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I need to power a halogen lamp with a constant power (5 V 900 mA) using a constant current switching driver IC. I will not use an LDO IC, since my power supply is 24 V and it will generate too much heat.

As I am looking for parts I notice that I see many "LED driver ICs," while my halogen lamp is actually resistive.

I want to make sure that there will not be any issue if I use a IC that is meant to be used for LEDs, to drive a resistive load (halogen lamp.)

  1. Is there any downside when driving my resistive lamp with a switching IC which is labeled as "LED driver"?
  2. As an alternative, can I use a simple buck converter in constant current mode to drive my lamp like this:

schematic

simulate this circuit – Schematic created using CircuitLab

As I haven't found many (or not at all) schematics that use buck converters in constant current mode, I’d like to know if the larger feedback loop will cause unexpected problems.

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    \$\begingroup\$ I do now or "I do not"?... "Will there be any issue" - there could be but how can anyone reading your post be sure? \$\endgroup\$
    – Andy aka
    Commented Sep 13, 2021 at 8:37
  • \$\begingroup\$ First one was typo mistake, and tried to fix the others, thanks \$\endgroup\$ Commented Sep 13, 2021 at 8:48
  • \$\begingroup\$ Low side buck, perhaps even without an inductor and capacitor? \$\endgroup\$
    – winny
    Commented Sep 13, 2021 at 8:55
  • \$\begingroup\$ You can choose constant power, constant voltage of constant current, but you can't set both current and the voltage simultaneously, as the lamp resistance will be low when cold and higher when it heats up. What are the lamp ratings? Usually these lamps are driven with certain voltage. \$\endgroup\$
    – Justme
    Commented Sep 13, 2021 at 9:00
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    \$\begingroup\$ Sure. The thermal mass of the lamp will low pass filter your incoming square wave. Adjust your switch frequency if needed. \$\endgroup\$
    – winny
    Commented Sep 13, 2021 at 9:05

1 Answer 1

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We use it to make spectrum analysis, and don't want the brightness to fluctuate.

OK, it was important to mention this.

Is there any downside when driving my resistive lamp with a switching IC which is labeled as "LED driver"? As an alternative, can I use a simple buck converter in constant current mode to drive my lamp like this:

From TPS54200 datasheet:

enter image description here

This is a very common schematic, and there are dozens of chips that will do what you want, look at the "LED Drivers" section on DigiKey and check "Step down" in the search criteria, along with suitable voltage and current. The chip is a buck converter, this one is synchronous so it has an internal MOSFET to replace the usual buck diode.

Your halogen lightbulb would replace the LED and get constant current.

There are some differences between "buck converter" and "buck LED driver":

LED driver will be designed for a low voltage on the sense resistor, so it wastes less power, but it will have lower accuracy which is not important in lighting applications. A buck converter will typically have a higher reference voltage for better accuracy, also lower drift, etc. But if you use a buck IC with say 1.2V reference voltage, then you'd have 1.2V on your sense resistor and 5V on the lightbulb, so the resistor R1 in your schematic would waste 24% of the power dissipated in the lightbulb. Not catastrophic, but not great. If you use a buck IC with a higher reference voltage, wasted power in the sense resistor will increase though. You can search for "minimum output voltage" which usually tells you what the reference voltage is.

In the LED driver case, capacitor Co smoothes the current in the LED (or lightbulb) which would otherwise be the sawtooth inductor current typical of a buck converter. In your schematic, C2 does the same job, but it is connected differently, one pin to ground instead of across the load.

It is expected a buck converter will have a nice output voltage transient response and keep the voltage well regulated when load current changes. LED drivers don't do that, because the load is a LED, which is going to draw fast changing current like a CPU would. So LED drivers use mostly simple control schemes like peak current or constant on/off time and the like. These are based on measuring the instantaneous inductor current and comparing it with some value, which means the current in the inductor and the resistor should be the same, which means the smoothing cap must be across the load, and not connected to ground. Whereas in the buck converter case, the output cap is part of the control loop.

These simpler control schemes mean a LED driver will have worse output regulation wrt input voltage than a buck regulator which will be specified for PSRR, etc. Except if you get a fancier LED driver that actually measures and regulates average current.

I don't think there will be any problems with a lightbulb instead of the LED.

If the LED driver allows a cap across the load, then it will be fine with 0V across the load at start-up, like a cold filament. Maybe check the datasheet doesn't say "maximum capacitor value..." with a small value.

For a buck converter, you should pick ont that is able to be set to an output voltage equal to its reference voltage, which is generally the case. Also, the feedback network formed by the lightbulb and the sense resistor will change its dividing ratio as the lightbulb heats up, so you should pick a dc-dc that doesn't need special compensation for a specific value of output voltage and feedback divider.

So basically, use whatever you want, both will work, but make sure the output smoothing cap is connected properly.

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