# How to control a FET by (preferably zener diode and BJT).

Discussion in 'General Electronics Chat' started by nikee, Dec 1, 2014.

1. ### nikee Thread Starter New Member

Jan 17, 2014
15
1
I want to supply a constant current of 1A to a variable resistive load. I have made my constant current source with LM317 and its working as expected. Now I want to put a FET in line and somehow break the circuit through FET if the supply voltage from constant current source exceeds about 4.7V (some sort of safety feature). I think it can be achieved by a zener and transistor but can’t put up a circuit. Can someone help me?

I don’t want you to draw complete circuit just give me an idea how to achieve this and I hope I will be able to draw the circuit.

Regards

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2. ### crutschow Expert

Mar 14, 2008
13,014
3,234
What does "break the circuit" mean? Do you want the FET to just limit the voltage or do you want the FET to totally shut the circuit down when the voltage exceeds 4.7V (and wait for a reset signal)?

3. ### MikeML AAC Fanatic!

Oct 2, 2009
5,450
1,066
To expand on Zapper's question: What happens when the voltage (presumably on the load side of the FeT; not the input side of the FET, as you drew it) exceeds 4.7V?

Does the FET then act like a voltage regulator, keeping the voltage across the load constant? If so you will have to heatsink the FET because it will be dissipating a substantial amount of power if the source of the FET is at 18V and the drain is at 4.7V. (18-4.7)*1 = 13.3W

Does the FET act like a switch, so that once the load voltage exceeds 4.7V, the FET snaps off, waiting for something to reset it for another cycle. This way, the FET doesn't dissipate more than a few hundred mW.

4. ### nikee Thread Starter New Member

Jan 17, 2014
15
1
@crutschow and MikeML: Break the circuit means FET is completely off. Actually the load is never expected to go beyond 4.5V and I am trying to give a fail-safe to never exceed this voltage. So the FET will act like a switch. I want to make the circuit such that if voltage on load side exceeds 4.7V (which means something is wrong in my load side), the FET should snap off and wait for me to restart the circuit(and a LED should start glowing to give me the failure signal), though it is also acceptable if the FET turns off at 4.7V but again gets on if somehow voltage return back to below 4.7V.

5. ### crutschow Expert

Mar 14, 2008
13,014
3,234
A P-MOSFET won't conduct down to 0V Vgs.
Do you have an 18V source available for biasing?

6. ### nikee Thread Starter New Member

Jan 17, 2014
15
1
Oh, its a n channel irfz48n mosfet. I draw it in microsoft word and messed up the pin. My apologies.

7. ### crutschow Expert

Mar 14, 2008
13,014
3,234
You didn't answer my question about what bias voltage you may have available.

8. ### nikee Thread Starter New Member

Jan 17, 2014
15
1
yes I have 18V available (i will tap it before the LM317). It may go upto about 20V not more than that.

9. ### crutschow Expert

Mar 14, 2008
13,014
3,234
Below is the simulation of a circuit that should do what you want. It uses two transistors connected to form a latch (an SCR could also be used) to shut off the MOSFET when the output voltage goes too high.
For simulation purposes I ramped the constant current source up to 1A and back. As you can see the output is shut off when the voltage reaches about 4.7V and the LED goes on (DLED current goes to 14mA). The circuit doesn't conduct again until the reset push-button is momentarily pressed.

Diode Dtest is there only for simulation purposes and is not needed in the actual circuit.

Edit: Below is a simple circuit that just limits the output voltage of the LM317 constant-current source. The simulation is shown for load resistances of 1 ohm (green traces) and 10 ohms (blue traces). As you can see the 10 ohm output is limited to about 4.7V. Thus for any load resistance greater than 4.7 ohms, the output voltage will be limited.

Edit2: As a third choice here's a circuit using two LM317s, U1 as the current limit, and U2 as the voltage limit. The plots show the output voltage and current versus load resistance.

Last edited: Dec 2, 2014
10. ### ian field Distinguished Member

Oct 27, 2012
4,415
784
Your best bet for a fail-safe on a current regulator, is either a set voltage shunt regulator or a crowbar protection circuit - depending on your exact requirements.

11. ### nikee Thread Starter New Member

Jan 17, 2014
15
1
@crutschow: Circuit 2 is exactly what I was looking for. Circuit 3 is also looking interesting. It seems to be a constant current constant voltage (CC-CV). Can I use it like a CC-CV battery charger type application?
Really really thank you for putting up the circuit for me. hope it will be helpful for other members also.

12. ### crutschow Expert

Mar 14, 2008
13,014
3,234
Both Circuit 2 and Circuit 3 are CC-CV circuits and certainly can be used for battery charging.

13. ### nikee Thread Starter New Member

Jan 17, 2014
15
1
@crustschow: thank you for your generous help. i will be putting up the circuit by tomorrow night. Will report the working in here.
@ ian: I am happy with the crustschow circuit so I will not explore other options, but still really thank you for your help and time. Much appreciated.

Regards.