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.

  • \$\begingroup\$ Not switching it on very hard, allowing significant voltage drop across it, could heat it up. Is T1 turning on hard (close to saturation)? \$\endgroup\$ May 18, 2015 at 21:42
  • \$\begingroup\$ 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? \$\endgroup\$
    – Kamil Szot
    May 18, 2015 at 21:45
  • \$\begingroup\$ 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). \$\endgroup\$ May 18, 2015 at 21:49
  • \$\begingroup\$ If you look at Figure 4 in the BUZ11 datasheet, your worst case of 2A @ 15V is outside the safe operating area. \$\endgroup\$ May 18, 2015 at 22:00

2 Answers 2


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.

  • \$\begingroup\$ 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. \$\endgroup\$
    – jjmilburn
    May 18, 2015 at 23:14
  • \$\begingroup\$ 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. \$\endgroup\$
    – Kamil Szot
    May 19, 2015 at 8:59
  • \$\begingroup\$ 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. \$\endgroup\$ May 19, 2015 at 12:18

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.

  • \$\begingroup\$ 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. \$\endgroup\$
    – ilkhd
    May 19, 2015 at 2:18
  • \$\begingroup\$ "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. \$\endgroup\$
    May 19, 2015 at 2:33
  • \$\begingroup\$ 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. \$\endgroup\$
    May 19, 2015 at 2:35
  • \$\begingroup\$ @ARMATAV: 1. Would you please use "@ilkhd" when replying- this way I'll get notifications. 2. upload.wikimedia.org/wikipedia/commons/thumb/1/18/… 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. \$\endgroup\$
    – ilkhd
    May 19, 2015 at 3:04
  • \$\begingroup\$ @ARMATAV electronics.stackexchange.com/questions/18885/… \$\endgroup\$
    – ilkhd
    May 19, 2015 at 3:22

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