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There's these really cool things called power modules, but they are all marketed as normal dc-dc converters. Not many as "led drivers" for currents more than 1 A. The basis behind this question is understanding if it's possible to convert a power module / any dc-dc controller IC for that matter of fact into what an LED driver can do (i.e. constant current).

I see the main difference between an LED driver and, say, boost converter is the fact that the former either is constant voltage or constant current output. But almost every single IC I've come across on the "normal converter" side, in its application note has the FB attached to a voltage divider, hence the constant voltage is accomplished there. Therefore, one can say the only unique aspect is the constant current output. But that is only achieved with the voltage from a sense resistor fed into a standard error amplifier, which compares it to some Vref, then fed into the FB pin. Which isn't even integrated in an LED driver IC, at most I've seen there being two pins for current sense with an internal error amplifier.

Essentially three sub questions arise from this:

  • Are the LED driver ICs just a marketing thing? When there are so many variations of boost/buck converters available and limited LED drivers at higher currents?
  • Can any boost converter IC be made to output constant current by adding in a current sense resistor? Any limitations?
  • If I take a boost IC module such as TPS61288 - link, in the example in the datasheet the FB pin is connected to a voltage divider. Am I right to assume that if I replace it with a current resistor and amp, fed into the FB pin, it should be able to regulate a constant current output?

I also understand that the current sense pins are probably designed to measure currents based on voltage drops across series resistor. They must be suited for high side measurement where high common mode voltages are present. I don't think that changes anything, though.

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2 Answers 2

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  1. No they're not just a marketing thing. The market for LED drivers for illumination, LCD backlights and such like is enormous so the chips are optimized for the purpose. Both buck and boost are made. Usually with things like open/short load protection, soft start, prevention of dim LED illumination when they are supposed to be off, and LED (as well as chip) overtemperature limiting.

  2. Generally so, and more easily if the load does not need to be grounded so you can sense current on the low side. The maximum output voltage in a boost converter is limited by the breakdown of the chip and you might want to add some protection against an open output which might otherwise destroy the chip. LED driver chips will usually have this feature. There will be limits on maximum/minimum output voltage vs maximum/minimum input voltage like any switching regulator (depending on topology, circuit design and inductor selection).

  3. That particular chip has a relatively low 0.6V reference so a sense resistor might not be considered to be ridiculously wasteful at a couple amperes. Adding an amplifier is going to affect stability and it may not be easy to stabilize the regulator under all input and load conditions. It might be better to use a rather more complex circuit with a controller chip rather than an all-in-one chip.

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  • \$\begingroup\$ For number 3, since it's 0.6V an amplifier probably isn't necessary right. But, how would you go about simulating stability - is that where bode plots come in? \$\endgroup\$
    – roaibrain
    Commented Apr 15 at 21:16
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    \$\begingroup\$ Or in time domain applying step changes in load (or input voltage) and seeing what happens (how oscillatory the response is). It's quite possible for the regulator to be stable under some conditions of input and load and unstable on others. \$\endgroup\$ Commented Apr 16 at 14:10
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Can any boost converter IC be made to output constant current by adding in a current sense resistor? Any limitations?

I'm going to answer this one first. Short answer, yes.

For any regulator (closed loop control), there must be a parameter to be fed back to the controller. This could be voltage, current, power, temperature... The controller has an error amplifier which accepts a reference and an output sample. So if you feed the parameter in the "format" that the error amplifier accepts, the regulator will try to keep it within range without problems.

The boost converter ICs accept the sample as voltage. If you want to regulate the output voltage then you can feed the sample with a voltage divider. Or, if you want to regulate the output current then you can feed the sample with a shunt resistor. Likewise, if you want to regulate the temperature you can feed the sample from a divider formed by thermistors and resistors.

So, yes, any buck or boost converter IC can be used for either constant voltage or a constant current. Of course, there are some limitations:

  • If a boost converter IC has internal MOSFETs and diodes, the voltage ratings shouldn't be exceeded when used as a constant current regulator.
  • If the reference voltage is high, there's going to be a good amount of lost power when used with high currents even if the total power is acceptably low. For example, if the reference voltage is 1.23V, the shunt resistor for 1A current should be 1R2, so the power lost across this resistor will be ~1.5W. To keep the losses minimum a lower shunt and an amplifier can be used but the response speed will be lower due to the presence of an amplifier. A high-BW amplifier may be needed for better response.

Are the LED driver ICs just a marketing thing? When there are so many variations of boost/buck converters available and limited LED drivers at higher currents?

Actually, no.

To keep the power lost by the shunt lower, LED driver ICs tend to have much lower reference voltage (e.g. 0.3~0.4V).

They are also optimised for constant current. For example, a CV boost regulator IC may turn itself off if the chip temperature reaches a level and then turn back on once the junction temperature decreases back to a normal level, but an LED driver IC tends to reduce the current gradually instead of turning itself off which is a more pleasant effect instead of having a constantly blinking thing.

Apart from that, there are some ICs that use the DC current as a control parameter (i.e. the voltage across a shunt resistor placed right on the return path) therefore regulate the DC current directly, and there are those that use the inductor current as the control parameter to regulate the output current indirectly. I will not dive into the details here but what they do is, basically, detect the zero crossings of the inductor current to keep the operation boundary conduction mode (a.k.a. critical conduction mode) so the output current will be half the peak of the inductor current. These don't have any oscillator inside, so the switching frequency is determined by mainly the choke. These are generally more advantageous as the efficiency is generally higher and the lost power is generally lower.

If I take a boost IC module such as TPS61288 - link, in the example in the datasheet the FB pin is connected to a voltage divider. Am I right to assume that if I replace it with a current resistor and amp, fed into the FB pin, it should be able to regulate a constant current output?

Generally, it's possible but the response will be slower as I mentioned earlier. And if you design an adjustable thing the effect can be more pronounced.

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  • \$\begingroup\$ "a lower shunt and an amplifier can be used but the response speed will be lower due to the presence of an amplifier" How would I go about calculating response time given an arbitrary shunt, amp and ref voltage? \$\endgroup\$
    – roaibrain
    Commented Apr 15 at 21:18
  • \$\begingroup\$ @roaibrain it's difficult to calculate, at least for me. As the gain of an amplifier increases its bandwidth decreases (gain-bandwidth product) so it gets worse for it to generate response to rapid changes in load (e.g. turning on and off or dynamic lighting such as dimming). Even if you keep the gain low and increase the number of cascaded amplifiers the response will be "delayed" further. Take a look at here. \$\endgroup\$ Commented Apr 16 at 6:54

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