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Need help understanding this circuit.

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Jimng

New Member
I'm learning electronics and don't understand why i can pull more than 1A with this circuit. It supposed to limit the current up to 600mA according to the schematic.

Please see attached, explain how this circuit works and how to limit the current to 300mA.

Thanks!
 

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RadioRon

Well-Known Member
I'm no expert, but as far as I can see, the switching regulator circuit, comprised of everything but R3, is capable of delivering up to 2A and has no hard limiting circuit that might be tweaked. The only thing that appears to be able to limit current is R3 which is nothing more than a voltage dropping resistor. The more current that gets pulled from P2, the more voltage is lost across R3. So, if someone were to put a 10 ohm load on the output in hopes of getting 0.9 amps at 9V, R3 would consume 3V and the output voltage would be only 6V. It is a crude sort of current limiting. You can pull more than 1A from the circuit, but when you do, your output voltage is probably close to 5.6 Volts. I think you misunderstand the meaning of the text on the schematic when you say "it is supposed to limit the current to 600mA". I think the meaning of the "9V/600mA" is more along the lines of ..."the circuit is pretty much useless as a 9V supply if you try to pull more than 600mA". Even 600mA is somewhat generous, and I think that the rating of 300~500mA is more realistic.

R3 may be included to prevent destructive failure of the circuit (for example explosion or fire) when the output gets momentarily shorted out.

If you truly want current limited to 300mA then a current limiting circuit will have to be added. Why do you want to limit to 300mA? Do you want a hard limit (circuit delivers 9V at 299 mA, but not more than 300mA at any voltage)? Or a soft limit (output voltage simply begins to decline at 300mA and gets to zero at, say, 500 mA)?
 

MrAl

Well-Known Member
Most Helpful Member
Hello there,

Yes there is no true current limit on this circuit, except for the fact that if you short the output and the base circuit can put out 2.7 amps then it will do so. If the load is such that it draws the voltage down to half of 9v (4.5v) then the output current would be around 1.4 amps. That's a lot above 300ma or 600ma.

So this circuit, as mentioned previously, will not limit current in the way that you want it to. You can add a transistor maybe and get some current limiting that would be much better than what you have now. If you are interested i'll post a circuit mod.
 

ci139

Active Member
the only way to limit output current here (see device datasheet) is to select high enough inductor ( i suppose /!\ )
e.g. where the current ramp up does not have time to reach "significant" values for higher average current output at the specified output voltage

if you however can pull more than specified ?max. -- then at what OUTP terminal voltage /-waveform . . . OR there might be some overhead to enable less pulsating (more smooth) output

to strictly limit your average output current to 600mA requires sophisticated comparing circuitry for E(nergy)."from switch(inductor)" to - what our members like to call it - a "ballast capacitor ???(in DC circuit)" and E."from inductor +capacitor to external load" that acts back to FB pin-4
 

Jimng

New Member
Thank you all for your feedback and suggestion.

@MrAI: I am very interested in your circuit mod. Please share.

Thanks,
 

MrAl

Well-Known Member
Most Helpful Member
Thank you all for your feedback and suggestion.

@MrAI: I am very interested in your circuit mod. Please share.

Thanks,

Hello there,

Here is the circuit with mod's.

Since your original circuit had a resistor on the output to act as current limit, this circuit does not attempt to get super precise current limit but works good enough for most purposes. You could state your application too.

One important note for this circuit:
The input ground is NOT the same as the output ground. This has some implications, but you should state your application too.

The 2 ohm resistor shown sets the current limit. 2 ohms will give about 300ma, 1 ohm about 600ma. You can trim if you like, but usually we dont need super accurate current limit often we just want to protect the circuit.

The PNP can be 2N4403 or equivalent, and the NPN 2N4401 or equivalent.

If someone would want to do a simulation of this circuit that would be great.
 

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dr pepper

Well-Known Member
Most Helpful Member
A very sensible circuit.
You could if component count is an issue and efficiency isnt just have the current sense resistor through a diode to the feedback junction, however the 1.2v required to turn on the diode and bias the feedback junction would mean the sense resistor and therefore the system would be inefficient, burning 1.5 watts when limiting.
 

MrAl

Well-Known Member
Most Helpful Member
A very sensible circuit.
You could if component count is an issue and efficiency isnt just have the current sense resistor through a diode to the feedback junction, however the 1.2v required to turn on the diode and bias the feedback junction would mean the sense resistor and therefore the system would be inefficient, burning 1.5 watts when limiting.
Hi,

Not sure how you got that value.

If the circuit limits at 0.3 amps, the power in the 2 ohm resistor is 0.18 watts.
The transistor only takes about 0.6v to turn on.
 

dr pepper

Well-Known Member
Most Helpful Member
I was thinking 600ma, the ic wants 0.6v to regulate, to prevent the circuit from interfering with the chip a diode in series would be required, making the voltage over the sense resistor 1.2v before it started to shut off the chip, 600ma over 1.2v is about 0.75 watts, so yep dunno how I got that either, doh.
Your design might well dissipate 0.18 watts its a more efficient circuit, I was suggesting an alternative using less parts in case the o/p was struggling with the complexity.
 

MrAl

Well-Known Member
Most Helpful Member
Hi,

Oh ok, well i guess then you'd have to post the circuit so we can see what you want to do with it.
 
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