Op-amp current protection for power supply

Thread Starter

Jon Sam

Joined Apr 6, 2016
42
Hello,

First off my english is not very good

I have a 12V 2A power supply and i use it for lot of stuffs like fans and all that, I'll put alligator clips on the cables but there is one problem :

There is no current limitation circuit in this power supply and the + and - touches sometimes making a short-circuit and i don't like that.

So i search on internet for a protection.

A LM358 Op-amp can sense current so i would like to make a protection circuit with a LM358 and a mosfet and a nor/nand gate (flip-flop)

What i would like to do :
When they will be a short-circuit the op-amp will detect a high current and will shut off, the only way to have back 12v again is to press on a switch

Can someone help me please

Thanks by advance

Regards
 

crutschow

Joined Mar 14, 2008
24,052
You don't necessarily need an op amp for the fuse.
Below is the LTspice simulation of a two transistor electronic fuse (one BJT and one MOSFET).
The simulated load is varied from 20Ω to zero and back to 20Ω to test the current limit.
The fuse triggers when the current increases to about 1.8A and the current then stays at zero until the reset switch momentarily closes at 30ms.
The P-MOSFET can be just about any that has a 20V or greater rating and an ON-resistance of 0.1Ω or less.
The PNP can be any small signal BJT with at least a 20V Vce and 100mA collector current rating.

If you have trouble with nuisance tripping on short load transient currents, you can add a small capacitor from the MOSFET gate (G) to ground. You then should also add a 10Ω resistor in series with the reset switch to limit the surge current through the switch.

The main disadvantage of this simple circuit is that it has a 0.35Ω resistor in series with the output and this will cause some voltage drop at higher currents.
If that is too much voltage drop for you then a more complex circuit with a op amp or comparator is needed.
Alternately if you can put the fuse circuit in front of the power supply regulator then the voltage drop shouldn't be a problem.

upload_2016-4-6_17-59-25.png
 

Attachments

Last edited:

Thread Starter

Jon Sam

Joined Apr 6, 2016
42
Thanks for your answer

I'm not an expert in electronics

For the voltage drop if it isn't too big it doesn't matter.

On the shematic there is :

Rld
R=abs(22-(time*1k))


What this resistor is for, i don't understand how to calculate it ?

On the down-right bottom there is :

Reset V2

What is it ? A capacitor ?

SReset and SW is the switch ?

Is there any similar part of FDS4465 ? Because it's smd part.

If i would like to change the current limit, what component will i have to change ?
Is this shematic is ok if it was a 5V power supply ?

Thanks !
 

crutschow

Joined Mar 14, 2008
24,052
........................
For the voltage drop if it isn't too big it doesn't matter.
The voltage drop is approximately Iout * R1(0.35).
On the shematic there is :
Rld
R=abs(22-(time*1k))
What this resistor is for, i don't understand how to calculate it ?
It is a dummy test load resistor. There is a note that states everything inside the dotted-line box is for "Test Only" (simulation purposes).
On the down-right bottom there is :
Reset V2
What is it ? A capacitor ?
It's a voltage source that controls the reset switch for simulation purposes.
SReset and SW is the switch ?
Yes, it represents the manual momentary Reset switch.
Is there any similar part of FDS4465 ? Because it's smd part.
As I stated, the P-MOSFET can be just about any that has a 20V or greater rating and an ON-resistance of 0.1Ω or less. You have to look at whatever electronic supplier you use to see what they sell.
If i would like to change the current limit, what component will i have to change ?
R1. The current limit is approximately 0.64V / R1.
Is this shematic is ok if it was a 5V power supply ?
The basic schematic is okay, but some of the part values may have to change, and the MOSFET would have to be a logic-level type that turns fully ON at a Vgs of -5V.
 

ErnieM

Joined Apr 24, 2011
8,011
Anything you add with a transistor in the path will make the output sloppy, more variation with load and the like. This is not really a problem with loads like fans but may be overly complicated anyway.

A nice simple circuit breaker is the way I would go.
 

Thread Starter

Jon Sam

Joined Apr 6, 2016
42
Thanks !

At first i wanted to work with a comparator and a flip-flop because i knew stuff about it and it's easy to find where i live.

As I stated, the P-MOSFET can be just about any that has a 20V or greater rating and an ON-resistance of 0.1Ω or less
Sorry, i didn't see.
An IRF540N is good ?

at a Vgs of -5V.
minus 5V ? i can't find any.

A nice simple circuit breaker is the way I would go.
What do you propose ?

Thanks again
 

crutschow

Joined Mar 14, 2008
24,052
.................
An IRF540N is good ?

minus 5V ? i can't find any.
An IRF540N is N-channel.
You need a P-Channel device which requires a negative Vgs to turn on (positive Vsg).
For 5V operation you need a logic-level device that fully turns on at a Vgs of -5V (voltage at which the Rds(on) resistance is measured as shown on the data sheet).
What electronic suppliers do you use?
 

crutschow

Joined Mar 14, 2008
24,052
So pick you favorite supplier and search for P channel MOSFETs of the desired voltage, package type, and ON resistance (at Vgs=-5V if you want to operate from a 5V supply).
 

crutschow

Joined Mar 14, 2008
24,052
The FDD4685 looks okay and will also work for a 5V supply (the data sheet shows the ON resistance for a Vgs of -4.5V) but it's a surface mount package.
Is that okay?

The FQP12P20 will not work as a switch at 5V, since it's ON resistance is only specified at a Vgs of -10V.
The FQP3P50 has too high an Rd(on) resistance of 4.9Ω.
 

crutschow

Joined Mar 14, 2008
24,052
Then where can you find Vgs = -5V ?
Data sheet, page 2, under ON CHARACTERISTICS, Static Drain−to−Source On−Resistance.
(They left out the minus sign which should be there to indicate the proper polarity.
Alternately it would be correct to say Vsg = 5V or +5V).
 
Top