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LDO Voltage Regulator Current Source Not Working As Expected.

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mattishere

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So everyone know famous LM317 can be configured as Current Regulator like this:

YvyY3.png


However the dropout voltage of LM317 is pretty high and so I am looking for some LDO chip to do the same thing. Here I got some RT9018 chip to do this:

https://www.richtek.com/Products/Linear Regulator/Single Output Linear Regulator/RT9018ART9018B.aspx

I have successfully configured RT9018 as Voltage Regulator like what it supposed to be. However I can't get it working as a current regulator. I connect the load exactly like what I do for LM317 but it just don't working. Is it not all the voltage regulator can regulate current or is there something I am missing? Thanks for helping.
 
You are missing the point that when you use the LM317 as a high-side current regulator, its ADJ pin is elevated above 0V (Gnd) by the voltage drop across the load resistance (E=I*Rl). The 9018 has other circuitry and inputs that must always be referenced to Gnd, so cannot arbitrarily be floated to the voltage across the load like the three-pin 317 can.

Also the minimum Vdd to Gnd pin voltage is 3V, so even if you could elevate its Gnd pin, the effective minimum DropOut voltage would be higher than a LM317 (3V vs about 2.5V).

You could use it as a low-side current regulator, however, by putting the load Rl in the Vin lead. The current would be V0/Rc, where Rc is tied between Vout and Gnd. This could be used to regulate the current through a 1W LED, for example.
 
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You are missing the point that when you use the LM317 as a high-side current regulator, its ADJ pin is elevated above 0V (Gnd) by the voltage drop across the load resistance (E=I*Rl). The 9018 has other circuitry and inputs that must always be referenced to Gnd, so cannot arbitrarily be floated to the voltage across the load like the three-pin 317 can.

Also the minimum Vdd to Gnd pin voltage is 3V, so even if you could elevate its Gnd pin, the effective minimum DropOut voltage would be higher than a LM317 (3V vs about 2.5V).

You could use it as a low-side current regulator, however, by putting the load Rl in the Vin lead. The current would be V0/Rc, where Rc is tied between Vout and Gnd. This could be used to regulate the current through a 1W LED, for example.

Thanks Mike,

It make sense. I think I have to look at another chip to do the job. However how can I know whether the chip is high-side current regulator or low-side one? Can it be determined by looking at the block diagram?
 
The regulator
...However how can I know whether the chip is high-side current regulator or low-side one? Can it be determined by looking at the block diagram?
The chip doesn't know. If the load is connected between Vcc and Vin on the regulator, the regulator acts as a low-side current sink. If the load is connected between Adj on the regulator and Gnd, the regulator acts as a high-side current-source.
 
The regulator

The chip doesn't know. If the load is connected between Vcc and Vin on the regulator, the regulator acts as a low-side current sink. If the load is connected between Adj on the regulator and Gnd, the regulator acts as a high-side current-source.

The problem is. I am choosing a LDO which can be run as high-side current-source. The date sheet usually don't showing a load but two resistor as they are supposed to be a voltage regulator. How can I know whether it can be a high side or low side one?
 
Here is the simpler '317 used as either a current source (high-side) or a current sink (low-side). The current through both LEDs vs the input voltage V(IN) is identical. Not a particularly good constant-current LED driver because the input has to get to 7.3V before either circuit regulates...

c317.gif
 
Here is the simpler '317 used as either a current source (high-side) or a current sink (low-side). The current through both LEDs vs the input voltage V(IN) is identical. Not a particularly good constant-current LED driver because the input has to get to 7.3V before either circuit regulates...

View attachment 96331

Thanks for your example of the LM317. However I a not going to use LM317 due to its huge dropout that's why I am looking for LDO chip like RT9018. I need to know how to determine whether the chip can be run as a high side current source. Can it be determined by looking at the block diagram? Or looking at other specification? If so how?

Thanks again.
 
Thanks for your example of the LM317. However I a not going to use LM317 due to its huge dropout that's why I am looking for LDO chip like RT9018. I need to know how to determine whether the chip can be run as a high side current source. Can it be determined by looking at the block diagram? Or looking at other specification? If so how?

Thanks again.

You obviously missed the point of the posting. It wasn't to show how good or bad the LM317 is a current-regulator, it was to show that it could either be used as a current source, or a current sink.

I already told you that the RT9018 can not be used as a current source in post #2

Why are you hung up on a current_source? Wouldn't a sink do as well?. There is no reason that the RT9018 cant be use as a current sink.
 
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You obviously missed the point of the posting. It wasn't to show how good or bad the LM317 is a current-regulator, it was to show that it could either be used as a current source, or a current sink.

I already told you that the RT9018 can not be used as a current source in post #2

Why are you hung up on a current_source? Wouldn't a sink do as well?. There is no reason that the RT9018 cant be use as a current sink.

Hello the reason that I can't use RT9018 as a current sink because I need the load directly connected to the GND due to its nature. That's why I need a high side current source. Thanks
 
the RT9018 can not be used as a current source in post #2
The "ADJ" pin functions very differently! The RT9018 ADJ pin has a voltage that is 0.8 volts above ground, while the LM317 ADJ pin is 1.25 volts below the output.
Read about the VDD pin.
 
Here is a high-side current-source that has Vdo=V(in)-V(load) of ~0.8V for a range of load currents.


326.gif
 
Rather than explain a bad choice, define what you need.
E.g.
100mV drop 1A current source or sink (hi or lo)
Max (Vin-Vo)*Io=Pd
 
Hi Mattishere,

Welcome to ETO

You are making a rod for your own back by trying to make an LDO constant current generator from an LDO voltage regulator because, by definition, you have the VRef plus the DOV in series with the input supply rail.

There is a proper way to do the job. The schematic below shows a rough and ready constant current generator of around 1A with a dropout around 1V. This illustrates the principle.

The proposed constant current generator can be configured as:

(1) High side +Ik
(2) Low side +Ik
(3) High side -Ik
(4) Low side -Ik

If you post the following data, I in return, will post a simple circuit, using a few cheap and readily available components, that will meet any reasonable requirements

(1) Input voltage range
(2) Value of constant current
(3) Accuracy of constant current
(4) Definition of load
(5) Maximum acceptable dropout

spec

ETO_LDO_REG_CONSTANT_CURRENT REGULATOR.png
 
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