Limit current spike upon short-circuit

[cava0035]

When the circuit goes experiences a short I notice a current spike to about 250mA, and lasts for about one second. My steady state current limit is 150mA, this is good. What can I add to my circuit to prevent the 250mA current spike and only allow to reach about 200mA?

Thanks.

 

[Hero999]

Please don't use Word for schematics.

I couldn't find a datasheet for the BTS414N you've used, is it any of the ones on the page linked below?
BTS4144B datasheet, BTS4144B datasheets, BTS4144B datenblatt, BTS4144B manual, BTS4144B data sheets, BTS4144B pdf - ALLDATASHEET.COM

 

[cava0035]

Hi Hero,

No more WORD, my mistake. The datasheet is attached.

 

[Hero999]

So you don't want the spike activating the overcurrent protection on the high side switch?

It's probably some decoupling capacitors causing the surge.

Please post the schematic for the circuit you're switching.

I like the high side switching IC by the way, especially the on chip charge pump: impressive stuff.

 

[cava0035]

Hero,

When I place a direct short to gnd, the current will spike to exactly 350mA in less than 1 second, then level-off to about 150mA. I don't want it to go above 200mA during the initial short to ground.

 

[cava0035]

Hero, I need to resend my circuit.

 

[Hero999]

If it's in Word format, copy and paste to MS Paint and save as a PNG file.

If it's a simple line drawing (only black and white) you can set the image attributes to black and white which will save space and give faster up/download speeds.

 

[cava0035]

Here it is.

 

[Hero999]

Right so it's a current sensor.

What's the load?

The load is what's causing the current surge and needs to be looked at.

Do you have a schematic for the load?

 

[JimB]

The current sensing resistor R11 is 270 ohms, you want to limit the S/C current to 200mA.
That will give a voltage of 54volts across R11 which will be dissipating 10.8 watts, is that correct?
If it is, what is U4 going to do with an input of 54 volts and a gain of 5 or so?

Something is not quite right here.

JimB

 

[Hero999]

The current sensing resistor R11 is 270 ohms

No, it's 270mΩ.

EDIT:
I didn't look at the circuit properly last night.

The BTS4141N already has current limiting built-in so the rest of the circuit is not needed.

Why do you need to limit the current?

You do know that it's not needed to protect the load as it will just draw as much current as it needs, don't you?

The purpose of a current limiting should be to protect the wiring and any semiconductors in series with the load.

 

[JimB]

No, it's 270mΩ.

Hmm... so it is!

JimB

 

[cava0035]

Hero,

Another part of this circuit you're not aware of is I'm using an ASIC.
This ASIC uses a 31Vdc bus. This 31Vdc bus powers sensors (PNP) as well as transmit/receive data.

With that said:

Case 1: the BTS4141N is using the Bus power to function, I limit the amount of current that can come from the Bus. Basically, the 150mA is what I want, but 200mA for less than one second is ok.

Case 2: I can power the BTS4141N from an auxiliary power supply. Therefore, the bus is being used to transfer data only to/from the ASIC to/from the Master. When this is the case, if the PNP sensor ever fails (shorts to gnd for example) the BTS4141N will self-limit. With this configuration I'm not concerned about the auxiliary power supply, and the Bus is not affected.

My test is: I short the output, of the BTS4141N, to ground and observe how high the current rises and for how long for case 1.

 

[Hero999]

Please post the whole circuit - we can't help without enough information.

Its possible to make a current limiting circuit using a P-channel MOSFET and a BJT but you might as well get rid of the BTS414N .

 

[cava0035]

Hero,

This is a plug in module. The circuit I provided is the focus of my question. If I power the BTS4141N via the Bus, I plug it into the assigned receptacle on the mother board. If I power the BTS4141N via the Auxiliary supply, I plug it into the assigned aux receptacle on the same mother board.

The remaining part of the design is confidential.

Thank you

 

[Hero999]

The remaining part of the design is confidential.

Then don't exepect too much help on Internet forums like this.

The quality and quantity of the advice is more often than not, proportional to the amount of information you provide.

As I said above, it's easy to make a current limiter with a P-channel MOSFET and a BJT. The idea is you use a current sense resistor and a BJT to switch off the MOSFET when the base voltage exceeds the turn on threshold, the thing is, you might as well get rid of the BTS4141 because it's easy to gate, therefore providing the same function as the BTS4142. I'll post a circuit if you're interested.