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Problem using LDO as a current source.

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smarthonghong

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Heys guys I have some question using a LD29150PT LDO as a current regulator. I know I can simply put a resistor between the adj pin and the output pin to regulated the current like what I do to the LM317 LDO.

View attachment 68491

However the same trick didn't work on LD29150PT. I am pretty sure my layout is correct as is is a very simply circuit only. Can someone please explain to me what I am doing wrong?
 
You have to use the adjustable version which has 4 pins, and the ground pin has to connect to circuit ground. Otherwise, the circuit looks like the LM317 circuit.

EDIT: I was wrong. You can't use this part this way. Sorry.:eek:
 
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Yeah, 29xx series can be a real pain, they have features to shut down etc if there is anything weird.

Personally I don't want that much brains inside a 3 pin regulator, I got burned once and decided to never use the mongrel things again... ;)
 
Yeah, 29xx series can be a real pain, they have features to shut down etc if there is anything weird.

Personally I don't want that much brains inside a 3 pin regulator, I got burned once and decided to never use the mongrel things again... ;)

Not even as a voltage regulator? If so, could you please elaborate?

I was about to start using them...
 
Hello,


Yes i'd like to hear more about this too.

This chip looks like it may function in either the adjustable version or the non adjustable version. However, it would not be used exactly like the LM317 would be used in a current mode application.

[See diagram]
For the adjustable version, it might work to tie the output to the ADJ pin, then use the GROUND pin as the 'new' ADJ pin as in the LM317.
For the non adjustable version, same thing except you dont have to tie the ADJ pin to the output because there is no ADJ pin, but still use the GROUND pin in place of the LM317 ADJ pin. There will always be some error current but that's typical using a standard regulator like the 7805 in current mode.

One catch is that it is probably better to use the non inhibit pin version because it may be hard to keep the inhibit pin input voltage at 2v or higher.

It's unfortunate that the data sheet shows a partially blocked diagram rather than a true schematic. This is what i liked about National Semiconductor when they did a data sheet they often showed the whole internal structure so you could make better guesses about stuff like this. With the current data sheet link, there's no way to be sure without actually bench testing it.

So connect the ADJ pin to the output pin, connect the output to the top side of the current sense resistor, connect the GROUND pin to the bottom side of the current sense resistor, connect load to bottom side of resistor and other side of load to power supply ground, connect power supply plus terminal to input of regulator, see what happens. This being an LDO regulator it most likely uses a PNP power stage which could cause significant error current Ierror, so check that too.
 
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Not even as a voltage regulator? If so, could you please elaborate?

I was about to start using them...

I can't remember which regulator it was but it was an "automotive" 29xx series. It was a lot of years ago but there were problems with shutting down, it expected certain capacitances and certain conditions of load and supply current rise times etc. It might have been that the problems were made worse by the app being quite low-current, but the things would just stop working and needed power down and restart.

If you are going to use them then I suggest you do a lot of testing...
 
LDOs generally use PNP transistors as the main pass element, so that they can operate down to the saturation point. But this means that they are more dependent on minimum input to ground (Vadj) then a NPN based device would be. The minimum Vin for the LD29150PT is 2.5 Volts.

They could probably be used as current sources, but I doubt if it could be done with an input to output difference much less than that 2.5V, unless external circuitry were used.
 
Just thought, have you tried connecting the GND pin of the regulator to the 'ADJ' pin? Also this reg requires an output capacitance (try, I dunno 22uF), so I would put that between 'out' and adj' (across the current set resistor). I'm assuming you are using the adjustable version here, because the fixed voltage version simply will not work that way. I also noticed it has an enable line which should be connected to VIN for normal operation.

@atferrari, I've never had a problem with these regulators, I believe they do require certain things to be adhered to, such as output capacitance, and minimum load (I used an LED) . I bleieve it was the MIC2941A series LDO's.

Sorry MrAL, onlyjust noticed your schem there!
 
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Here's a simulation of a highly simplified LD29150 model, configured as a current source per MrAl's scheme. I set the beta on the pass transistor at 50, so the quiescent (GND pin) current is ≈2% of the output current, as in the datasheet typical specs.
Of course, it doesn't model stability or minimum voltage.
 
Hi again guys,

Roff (Ron):
I think you did a nice job of modeling this circuit so we can get an idea how it works. On the face of it your simulation shows that is does look like it could work similar to the way other voltage regulators work when using them in current mode. So maybe this is just the case of a newish part that we just arent that familiar with yet so we all have some doubts.

As to the minimum input voltage requirement replies:
All of these type of regulators will have some minimum input voltage requirement. In fact, every current regulator has this requirement as we can not expect to power a load that needs 5 volts with a current regulator that only has 2 volts input. We call it a "current source" anyway but it's not a true current source it's really just a current regulator.
With this in mind, we might note that using a voltage regulator chip that normally regulates at a voltage Vout we need an input that is at the very least:
Vin=Vout+Vload
and there is no getting around that (without a boost converter of course).
Add to that the regulator drop out voltage and we end up with:
VinMin=Vout+Vload+Vdo

So applying this to the 7805 regulator for example with a 10 ohm sense resistor and a 10 ohm load, we have (approximately):
Iout=Vout/Rs=0.5 amps
Vload=0.5*10=5 volts
Vout=5 volts (just the voltage of the regulator chip itself)
Vdo=2 volts (just a characteristic of the regulator chip itself)
so we end up with:
VinMin=5+5+2=12 volts minimum input voltage.

So we see in this case we needed a lot more input than we got as output, and the power dissipation is going to be a big concern too, but all these numbers are just typical when using a voltage regulator as a current regulator.
The other concern is the error current present in the ground line, which can be significant. The change over various extremes has to be looked at. This means this chip probably wont be good for doing low output currents, but then these setups rarely are good for that (op amp and transistor the better choice there).

I think the max drop out voltage for the chip we are talking about in this thread is 0.7v, but the min input voltage for ANY output is something like 2.5 volts. I would hope that for any use of this chip there was an input voltage greater than that.
 
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