Current sense and efficient LED driver?

The circuit was meant as a short-circuit snap-off. Whether it'll do what you want, I guess just try it (or simulate). ...Jim Thompson

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| James E.Thompson                                 |    mens     | 
| Analog Innovations                               |     et      | 
| Analog/Mixed-Signal ASIC's and Discrete Systems  |    manus    | 
| San Tan Valley, AZ 85142   Skype: Contacts Only  |             | 
| Voice:(480)460-2350  Fax: Available upon request |  Brass Rat  | 
| E-mail Icon at http://www.analog-innovations.com |    1962     | 
              
I love to cook with wine.     Sometimes I even put it in the food.

I guess I need to get myself a simulator ;-) or spend the time to analyze it fully.

An LED-driving circuit which might interest you is at:

It works very efficiently and the two prototypes have given a couple years of good service under practical conditions without showing any sign of problems.

The inductor is wound on a ferrite ring core as a hank of four wires twisted together (this if very important because they must be closely coupled). The FET is chosen for as low an 'on' resistance as you can afford.

The voltage across the three paralleled windings L2 - L4 is the sum of di/dt and the i*r losses, whereas L1 only generates the di/dt voltage. Therefore the emitter voltage of Tr2 (relative to the +ve rail) is proportional to the current through the LEDs and is independent of the rate of change. The unavoidable winding resistance is used as a current-sensing resistance, thus avoiding any extra losses that would occur in a separate current-sensing resistor .

DL2 is used as a Zener to set the operating current (and is included in the main lamp cluster). DL1 is mounted beside the on/off switch and is there to assist in finding the torch during a blackout.

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~ Adrian Tuddenham ~ 
(Remove the ".invalid"s and add ".co.uk" to reply) 
www.poppyrecords.co.uk

Is there a problem with battery polarity in that circuit?

A moments thought before I go get a haircut... the pulse of current charging the cap will trip the circuit... so it may do some bang-banging ;-)

Maybe a small bypass cap across the sense resistor?

...Jim Thompson

--
| James E.Thompson                                 |    mens     | 
| Analog Innovations                               |     et      | 
| Analog/Mixed-Signal ASIC's and Discrete Systems  |    manus    | 
| San Tan Valley, AZ 85142   Skype: Contacts Only  |             | 
| Voice:(480)460-2350  Fax: Available upon request |  Brass Rat  | 
| E-mail Icon at http://www.analog-innovations.com |    1962     | 
              
I love to cook with wine.     Sometimes I even put it in the food.

How about this...

Modify to suit your supplies. Note that VCC > VFLED + VBE

...Jim Thompson

--
| James E.Thompson                                 |    mens     | 
| Analog Innovations                               |     et      | 
| Analog/Mixed-Signal ASIC's and Discrete Systems  |    manus    | 
| San Tan Valley, AZ 85142   Skype: Contacts Only  |             | 
| Voice:(480)460-2350  Fax: Available upon request |  Brass Rat  | 
| E-mail Icon at http://www.analog-innovations.com |    1962     | 
              
I love to cook with wine.     Sometimes I even put it in the food.

A nice LED regulator circuit is the LM3404HV. You need a couple resistors and capacitors, a Schottky diode, and an inductor. It runs stone cold delivering 300 mA to a string of lighting LEDs. I built a light for our pantry/laundry area using this chip and have been happy with it.

Jon

but what the point? it doesn't gain anything

add an inductor and a diode, and you might be able make it work like an upside-down buck converter

-Lasse

If you're looking for power efficiency an analog method isn't going to get you the best results.

A simple constant current would be the best analog solution if you're worried about power! As for regulation, a simple constant current may not be as well..

I am not sure what you consider a simple constant current? Maybe you are referring to a resistor ? Which is what I would assume here.

Voltage all over the place

+-------+----------------+ + | --- | - | + + | V -> === - LED GND + | | | + |/ +----------+| Bias voltage here |> + | + .-. | | | | 100 '-' | === GND

with this, you need only to bias the BASE with a voltage that would appear across the 100 ohm R to match your constant current.

Normally a voltage divider at the base is enough for this from a known stable source.

for example, if I wanted 20mA on the led I would need 2 volts on the emitter. Of course, you need to add the Vf of the base to emitter for this which is ~ 0.7 depending on the temperature at hand. So, in all, we need 2.7volts at the base to hold constant 20mA for that LED or LEDS in series.

Something to think about. Jamie

It does what the OP asked.

With an inductor it is certainly more efficient. ...Jim Thompson

--
| James E.Thompson                                 |    mens     | 
| Analog Innovations                               |     et      | 
| Analog/Mixed-Signal ASIC's and Discrete Systems  |    manus    | 
| San Tan Valley, AZ 85142   Skype: Contacts Only  |             | 
| Voice:(480)460-2350  Fax: Available upon request |  Brass Rat  | 
| E-mail Icon at http://www.analog-innovations.com |    1962     | 
              
I love to cook with wine.     Sometimes I even put it in the food.

No: charging a capacitor from another capacitor (or a constant voltage source) conserves charge, not power. You need an inductor (or constant current source -- as in, the source itself generates current, not that you're using a lossy circuit to mimick that characteristic) to charge the capacitor without losses.

Here's, more or less, an inductive (with feedback) version of what Jim posted... the secondary winding basically saves a transistor.

Efficiency is even better if switching is sharpened (more loop gain), the current shunt voltage reduced (it depends on a Vbe as shown), and base drive minimized (e.g., use a MOSFET). All of this costs transistors of course, but a nice circuit can still be made in under 10 transistors.

Tim

--
Seven Transistor Labs 
Electrical Engineering Consultation 
Website: http://seventransistorlabs.com

It's no more efficient than a simple resistor or current limiter. Less, actually.

--

John Larkin                  Highland Technology Inc 
www.highlandtechnology.com   jlarkin at highlandtechnology dot com    

Precision electronic instrumentation

Hey, PAL, you really need to seek treatment for that HUB problem before you suffocate >:-} ...Jim Thompson

--
| James E.Thompson                                 |    mens     | 
| Analog Innovations                               |     et      | 
| Analog/Mixed-Signal ASIC's and Discrete Systems  |    manus    | 
| San Tan Valley, AZ 85142   Skype: Contacts Only  |             | 
| Voice:(480)460-2350  Fax: Available upon request |  Brass Rat  | 
| E-mail Icon at http://www.analog-innovations.com |    1962     | 
              
I love to cook with wine.     Sometimes I even put it in the food.

On a sunny day (Sat, 08 Feb 2014 21:52:40 -0800) it happened John Larkin wrote in :

Right I am glad somebody noticed.

Without any 'L' there are just resistive losses, could as well put the LED in a JFET drain, no stroboscopic effects: + | --- \ / -> LED --- | | d g |--- -->| BF245 | |--- s | | ---------- GND

Pick JFET for desired current.

JFET is about flat Id versus Vds above some voltage For this JFET Id is 10 mA flat above 4V Vds at Vgs = 0V BF245A-B-C_2.pdf datasheet page 5 fig 4.

Beat this for components.

He really wants a switcher with a current transformer to sense I, if he wants efficiency.

Sort of - it depends which convention you use (both were used in the past). I normally mark the circuit with "+" and "-" to avoid confusion but on this occasion I forgot to do it - and I do normally use the "wide thin line = +ve" convention, but got it wrong this time. I'll correct it and upload the circuit again.

To avoid any further confusion, the positive rail is at the top.

--
~ Adrian Tuddenham ~ 
(Remove the ".invalid"s and add ".co.uk" to reply) 
www.poppyrecords.co.uk

What you did was a truly stupid circuit. You do that a lot lately.

An LED and a resistor, or a simple current limiter, is more efficient than the howler that you posted.

For more efficiency, put some LEDs in series and reduce the current. To really do it right, you need an inductor, but inductors seem to confuse you.

--

John Larkin                  Highland Technology Inc 
www.highlandtechnology.com   jlarkin at highlandtechnology dot com    

Precision electronic instrumentation

The Supertex depletion mosfets seem to be more repeatable than the typical jfet. Jfet Idss specs can be 10:1.

If you've got the voltage, put a few LEDs in series; efficiency goes up linearly with N.

Next step, proper switcher.

--

John Larkin                  Highland Technology Inc 
www.highlandtechnology.com   jlarkin at highlandtechnology dot com    

Precision electronic instrumentation

The capacitor isn't charged through the sense resistor, only

*dis*charged through the resistor and LED.

Though, it is charged through R4, which in your original diagram is

3.3k. That means ~1mA. Hmmm...

Perhaps I need another transistor which is driven by the voltage out of Q1, but connects ground to C1.

I think this is the fundamental concept I'm missing. Is there somewhere I can read up on this?

I think you forgot to post the link.

Well, so the full application is an 8x8 RGB LED display matrix, which is run off of 1 common anode per column, and 24 cathodes per row, so I'd have 24 of these circuits.

Another wrinkle is that they will not be running at 100% duty cycle, so that I can display different patterns on the display.

Yet another wrinkle is that I want to try to cram all of this onto a double-sided 50x50mm PCB. Along with some additional control circuitry.

It may be impossible, but I was just trying to see what could be done. My first design used resistors and a 74HC164. My improved design used a constant-current sink with shift register (TLC5916). The inefficiency comes from the fact that the voltages required differ for each of the colors of LEDs.

Thanks for the information!

Daniel.