# LED Current regulator basics

Discussion in 'General Electronics Chat' started by Bosparra, Apr 29, 2010.

1. ### Bosparra Thread Starter Member

Feb 17, 2010
79
3
Hi,
I am a hobbyist with mainly digital experience. I am trying to improve my analog knowledge.

I have experimented with a couple of current regulating circuits to drive 6 bright white LED's in this schematic:

Through experimentation, I found that I get 40mA, that is 20mA through each of the strings of LED's, when Vgs is 4.2V which is what the diode and the 3.6V zener is there for. This all works fine, but at 10V the current drops to about 25mA and at 14V it goes up to 50mA. This doesn't seem like my circuit is regulating the current very well.

Now, my question is, what level of current regulation can be achieved in practice? I understand that there will be limits, but I am not sure when to judge whether my current regulation is as good as it can be. All I know is that my circuit sucks at it.

Secondly, what role does the 100 Ohm resistor play in the effeciency of the regulator? Surely, if the current regulator does it's job properly, then it is not needed? Bill Marsden commented in the other thread, that the maths would explain everything, I've done all of that, but clearly I am missing something fundamental. Here's what I've got:
1. The 100 Ohm resistor drops 2v at Vcc of 12V.
2. The 3 LED's drop 9.6V (@3.2V each) altogether.
3. That leaves a Vds on the jfet of 0.4V
4. This would leave me with a lower Vcc limit of 10,6V before the jfet starts shutting down, right?
5. The reason for the current increasing with Vcc is because Vds on jfet is increasing which lowers it's resistance, right?

According to mu understanding a jfet is an excellent current regulator, what am I missing?

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2. ### Wendy Moderator

Mar 24, 2008
20,772
2,540
JFETs can be used for current regulation, but in many way I don't think they are as predictable as BJTs. I could be wrong about this, but I've see quite a bit of variation among similar devices.

The ACC book has an experiment on them. Based on that, I suspect you need a source resistor.

Regular BJTs do a fair job by themselves.

I would not recommend this design, but it is a fun concept.

This design is much more practical. The diodes make a crude voltage regulator, which keep a constant voltage on the emitter resistor, which sets up a true constant current on the collector.

The big advantage with this concept is the emitter resistor can be kept small, which means its wattage is also kept small. The transistor dissipates all the heat. It is a odd fact that transistors have matching heat sinks, and in general are designed for more wattage than small resistors.

The insertion drop is also kept low, as the resistor drops 0.6V, and the rest of the power supply voltage is used for the LEDs.

Last edited: Apr 29, 2010
3. ### kkazem Active Member

Jul 23, 2009
160
27
Hi,
After reviewing the 2N7000 Datasheet, I can see why you're not regulating. The 2N7000 fet wants to be regulating at about 300mA with a 4.5V VGS and that;s over a VDS (drain to source voltage) from about 2.5 VDC on up to 60 VDC. But your circuit doesn't allow the drain to conduct anywhere near 300mA.

What you can do is connect a pot (1K pot up to a 100K pot would work) with the outer legs across the 2 diodes in the gate ckt, then disconnect the gate from your current diode ckt and instead, connect the gate to the wiper (middle pin, adjustable pin) of the pot. Now, you can adjust the gate voltage to set any current lower than 300mA that you want, although as you get very low in current, the accuracy will go down, but I think it will work ok for you by setting the gate-source voltage to about 3 Volts DC. You can measure it monitoring the voltage across one of the 100 ohm LED ballast resistors, especially if they both track with about equal voltage as you adjust the current. And if they don't track, you can simply measure both of them, one-at-a-time, and add the two currents together to get the total drain current.

Regards,
Kamran Kazem

4. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,990
1,115
Hey but 2N7000 is not a JFET. 2N7000 is a MOSFET
And you can try this simple current source

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5. ### kkazem Active Member

Jul 23, 2009
160
27
Hi,
Bill has a good circuit there, but I think that the modified version of your circuit using a pot to set the current is simpler, and actually has far lower ballast resistors at 100 ohms per 3-LED leg than his 360 ohm per leg.

And, Jony130, yes, it is a MOSFET as I originally said, and Bill's circuit was using a JFET. But you're circuit has no Re if you look carefully! I think you may have meant to use a PNP instead of an NPN for the bipolar.

But there is no need. MOSFETs have awesome and very linear current regulation when used in this manner over a wide range of VDS voltages. All one needs do is look at the datasheet (attached on pdf), page 3, Figure-1 tells the whole story.

Regards,
Kamran Kazem

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6. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,990
1,115
Well yes, my mistake is should be R2 instead of Re.

7. ### Bosparra Thread Starter Member

Feb 17, 2010
79
3
Thanks for the quick replies!

Ok, I think I understand why my circuit is not working as expected, but I am still confused about a couple of things:
1. How is a pot on the gate going to keep the gate-source bias constant when the input voltage goes up or down? Doesn't that have the potential to drive the fet into an undesirable 'zone'?
2. In Bill's BJT suggestion, the base emitter bias will remain constant, leaving only the collector-emitter voltage to regulate the current, right? What stops it from going into saturation resulting in high current flow, when Vcc goes up?

I think my main confusion is the principle that the current can stay the same, regardless of input voltage fluctuations.

8. ### Wendy Moderator

Mar 24, 2008
20,772
2,540
A common collector is a voltage follower, and the diodes uses for regulation are not going to change their value much if Vcc goes up, the current through the diodes will go up a bit, but not the voltage. If need be the diode voltage regulator can be replaced with something better.

If the diodes don't change their value, then the emitter won't change its value, so neither will the current. If Vcc does go up, the wattage dissipated by the transistors also goes up (not the current).

A BJT is a natural constant current source.

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