Simplest current regulator

There are a lot of white LED driver integrated circuits around. A quick session with Google threw up this

which might do what you want - the current sense resistor drops 0.2V.

You could do almost as well using an National Semiconductor LM10CLN to control a PNP series transistor

http://www.ortodoxism.ro/datasheets/nationalsemiconductor/DS005652.PDF

The LM10 includes a 200mV voltage reference. The output stage can't handle 300mA so you've got to drive some kind of buffer - a MOSFET with a low drop-out voltage would do, but a decent-sized PNP transsitor would probably be cheaper.

The LM334 only requires a 64mV drop across the current sensing resistor - the reference voltage is temperature dependent (see the graph at the bottom of page 3 of the LM334 data sheet), but probably won't move enough to worry you if you use it indoors.

http://www.ortodoxism.ro/datasheets/nationalsemiconductor/DS005697.PDF

-- Bill Sloman, Nijmegen

Min input is 4V, which is a bit high. And it's more complex than I was hoping for...

Getting closer!

Ah, now that looks interesting - 64mV is excellent, and it's a

4-component circuit (plus an extra R-C which I reckon it would probably work without...)

I think I might even have an LM334 somewhere... rummage... no, it's a '336. Oh well.

Does anyone know why we can't buy discrete zenners (or similar) down to

65mV ?

Many thanks.

Phil.

except it turns out that I would need a 0R22 current sense resistor. I can only find that value as a huge 3W ceramic-clad wirewound. Do resistors of that sort of value exist in "normal" packages? It will dissipate 20mW in my circuit.

Phil.

No, you need the RC to keep it from oscillating, with the extra loop gain added by the LM3906 pnp transistor.

In a pinch, you can use a higher-value sense resistor, and attenuate its output to 64mV with 2 more resistors.

Read the datasheet. The LM334 will not give you 300mA and requires around a 1V overhead at currents as low as 5mA.

I have done. Have you looked at the schematic that Bill linked to?:

Use an additional PNP transistor to increase the current. The 1V requirement is not a problem in this configuration.

Phil.

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You could use a small incandescent bulb in place of your series resistor. These have a positive temperature coefficient and thus will tend to keep the current constantish. There are several available for flashlights/torches which work at a volt or so - you'd need to experiment and may need to parallel a couple of bulbs to get the necessary current. An incandescent lamp run at below its rated voltage will last pretty much for ever.

The regulation isn't good, but is better than a plain resistor and very simple. I use this method for regulating a 9.6V LED device from a 12V battery on my bike. A side effect is that the glow from the bulb provides a useful battery level indication.

Cheers

--
Keith

Bob Pease did that circuit some time ago for Electronic Design magazine.

THIS circuit doesn't have a low tempco. Its output current increases at +0.33%/°C. But that's plenty good enough for many cases?like in a flashlight! If you need a better tempco than that, we know several ways to do it. Still, this will let you regulate the current into a red LED down to 2.1 V of supply voltage, or a white one down to 4.2 V?MUCH better than 7.4 V.

OH?I almost forgot to say?the LM334 sometimes needs a series RC damper. My first guess was 2 µF and 22 across the base-emitter of the PNP. Actually, this circuit didn't oscillate or ring badly without an added capacitor, but the noise was a bit quieter when I added a 2- or

10-µF electrolytic.

If you really want a low tempco, use copper wire (magnet wire) for the resistor?that will cancel out nicely. You'll need 6 ft. of #34 gauge, or 10 ft. of #32.

It depends on what you mean by a "normal" package. Farnell lists a number of 0.22R resistors in surface mount packages which are relatively easy to adapt to through-hole circuit boards.

They do a Phicomp 0,22R part under order code 940-3248 for about $1, a much smaller - 0402 - Tyco part under order code 886-7283 which will dissipate 100mW. The Tyco part is about half the price, but you can't order fewer than 25.

There's also a 2W Tyco surface mount part, order code108-6266 which less than a dollar and availalbe in a minimum qunatity of five.

You can also buy small conventional 1R axial lead resistors - four or five of them in parallel don't take up too much space and would do the job.

As Default has pointed out, you can also use enamelled copper wire (transformer wire) to make a temperature sensitive resistor that will track with the LM334 reference voltage though he talks about using relatively thick wire. I'd go for 0.1 mm diameter wire myself - I could wind that without breaking it - where you'd only need 10cm (four inches) of wire to get 0.22R. Farnell doesn't stock 0.1 mm wire but it does sell 38 SWG copper wire which isn't quite as thin, but you'd still only need 25cm (10 inches). You might want to use a non- inductive winding scheme.

-- Bill Sloman, Nijmegen

Zener knees get soft below 4V or so. If you want a 65 mV reference, you can always use a jFET with gate/source connected as a current reference, into a suitable load resistor.

You can also use a resistor divider to drop the 200 mV reference to something more convenient in terms of LED current sense resistor dissipation.

Can someone show me how I could use a potential divider to get a lower reference voltage than a zenner can provide in this sort of current source?

\\ / \\ Load / | | | | |/ +------| | |\\>

_|_| | | ^ \\ /_\\ / R | \\ | / | |

--+--------+

V_zenner = V_be + IR

If I want R to be small, I need V_zenner to be just a little more than V_be.

Or I could put a regular diode in series with the zenner, and have V_zenner equal the voltage across the sense resistor. Or use a second trasistor.

Of course I can use a zenner with a potential divider in an op-amp circuit, but I need help to use it in something simpler.

Thanks for all the useful suggestions.

Phil.

Have you considered using niads or NiMH. Secondary cells have less change in voltage over the discharge curve. Note that an alkaline really needs to be used from 1.5V to 0.5V to get the full capacity. For nicads, this range is 1.25V to 1V. The nicad will probably start a bit higher than 1.25V, but will drop very quickly.

Any LDO can be turned into a current regulator. You make the load be a resistor, then put the leds in series with the power supply. Of course, this is not every efficient, especially for one LED.

I'd really like it to work with both, hence the need for a regulator of some sort.

But if you did want to work only with nicads, it would still be difficult to use just a series resistor; the difference between the nominal 3.6V from the cells and the nominal 3.4V for the LED is small, making it sensitive to variations in both.

Although the LM334 circuit and other things we've discussed are interesting, I'm actually quite happy with my single transistor circuit; I have superglued the transistor on to the back of the 20mm aluminimum disc that's the LED's heatsink. I'm just concerned about potential variability in transistor gain, and wonder if anyone has any tips about how to get consistent gain, e.g. are there any particular parts that have more repeatable gain from batch to batch. It needs to handle 300 mA and about 400 mW, with a somewhat-elevated ambient temperature; I'm currently using a ZTX549.

Phil.

Why not use an LED as the zener? The forward voltage on an LED has about the same temp-co as the EB drop of a transistor.

Well, I'd think an LM10 and pass transistor is fairly simple (half a square inch of circuit board), but there's another option that hasn't been discussed yet: a current mirror.

Just run a resistor from V+ to base of two transistors, both emitters to GND, and connect collector 1 to base, and LED from V+ to collector 2.

The trick is, the transistors have to be on a common heat sink and the first has to have much smaller emitter area than the second (the collector current is scaled by the emitter area, all else being equal). LED current is N*((V+) - Vbe))/R and the 'N' represents the area ratio.

ICs use this kind of trick all the time, but buying transistors in onesies it's hard to know what N is going to be.

Right, my thoughts exactly, but he really needs to go to MOSFETs, transistors will waste a lot of current saturating...it seems ridiculous: View in a fixed-width font such as Courier.

. . . . .---------------+------+------. . | | | | . | R1| | | . | [560] | | . ---BATT | |< | . - +----| Q1 | . | | |\\ | . | | | | . | | | | . | | | | . | | | | . | R2 | | |< . +---+--[100]-----------+----| Q2 . | | | |\\ . | R3| | | . | [10k] | | . | | | D1 | . | | R4 | 1N914 | . | +--[910]----|-------|

You can buy dual transistors - Farnell list a bunch, of which the PBSS4240Y looks good - and mimic the effect of emitter area by arranging the Vbe for one transistor to be higher than that of the other. At room temperature 86mV of extra bias gives you a 30:1 current ratio.

You'd use a 300R resistor to draw some 12mA from the battery; 10mA of that would go into the collector of the Vbe-setting transistor, 1mA would go through an 82R resistor to the base of the Vbe-setting transistor, and from there most of that 1mA would go on to the other terminal of the battery through a 680R resistor across the Vbe of the Vbe-setting transistor, while the remaining 1mA would provide the base current for the other half of the dual transistor, which would be sinking roughly 300mA from the battery via the white LED.

The junction of the drive transistor would be dissipating of the order of 300mW, so it would be warmer than the junction of the Vbe-setting setting transistor (which would be dissioating only about 8mW); in a dual transistor both junctions should be on the same substrate and physically close, so the difference ought to be small.

If I were building the circuit, I wouldn't start of with a 82R and

680R to bias the base of the Vbe-sensing transistor, but rather with a 200R pot in series with 560R, and use the wiper of the pot to drive the base; you'd begin with the pot set to give you 0R and 760R, then move the wiper down towards 82R and 678R while monitoring the current drawn through the LED.

The circuit would be a bit crude, but it would work.

-- Bill Sloman, Nijmegen

I would consider using a 4th battery. You will be barely using the capacity of the alkaline cells.

Your circuit goes up a zener, then down a vbe, so you will get a vbe of shift over temperature. You could go up a zener, then buffer it with two transistors, i.e. up and down a vbe. But zener voltages are too high, so you will end up getting a bandgap reference. However, at that point, you can just use a LDO.

There is a circuit called the "peaking current reference."

I've used in in situations where the spec calls for no trim. I'm not all that impressed with the performance, but it is better than moving a whole vbe over temperature.

Have you considered going to Harbor Freight and getting a LED flashlight for $5? Seriously, I always buy stuff off the shelf if it suits my needs. You can spend a lot of time reinventing the wheel.

Is this what you're talking about, a Vbe multiplier with current source setting???: View in a fixed-width font such as Courier.

. . . . .------+--------------------. . | | | . | | --- . | [300] \\ / . | | --- . | | | . | | | . | | | . | | |/ . | +-------+----------| . | | | |>

. --- | [82] | . - \\| | | . | |-----+ | . |