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I built the circuit suggested by @MicroservicesOnDDD in their answer to this question: Low Overhead Constant Current LED Driver and had a play. Here's the circuit in the linked question:

enter image description here

The most LED current I could get by adjusting R9 was 100mA. I've got a 1W power LED (forward voltage around 3.2V) and so I want to get the current up to around 325mA whilst minimising losses.

I'm driving it from either a lithium ion 3.7V battery (maximum 4.2V at full charge) or 3 NiMH in series (maximum 4.5V at full charge.)

Any suggestions on modifying the circuit?

I've tried to add another PNP so it forms a Darlington pair with the existing PNP transistor and keeping the LED at the joined collectors but am not making progress.

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    \$\begingroup\$ You can't minimize losses with a linear circuit. Linear circuits work by burning off excess voltage from the supply as heat. Losses remain unchanged for a given input and output. So you should forget about that objective if you want to keep using this style of circuit. \$\endgroup\$
    – DKNguyen
    Commented Jan 25, 2022 at 15:18
  • \$\begingroup\$ There are many circuits 'out there' to acheive this that are very inefficient, having lots of components each consuming power. Those are the losses I'm trying to minimise. \$\endgroup\$
    – Mark Highton Ridley
    Commented Jan 25, 2022 at 15:56
  • \$\begingroup\$ @MarkHightonRidley If staying simple and linear, I'd pair up a BJT (for measuring the current) and a FET (for controlling the current.) About 700 mV minimum total overhead. So it may work down to 4 V. No opamp needed. \$\endgroup\$
    – jonk
    Commented Jan 26, 2022 at 7:31

2 Answers 2

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Your circuit will not work at the end of the discharge cycle: there isn’t enough voltage to light the LED. This is regardless of whether you use a BJT or FET (although the FET will be sightly better having only Rds(on) vs. Vce(sat)) or sense using an op-amp.

Even then, op-amp selection is slightly tricky: it needs to work at the low voltage, and permit inputs close to ground for low-side sensing. A low-voltage version LMV321A could work.

But, this still doesn’t fix things as the battery falls below 3.2V: your LED will get dim.

Instead, consider a buck-boost converter. This extends the voltage range and gets good efficiency. Here’s one that uses a charge pump: http://www.sg-micro.com/uploads/soft/20190626/1561535688.pdf

Another source: https://assets.maxlinear.com/web/documents/sipex/datasheets/sp6686.pdf

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  • \$\begingroup\$ Thanks for that point, @hacktastical. My design choice is that I'm happy for the LED brighteness to fall off a lot as the battery voltage drops to below around 3.5V as both types of batteries will have delivered most of their capacity by then. \$\endgroup\$
    – Mark Highton Ridley
    Commented Jan 25, 2022 at 15:43
  • \$\begingroup\$ Li-ion will have only delivered 50% of its mAh capacity. It can be safely drawn down to 3.0V. See: batteryuniversity.com/article/… The story is similar for NiMH. \$\endgroup\$
    – hacktastical
    Commented Jan 25, 2022 at 15:53
  • \$\begingroup\$ I stand corrected, thanks @hacktastical! I've been learning so much in recent weeks, not all of it has stuck correctly ;) \$\endgroup\$
    – Mark Highton Ridley
    Commented Jan 25, 2022 at 15:59
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    \$\begingroup\$ The buck-boost devices shown use a 33 mOhm sense, and 50~150mV feedback voltage. They sense current, not voltage. \$\endgroup\$
    – hacktastical
    Commented Jan 25, 2022 at 16:29
  • \$\begingroup\$ I found the LMV321A is only available in packages I can't work with @hacktastical - I'm building on a stripboard. I've done lots of hunting and found the LM4562NA/NOPB, datasheet here: ti.com/lit/ds/symlink/lm4562.pdf It's rail-to-rail and seems to be a good alternative. Do you mind having a look before I order some? Thanks! Mark \$\endgroup\$
    – Mark Highton Ridley
    Commented Jan 26, 2022 at 13:01
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TL,DR: Opamp that controls a MOSFET

If you have low headroom voltage, that means you cannot use a large resistor to sense the current and provide feedback.

Instead you use a small resistor like 100 mOhm. That will drop only 32 mV at 320 mA. Use an op-amp to regulate the voltage drop across this sense resistor to be 32 mV.

schematic

simulate this circuit – Schematic created using CircuitLab

You can realise the 32 mV reference with a resistance divider depending on your accuracy needs. Use an opamp that contains the low supply in its common-mode input voltage range.

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  • \$\begingroup\$ Thanks, @tobalt, itlooks like it will do as I want. I'll get hold of an appropriate op amp as suggested and give it a try. I still would like to know how to adjust the original circuit for my own learning :) \$\endgroup\$
    – Mark Highton Ridley
    Commented Jan 25, 2022 at 15:29
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    \$\begingroup\$ @MarkHightonRidley also use a MOSFET with low enough turn on voltage. so that you can turn it on with the available output swing of the opamp. Basically just use a rail-to-rail input-output opamp and MOSFET with maximum of around 2.5V gate threshold voltage. As you have spice, you can check how different MOSFET will cope. \$\endgroup\$
    – tobalt
    Commented Jan 25, 2022 at 15:32
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    \$\begingroup\$ Thanks for your guidance @tobalt and hacktastical. I've now got a properly working circuit. I used a MCP6021 PDIP package op amp (perfect for stripboards) and an IRLB8721 MOSFET. For the reference voltage I used a resistor divider (150kΩ and 1.2kΩ). \$\endgroup\$
    – Mark Highton Ridley
    Commented Feb 2, 2022 at 9:45
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    \$\begingroup\$ @MarkHightonRidley Thanks for reporting back! I wish many more people did that :) Usually, people only return if things don't work out as planned. \$\endgroup\$
    – tobalt
    Commented Feb 2, 2022 at 10:30

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