Can a LM723 Be Used To Regulate Current Between .25ma And 1ma?

[VicMac]

I want to use a 40 volt battery to supply current to perform electrolysis. The circuit uses 5 NiMh 9 volt rechargeable batteries in series and a current regulating diode (I bought 4 X Siliconix J500 0.24mA current regulator diode from a popular auction site). I have looked at the LM723 datasheet, but I'm unable to decipher it well enough to be able to tell first, if it can be used this way, and secondly, what the circuit would look like.

Would this be a good idea? Should I just buy more diodes of the .5ma , .75ma, and 1ma that I want to experiment with? Each lot of 4 current regulating diodes is costing me about $15, so if I could learn to use the LM723 (less than $1.00 each delivered) instead it would be way easier on my wallet.

Thanks for your help and time!

 

[Boncuk]

Hi,

you can't use an LM723 since your supply voltage (5 * 9V = 45V) exceeds the maximum continuous input voltage of 40V.

If you reduce the maximum supply voltage to 36V (4 * 9V) you might get down to that current using an LM317L wired as constant current source.

It can put out a minimum current of 1mA. At a higher input voltage than 9V that low output current may not be reached.

You might be lucky getting the output current down to 250µA adding a 5K resistor in series with the 1.25K resistor - being well beyond the chip's regulation curve.

Boncuk

 

[VicMac]

Regulate Current Under 1ma

Hi Boncuk,

Thanks for your reply. In researching my problem today, I found a diagram that

suggested using an LM317T with a 27 volt input and a 125 Ω resistor between

Output and Adjust, but they allow a higher current than I want to use. The resistor

you suggest is 10 times greater.

How large a wattage would the 1200 Ω resistor have to be?

Could a .1w trim pot (maybe 2k) be used for the resistor between Output and Adjust?

Thanks for your help!

Victor

 

[Boncuk]

Hi Victor,

you can't use the LM317T for your project.

It's minimum current output current is 10mA!

As you can see from the simulation the output voltage is approximately 1mV at a load resistance of 1Ω. The feedback resistor always causes a voltage drop of Vref (nominal 1.25V). It's power dissipation is 1.25V*0.001A which equals 1.25mW. Use any standard 1/4W resistor.

Of course the value of the resistor in my example must be 10 times higher than using the LM317T at an output voltage of 10mA. The current of 1mA was given in your original thread. Using a feedback resistor of 125Ω the LM317L also supplies 10mA of constant current.

To determine the feedback resistance the formula is: Rref(Ω)=Vref(V)/Iout(A), hence Rref=1.25V/0.001A, Rref=1.250Ω

Here is another example with the load resistor being 1KΩ.


Boncuk

 

[colin55]

Just use a transistor in a constant-current arrangement.

Here is a simple circuit:
**broken link removed**

Take 4k7 to positive rail.
For 1mA R = 680R
For 0.25mA R = 2k7
Use a pot with a stopper resistor.
The LEDs are the load
Use any voltage up to that allowed by the transistor.

 

[carbonzit]

Just use a transistor in a constant-current arrangement.

Here is a simple circuit:

[snip]

Interesting li'l circuit.

So generalizing things, is this what one would use (or something like it)?

**broken link removed**

Couple questions: what are the base diodes for? and why two? Without any other connections, where does the base derive its bias?

What's the usable range of currents with this, approximately?

Substitute practically any small-signal NPN? (We don't do that BCxxx thang here in the U.S. Give us good old 2Nxxxx any day!)

Thanks for your contribution.

 

[colin55]

Take 4k7 to positive rail.

As the voltage on the posistive rail increases, the max current will be less because the limiting factor will be the milliwatts dissipated by the transistor. Max for a small transistor will be about 200mW.

 

[carbonzit]

Like this? (I updated the picture in my previous reply so as not to clutter the thread.)

Still wondering why the two diode drops?

I'm going to try this shortly, will report results.

Thanks again.

 

[ModemHead]

Still wondering why the two diode drops?

The two diode drops create a 1.4V reference at the base. A 0.7V base-emitter drop leaves 0.7 across the resistor, which fixes the emitter current. The emitter resistor then acts as a negative feedback mechanism. When the current increases, the voltage across the resistor increases, thus decreasing the base-emitter voltage, which causes Q1 to conduct less, decreasing the current.

 

[carbonzit]

@Colin: Built the circuit, tested it. Works as advertised. Thanks again. I'm going to use this to build a quick-n-dirty LED tester.

@ModemHead: Thanks for the explanation. If I study that long enough, then transistor action will really sink in. I've almost got it ...