$Rds_{on}\$. for that part; even if your \$V_{gs}\$ is above the specified \$V_{th}\$.

They key point in this is knowing that the gate-source voltage is not enough to guarantee that the resistance of the MOSFET is going to meet the specified on-resistance on the datasheet.

If you have higher resistance than you expected, that might affect your power calculations; e.g. you might be dissipating more heat in the MOSFET than you otherwise expected.

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jjmilburn
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  • This is incorrect. Reread the answer you linked. This is not a BJT and "linear" can mean 2 different things for a MOSFET. If you look at the graph again, you'll that MOSFET can have lower resistance in the linear mode than in active and linear can be actually more desirable than active mode. – ilkhd May 19 '15 at 02:18
  • "Linear region" means a single thing for FETs. It's the part of that graph where you are operating like a variable resistor instead of a switch. In no part of that graph does it have lower resistance than in it's active region for a certain saturation. That's the drain current you CAN have at max, not the drain current you are passing. It's getting worse, not better, as you go back into the linear region. That's why they heat when you switch them super slow at middling power levels. – ARMATAV May 19 '15 at 02:33
  • Figure 5 and 7. Show me where that resistance gets better as you go DOWN in Vgs. Active and linear are used for totally separate things. – ARMATAV May 19 '15 at 02:35
  • @ARMATAV: 1. Would you please use "@ilkhd" when replying- this way I'll get notifications. 2. http://upload.wikimedia.org/wikipedia/commons/thumb/1/18/IvsV_mosfet.svg/400px-IvsV_mosfet.svg.png When you increase Vgs, you move _out_ of active region. If you have Vds 1v and Vgs-Vth 1v you are in the active mode but you are able to draw only 5 units of current; now set Vgs-Vth 7v, Vds 1v; we are up linear mode, and Id now 15 units. Counterintuitive? Yes. Also reread the definition above for linear and active mode if you think I am mistaken. – ilkhd May 19 '15 at 03:04
  • @ARMATAV http://electronics.stackexchange.com/questions/18885/mosfet-as-a-switch-when-is-it-in-saturation – ilkhd May 19 '15 at 03:22
  • Is there something I could add to clarify the answer? I do think that the 'linear' definition as I have described it above is correct; perhaps it would be clearer if I called it the 'triode' region (same thing)? @ilkhd When you increase Vgs, I think you only move out of active region *if* in so doing your Vds is less than Vgs - Vth, which is part specific. – jjmilburn May 19 '15 at 17:40
  • @jjmilburn "I think you only move out of active region if in so doing your Vds is less than Vgs - Vth, which is part specific." - It is absolutely not part specific, it is circuit specific; if you have with same part Vgs - Vth above the Vds (which is usually very low anyway in switch circuits, where the load is between drain and power rail) you are in the linear mode; this is in vast majority of use cases. This is why MOSFET driver ICs exists at the first place - apply as much Vgs as possible to open MOSFET as hard as it can. – ilkhd May 20 '15 at 03:09
  • @jjmilburn "Active mode", when applied to MOSFET means that MOSFET acts as Voltage Controlled Current Source; very similar to a BJT in unsaturated mode, a Current Controlled Current Source - AFAIK vacuum triode works differently. Irrespectively of the particular circuit, if you are using MOSFET as a switch you want to be in the "MOSFET-linear" mode. I know, it sound like a bad joke but MOSFET theory uses words opposite way. – ilkhd May 20 '15 at 03:20
  • @illhd Vth is part specific, I should have said. I thought it would be obvious that the circuit conditions driving the gate are circuit specific; it seems I am unable to update my comment to clarify though. Regarding the 'triode' region -- in the linear mode, it is in the 'triode' region. They are synonymous, see slide 6 here - http://users.ece.gatech.edu/~alan/ECE3040/Lectures/Lecture24-MOS%20Transistors.pdf Agreed that a vacuum triode is a different device; I have no experience with those. For MOSFETS: Linear=Triode; Variable Resistance. But Saturation=Active; use as a switch – jjmilburn May 20 '15 at 16:44
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If you look at the data sheet https://www.fairchildsemi.com/datasheets/BU/BUZ11.pdf you'll see the thermal resistance to ambient, which for a non-heatsunk TO-220 (such as you are using) is in the vicinity of 75 deg/W, if the transistor is dissipating 2 watts the chip will get to ~175 C (2 x 75 + 25 ambient), which is more than enough to do damage. If you were pulling 2 amps, a 1-volt drop on the transistor will produce this much power. If you look at the Output Characteristics (fig 5) this suggests a Vgs on the order of 3 volts. As ijmilburn has pointed out, you need more than this to turn the FET fully on.

WhatRoughBeast
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  • Great clarification on the need to consider package type for thermal resistance -- can make a big difference particularly when switching 'the same part' from a larger footprint to a smaller one. – jjmilburn May 18 '15 at 23:14
  • So basically I was passing 2A drain-source, at the same time having gate-source voltage at round 3V which opened the transistor but just barely, so the voltage drop across it was 1V. – Kamil Szot May 19 '15 at 08:59
  • You've got the idea, but the numbers are not firm. The voltage, combined with the current, provides enough power to the FET very hot, and would explain what you saw. It might be a bit more or a bit less, though. I was just using ballpark numbers. – WhatRoughBeast May 19 '15 at 12:18