# Role of the capacitor in a circuit controlling a heating element

#### LordOfThunder

Joined Jun 27, 2018
80
Hello!
I am struggling to understand part of a circuit. The purpose of this circuit is to start and stop the heating of a heating element using a PIC microcontroller.
I more or less understand the role of the two MOSFETs but I do not understand why we need a capacitor (C41) there.
By simulating the circuit in time, I see that even if the GPIO pin disables the circuit the current through the heating element does not go to zero immediately but with a time constant that should be more or less consistent with tau = C41 * R39 = 1 ms.

So my main question is: what is the role of the C41 capacitor?
My secondary question is: how can we explain qualitatively the voltage at Q9_GATE?

Thank you

Last edited:

#### AlbertHall

Joined Jun 4, 2014
12,293
When the MOSFET is switched off, the 600mA constant current must flow through the 8.2V zener diode. The BZV85C8V2 has a maximum power rating of 1W but when passing 600mA the power will be nearly 5W. The zener is going to have a very short life.

#### ronsimpson

Joined Oct 7, 2019
2,797
What is the constant current source. In real life they have a maximum voltage limit, but in SPICE they do not and I see why you need a Zener.
Q9 turns on/off by the voltage from Gate to Source not Gate to Ground. Many people do not understand that.
R39 & R42 makes a voltage divider. So the voltage from Gate to Groud is 10/11 of the voltage across D1.
V=8.2 x 10/11 (most of the input voltage is across R39)
C41 causes Q9 to act slowly. You might not need that function. It charges up at C41 X R42 and discharges at C41 X R39. (t=RC)
Usually when driving a heating element the switch is turned on/off every 10 minutes. If you want to turn in on 100 times a second remove the capacitor.

#### LordOfThunder

Joined Jun 27, 2018
80
@AlbertHall The zener diode is not present in the real circuit. It is there only to simulate the behavior of the real current source. If I do not put the zener there the circuit does not work as intended.

#### ApacheKid

Joined Jan 12, 2015
1,273
Hello!
I am struggling to understand part of a circuit. The purpose of this circuit is to start and stop the heating of a heating element using a PIC microcontroller.
I more or less understand the role of the two MOSFETs but I do not understand why we need a capacitor (C41) there.
By simulating the circuit in time, I see that even if the GPIO pin disables the circuit the current through the heating element does not go to zero immediately but with a time constant that should be more or less consistent with tau = C41 * R39 = 1 ms.

So my main question is: what is the role of the C41 capacitor?
My secondary question is: how can we explain qualitatively the voltage at Q9_GATE?
View attachment 235331
View attachment 235332
Thank you
What analysis tool is being used here?

#### LordOfThunder

Joined Jun 27, 2018
80
@ronsimpson
What is the constant current source. In real life they have a maximum voltage limit, but in SPICE they do not and I see why you need a Zener.
Q9 turns on/off by the voltage from Gate to Source not Gate to Ground. Many people do not understand that.
R39 & R42 makes a voltage divider. So the voltage from Gate to Groud is 10/11 of the voltage across D1.
V=8.2 x 10/11 (most of the input voltage is across R39)
C41 causes Q9 to act slowly. You might not need that function. It charges up at C41 X R42 and discharges at C41 X R39. (t=RC)
Usually when driving a heating element the switch is turned on/off every 10 minutes. If you want to turn in on 100 times a second remove the capacitor.
Thank you for the explanation. The constant current source is a LT3741IUF#PBF
We want to turn on the heating element for 10 seconds and then turn it off.

#### LordOfThunder

Joined Jun 27, 2018
80

#### DickCappels

Joined Aug 21, 2008
10,104
If C41 exists in real life it might be there to suppress a "flyback" pulse on the input lead because of its inductance when Q9 is turned off. OR, maybe the designer just wanted to smooth the input current wavrform for some reason.

#### LordOfThunder

Joined Jun 27, 2018
80
If C41 exists in real life it might be there to suppress a "flyback" pulse on the input lead because of its inductance when Q9 is turned off. OR, maybe the designer just wanted to smooth the input current wavrform for some reason.
Yes C41 exists in real life. Just to make sure that I understand, could you please explain in further detail what is a flyback pulse?

#### DickCappels

Joined Aug 21, 2008
10,104
A long wire has inductance associated with it. When current flows in the wire it builds up a magentic field around the wire. If the wire goes open circuit the magnetic field rapidly collapses causing voltage across the wire. The collapsing field may contain a tiny bit or a whole lot of energy. Even a tiny bit of energy might cause radio interference. Larger amounts can destroy semiconductors. The capacitor keeps the voltage developed by the collapsing field from getting very large.

The capacitor also reduces the change in the incoming power supply voltage when the circuit switches on and off. It could be that this smoothing capacitor is there to prevent the switching circuit from interfering with other circuits on the same power supply.

Last edited:

#### wayneh

Joined Sep 9, 2010
17,475
The capacitor also reduces the change in the incoming power supply voltage when the circuit switches on and off. It could be that this smoothing capacitor is there to prevent the switching circuit from interfering with other circuits on the same power supply.
That's my bet, that it's there for reasons not directly shown on the schematic. It wouldn't be too risky to remove it, in my opinion, and then its purpose might be revealed.

#### ronsimpson

Joined Oct 7, 2019
2,797
The constant current source is a LT3741IUF#PBF.................. 10 seconds and then turn it off.
The LT3741 has a enable pin. Use it to turn on/off the output. I think it is TTL compatible so drive it from your computer. Forget the MOSFETs and all.

#### LordOfThunder

Joined Jun 27, 2018
80
@DickCappels thank you for the clarification. Now I understand much better
@ronsimpson I am afraid we cannot do that. The reason is that we are using one current source to drive two wires sequencially. I mean that first we heat one wire for ten seconds and then we heat another wire for ten seconds. So we need a way to switch the output of the current source selectively. This design is due to power and space restrictions.