Overcurrent Protection using BJT

Thread Starter

sunney

Joined Sep 3, 2015
23
Hi,

I am trying to understand a circuit which is a part of current loop and it is protected for current > 100mA.
I tried simulating it using spice model (from mfgs) on LTSPICE (attached is .asc file and .lib files)

R_Load (or sense resistor), voltage across R_Load is fed to signal conditioning circuitry for decision making (which is not included).

My question/doubts are:
1. Is MOSFET working in linear region here rather than being in cutoff once overcurrent limit reaches?
Reason: Observed in simulation results
2. I have included two circuits in .asc file, one with current source (PWL - 0 to 100mA) and another with Voltage source to understand the functionality
2a. Current source to understand the functionality as this circuit is used in current loop.
2b. voltage source to check for short to battery requirement (28V)
3. As per my understanding, once the voltage across R1 (or R3) i.e. 5.6 ohm reaches above 550mV (at ~100mA). Q1 (or Q2) should TURN ON and U1 (or U2) the MOSFET should TURN OFF, is this correct?
4. Is base resistor required for Q1 (or Q2)?
4a. If base resistor is not used, is there a chance of BJT being damaged, since no base resistor is there?
5. The way I am doing the simulation is wrong?

1631012550180.png

Thanks in Advance
 

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Last edited:

ericgibbs

Joined Jan 29, 2010
18,766
hi sunney,
This is what your asc file shows.
BTW.
You didn't include the asy MOS model, so I created a basic model asy.

Update:
In I_limit U3 is dissipating 7Watts. 2nd image


EESP_ 755 Sep. 07 12.02.png


ESP_ 757 Sep. 07 12.09.png
 
Last edited:

Ian0

Joined Aug 7, 2020
9,678
What you have is feedback, so that the circuit will act as a constant current source so that if the load demands more than 100mA, the circuit limits it to 100mA. When that happens, you are absolutely correct that it will operate in the linear region and get warm.
What do you want to happen if the current exceeds 100mA? If you want the MOSFET to switch off, you have to ask yourself when would you like it to switch back on again? Do you want a push-button switch so that it has to be manually reset? Do you want it to time out and switch back on after a fixed period of time?
 

Ian0

Joined Aug 7, 2020
9,678
And, yes, you should use a base resistor for Q1/Q2, otherwise, if there is any delay in switching off the MOSFET all the current goes through the base-emitter junction of the transistor. I also think that 470k is rather large for a gate resistor. I’d suggest 4.7k.
 

ericgibbs

Joined Jan 29, 2010
18,766
hi sonney.
It is important to note that the Current source as set by you the simulation is 100mA maximum!

If the Source could provide more current the circuit would NOT limit at 100mA

Increased to 200mA

E
ESP_ 758 Sep. 07 12.32.png
 

ericgibbs

Joined Jan 29, 2010
18,766
hi sonney,
Double checking the actual circuit, I would say there is a problem with MOS Spice model.?

The Vgs goes close to 0V , but the current is not being controlled.?
Post your version of the asy file.

E
ESP_ 760 Sep. 07 12.46.png
 

ericgibbs

Joined Jan 29, 2010
18,766
hi @sunney
Just realised that the Current source in LTS will 'force' a set current' thru the MOS.!!!

So I have replaced the current source with a Voltage source ramp, all OK.
E
ESP_ 761 Sep. 07 13.05.png
 

Thread Starter

sunney

Joined Sep 3, 2015
23
What you have is feedback, so that the circuit will act as a constant current source so that if the load demands more than 100mA, the circuit limits it to 100mA. When that happens, you are absolutely correct that it will operate in the linear region and get warm.
What do you want to happen if the current exceeds 100mA? If you want the MOSFET to switch off, you have to ask yourself when would you like it to switch back on again? Do you want a push-button switch so that it has to be manually reset? Do you want it to time out and switch back on after a fixed period of time?
Hi Ian0,
1. So when it the load demands more for what ever reason, the loop current will be limited to 100mA however the MOSFET will continue to get warm or per say it will dissipate more power in terms of heat. It is just matter of power dissipation capacity of MOSFET how much it can dissipate or how long it can sustain based on FSOA (safe operating area). correct?
2. I do not want to turn off the MOSFET instead keep it the way it is right now i.e. operate in linear region. For manual reset or switch back after fixed period of time, it needs extra circuitry. So I will leave it to continuous current limiting mode only rather than latching to turn OFF state.
3. On base resistor, as long as MOSFET is working in linear region, the V.BE across 5.6 ohm will always be ~ 560mV because of MOSFET being in linear region (or variable resistor). Unless MOSFET drain-source terminal become short and let all the current exposed to base of BJT. I am not sure whether I am making sense.
4. Also, when I check for STB condition (VSTB = 28V) and simulate it, I can see similar behavior, in this case the current become constant to 89mA and VDS kind of follow VSTB ( I am using PWL voltage source 0 28V in 50ms). However the power dissipation in both is what confusing me? I think I need to have equations for MOSFET being in linear region to understand it in more details.
 

Thread Starter

sunney

Joined Sep 3, 2015
23
hi @sunney
Just realised that the Current source in LTS will 'force' a set current' thru the MOS.!!!

So I have replaced the current source with a Voltage source ramp, all OK.
Hi ericgibbs, I was confused in that too earlier, no matter what is the overcurrent protection you have put it will force all the current from lower limit to upper.

thanks
 

DickCappels

Joined Aug 21, 2008
10,152
I usually use a small resistor in series with the base, but this is from habit. In simple circuits the current loop has always been fast enough to prevent damage to the transistor. How much faith do you want to place in the response time of the circuit?
 

ericgibbs

Joined Jan 29, 2010
18,766
hi sunney.
This is how would expect this circuit look, the Load is the 1R

Try different R_load values..

Where did you get the schematic.?
E


Added a stepped load.

ESP_ 762 Sep. 07 13.36.png

ESP_ 763 Sep. 07 13.41.png
 
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