current limiter for a 48v supply (50ma)

[benbiles]

current limiter for 48v supply ( 50ma )

Hi I made this power DC - DC convertor form this design,



Its maxamum power output is supposed to be 50ma but it seams to be able to supply more and start to
cook some of the components if you draw more !! no current limiter I assume ?

So I was thinking I could add a current limiter? does anyone know a good way of doing this whilst keeping noise to an absolute minimum ? Or another quiet 48v power supply IC that would be good for phantom power?

I'm after around 250ma - 300ma from a 9 - 16v supply but dont want the thing to catch fire if someone shorts an XLR cable !!!!

Also, In cicuit above where would you add a protection fuse? on the output ?

ben

 

[KlausST]

Re: current limiter for 48v supply ( 50ma )

Hi,

What precision do you need.

For low precision a bjt ant two resistors are sufficient.
* use a shunt (550mV at 50mA)
* pnp. Emitter to left side of shunt, base to right side of shunt (a current limiting resistor in the base line is recommended for fast changing load current), collector with series resistor to feedback pin. Calculate the resistor according regulator datasheet to regulate voltage down to a special value.

This circuit has limitations. The limiting current is temperature dependent.
It does not work down to 0V, so it is no short circuit protection.

For better voltage regulation (not during current limiting) you should connect your regular feedback resistor at the right side of the shunt.

To improve output noise you could use a RC or LC low pass filter.

Klaus

 

[benbiles]

Re: current limiter for 48v supply ( 50ma )

Hi Klaus,

I don't think its massive precision I need, more curcuit protection I suppose..

I'm surprised the IC i'm using does'nt have in built in current limiter since it says maximum output is 50ma
maybe I'm interpreting that incorrectly. maybe it should say maximum current before it goes into overdrive and self distructs !! Think I'm going to check the data sheet again.

the opperatiing temp range I need is -50c to 50c ( approx )
so far all other componants can do -40c. It will actually be used in these tempretures too!

Maybe its just a fuse I need. I don't want to introduce any ripple or noise since its phantom power , driving condesor mics etc..

what does the shunt do? never used one..

Ben

Hi,

What precision do you need.

For low precision a bjt ant two resistors are sufficient.
* use a shunt (550mV at 50mA)
* pnp. Emitter to left side of shunt, base to right side of shunt (a current limiting resistor in the base line is recommended for fast changing load current), collector with series resistor to feedback pin. Calculate the resistor according regulator datasheet to regulate voltage down to a special value.

This circuit has limitations. The limiting current is temperature dependent.
It does not work down to 0V, so it is no short circuit protection.

For better voltage regulation (not during current limiting) you should connect your regular feedback resistor at the right side of the shunt.

To improve output noise you could use a RC or LC low pass filter.

Klaus

 

[KlausST]

Re: current limiter for 48v supply ( 50ma )

Hi,

I didn´t read datasheet, but I´m ver sure the IC has a built in current limiter.
The limiter is to safe the switch from damage and to prevent the coil from saturating.

Since this is switch current, the output current depends on output voltage.
It says 50mA, because the circuit ensures to deliver 50mA. But it is not a true output current limit.

Klaus

- - - Updated - - -

Hi again,

what does the shunt do? never used one..

A "shunt" is just another word for a resistor in series with the load. It´s for current measurement.

Klaus

 

[benbiles]

Re: current limiter for 48v supply ( 50ma )

Hi, I put a 10k resistor in series with this meter but don't know how to read it ?

I selected a 10ma scale & x10 Ohms as in the pictures.

what is the power output at 48v with these readings and the 10k resistor , not sure how to calculate. I know ohms law, but not very good at readng this rubbbish meter !!

I think I must have just shorted the output by mistake and thought it was driving too much power..
the first resistor I tried got really hot, must have been a .25 watt or something.

Also I was wandering if two Schottky diodes in series would increase there voltage rating? or achieve nothing ?

The diode is rated to low at 40v !! so it gets hot under load ..

Hi,

I didn´t read datasheet, but I´m ver sure the IC has a built in current limiter.
The limiter is to safe the switch from damage and to prevent the coil from saturating.

Since this is switch current, the output current depends on output voltage.
It says 50mA, because the circuit ensures to deliver 50mA. But it is not a true output current limit.

Klaus

- - - Updated - - -

Hi again,


A "shunt" is just another word for a resistor in series with the load. It´s for current measurement.

Klaus

 

[KlausST]

Re: current limiter for 48v supply ( 50ma )

Hi,

to measure a resistor´s value when it is in a circuit gives no valid results.
Please read the instruction manual of your meter on how to measure resistors, voltage and currents.

***

what is the power output at 48v with these readings and the 10k resistor , not sure how to calculate. I know ohms law, but not very good at readng this rubbbish meter !!

You can´t measure power with a meter with "ohms setup"
With your meter you can´t measure power at all.

The diode is rated to low at 40v !! so it gets hot under load ..

Use an appropriate diode. It most probably gets hot because of the current.

***
I don´t know what you are trying to do with your meter at all.

Klaus

 

[benbiles]

Re: current limiter for 48v supply ( 50ma )

ok, i see, that explains it then ! what would I need to mesure maximum power output of a power supply?

do you think its better to have some kind of a trip on a timer for short cuircit protection ? would still need something that can mesure the load .. maybe the fuse is the easiest answer here, but its likely this power supply will enctcounter short cicuits !

Yes, already ordered diode rated at 100v, thought it might be possible to double them up as a temprary fix, guess not.

Hi,

to measure a resistor´s value when it is in a circuit gives no valid results.
Please read the instruction manual of your meter on how to measure resistors, voltage and currents.

***

You can´t measure power with a meter with "ohms setup"
With your meter you can´t measure power at all.


Use an appropriate diode. It most probably gets hot because of the current.

***
I don´t know what you are trying to do with your meter at all.

Klaus

 

[KlausST]

Re: current limiter for 48v supply ( 50ma )

Hi,

ok, i see, that explains it then ! what would I need to mesure maximum power output of a power supply?

You can´t measure it. You will find the information in the datasheet or printed on the supply.

do you think its better to have some kind of a trip on a timer for short cuircit protection ? would still need something that can mesure the load .. maybe the fuse is the easiest answer here, but its likely this power supply will enctcounter short cicuits !

A current limit (like I described above) is a good way, as long as nothing is overheating.
A fuse, maybe a resettable "polyfuse" is also a good solution.

Klaus

 

[benbiles]

Re: current limiter for 48v supply ( 50ma )

ok, i'll keep it simple and add a 50v 50ma resettabe polyfuse on the output.
hope they self reset!

ben

Hi,

You can´t measure it. You will find the information in the datasheet or printed on the supply.


A current limit (like I described above) is a good way, as long as nothing is overheating.
A fuse, maybe a resettable "polyfuse" is also a good solution.

Klaus

 

[DWard]

Re: current limiter for 48v supply ( 50ma )

ok, i'll keep it simple and add a 50v 50ma resettabe polyfuse on the output.
hope they self reset!

ben

Polyfuse will reset by removing the power. Once it cools off it will reset. But it will not simply reset itself when the current drops below the trigger threshold - it will remain in a high resistance state until the power is removed and it cools down.