# Why is this switching transistor heating up?

I am building control board for fan cooling.
It runs from 12V power source, and it is controlled by input analog signal 0.3V to 1.2V which just controls speed of fan.
Problem is that switching transistor Q2 gets hot.
I tried to use opamp in circuit and then comparator. It takes longer time to heat up with comparator, but it gets hot as well.
I switched from opamp to comparator to minimize switching losses in mosfet.

How could I minimize heat dissipation in that transistor?

• At what frequency is it operating? Maybe try a Schmitt-Trigger instead of the comparator. But only with a small hysteresis.
– Ben
Jan 1, 2017 at 13:56
• I runs at 50kHz (input control signal 0.3V) up to 100kHz (input control signal 1.1V). Could you explain what do you mean by replacing comparator by schmitt-trigger? Jan 1, 2017 at 14:08
• @Chupacabras -- try whacking a 1meg resistor from the output to the positive input of the comparator to add a bit of hysteresis. Jan 1, 2017 at 14:42
• I tried many modifications including this one. It lowered switching frequency, but mosfet Q2 was heating too. Jan 1, 2017 at 14:58

This does not work as a switch. Instead, forms a current source or a linear voltage regulator (You can explain how by yourself, but detailed explanation is below). According to the notes on the schematic, the voltage across D and S of Q2 will be 12-3.0 = 9V at worst. If you multiply this with load (fan) current then you'll find the power dissipated ($P_D$) by Q2. Multiply $P_D$ with $R_{th j-a}$ of AO3401 which is given as min. 100 in the datasheet and you'll find temperature rise. This may explain the excessive heat. You can verify this by applying maximum control input voltage (1.2V) and seeing that Q2 does not heat up.

Now let me explain how this works as a linear regulator (According to the schematic in your question):

1) At the time of energizing the circuit, (assuming control input is 0) the output of comparator will be 0 due to pull-down resistor (R2). So, comparator output is low --> Q1 is off --> Q2 is off --> No load current/voltage --> Voltage across R2 is zero --> Output remains low.

2) When the control voltage is applied, comparator will attempt increasing its output to 12V. When this output voltage approaches/reaches Vbe threshold of Q1 (neglecting 100R tied to emitter), CE resistance of Q1 starts to decrease. Thus, G-S voltage (so, DS resistance) of Q2 starts to decrease, leading load current (so, load voltage) to increase.

3) This load voltage is divided by 1+(R8+R3)/R2=1+90k/10k=10 and fed back to negative input of comparator ($V_{in-} = V_L/10$). When this FB voltage (i.e. voltage across R2) reaches and exceeds the voltage on the positive input terminal (i.e. control voltage), the comparator attempts decreasing its output to 0.

4) Comparator output starts to decrease, so Vbe of Q1 starts to decrease, leading CE resistance to increase and forces Q2 to increase its DS resistance. This results in decreasing load current (and so, load voltage). This voltage is divided by 10 and fed back to negative input of comparator ($V_{in-} = V_L/10$).

5) Now the voltage at negative input is lower than positive input, so the comparator will attempt increasing its output to 12V. Output start to increase and the cycle begins afresh from (2).

Consequently, the voltage across the load will be 10 times control voltage: $V_L = V_{ctrl} \cdot 10$ and the voltage across MOSFET is $V_{DS} = 12V - V_L$. We don't have any information about your load, so it's quite hard to guess how much the load current is. Anyway, the power dissipated by MOSFET will be $P_D = V_{DS} \cdot I_L$.

I made a simulation on Proteus 7. You can download from here and here's a screen shot:

(I used LMV393 because LM393 is not defined in Proteus, but LMV393 is the low-voltage version of LM393).

Let's assume your fan current $I_L=50mA$ @ $V_L=5V$. So MOSFET's power dissipation will be $P_D = 7V \cdot 0.05 = 0.35W$. Multiplying this with $R_{th(j-a)}=100$ will give a temperature rise of $\Delta T = 100°C/W \cdot 0.35W = 35°C$. Assuming ambient temperature is 24°C, so MOSFET's final temperature will be 24+35=59°C.

Hope this explanation is enough and useful for you.

• Jan 1, 2017 at 10:38
• Explaining how the originals poster''s switch is not a switch is well within scope. Jan 2, 2017 at 16:13
• @ScottSeidman My explanation about that was in the comments until Dave has edited this answer. Looks like he has deleted all. There's no point in re-posting. Jan 2, 2017 at 16:57
• @Rohat, I do not see explanation why it does not behave like a switch. Can you explain that? Because that is what I am asking. Jan 3, 2017 at 20:05
• You are right. I returned to this after some time. I modified the circuit to simple buck convertor. My first thoughts about my original circuit were not correct and naive. Dec 25, 2017 at 7:53