Hi all,

I'm currently developing a project where I need to limit the current delivered to the load at 350mA, with a supply voltage of 4.2V.

I've made some researches on commercial products and self-made circuits, but all of them consume around 500mW power.

As I'm trying to do a portable low-consumption solution, I would need a current limiting circuit that would consume a maximum of 100mA.

Does anyone have any idea?
Thanks in advance.

 

hi,
As it will most likely be a series CL source, to achieve 350mA thru a series load would mean drawing at least 350mA from the 4.2V source.???

If the load is CL to say, 100mA is there any requirement on what the voltage must be across the load at 100mA.?

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Eric " Good enough is Perfect "
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Thanks for your reply.

I'll try to explain what I intend to do.

I have a 4.2V power supply, that should deliver 350mA to a Rload.
The load value is set to be 12Ω.

My problem is that there can be short circuits on my load system which will decrease resistance and therefore increase the current.

My goal is to prevent the current increase when these sc happen.

Unfortunately all the circuits I found from my researches to limit the current have too high power consumption, which makes my project not viable.

With these values I currently have P=1.5W. I was hoping you could help me find a current limiting circuit that would increase this value as less as possible.

Thanks

 

What is "too high" power? What is the highest power that can be tolerated? What are the other solutions you've researched? What happens when the output is shorted? Why must it be current limited?

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You don't need a quadraphonic Blaupunkt -- you need a curve
ball.

 

Hi there,

- to high power is something greater than 100mW;
- so this is my limit power goal for the limit current circuit, since it will be added to the rest of the system.
- when the output is shorted the resistance just falls.

The load is a heating device that increases heat with current, that's the reason why I have to limit it to 350mA.

 

A 0.5 ohm resistor in series with your load will consume only about 60mW. It's voltage at 350mA will be 0.175V. You have 40mW headroom for an amplifier to produce a voltage from this signal sufficient to control your output device. That seems doable to me. You'll have to increase your supply voltage to make up for the 0.175V across the series resistor.

What do you think, Eric?

__________________
You don't need a quadraphonic Blaupunkt -- you need a curve
ball.

 

I'm concerned that the ampop may consume some power...

 

Find a low power opamp or comparator.

__________________
You don't need a quadraphonic Blaupunkt -- you need a curve
ball.

 

Does one end of your load need to be connected to supply- (gnd?), or can that path be broken to insert a current measuring resistor (shunt)?

Would a shunt have to be inserted into the high side?

__________________
Mike ML.

 

Yes I could put a current measuring resistor after the load...


Thanks you guys... I'm going to search for some opamps and cmr and try to insert it in the circuit.
The prototype may take a while to develop, but I'll post here the result... or maybe some more doubts.

Many thanks to you all.

 

Ok, you have options. First, if you put a low-value (few 10s of mΩ) in series with the load (shunt), you then have to compare the differential drop across the shunt to a threshold, and if a limit is exceeded, shut down the regulator...

Many ways to do this, including using my favorite "high-side-current-monitor" IC called a ZXCT1009. Look it up, and maybe it will give you some ideas. Low-side monitoring could be done with an opamp whose input common-mode range includes ground, like an LM358. The power use by the opamp will be tiny compared to your load.

__________________
Mike ML.

 

Let me start by saying I work at Maxim. My suggestion for this is to get a current limited switch. These are little chips that are designed to protect ports on PCs. They are meant to work from 5V but most will do fine at 4.2V and less. Here is a link to a list of parts.

Maxim - Parametric Search - Product Table

It looks like none are exactly 350mA, but some are adjustable. Doing current limiting with no headroom is a PIA so it's a place where an IC really makes sense. Also, don't tell anyone I said this ;-) but Maxim is pretty liberal with free samples, or there is always Digikey.