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In my case, a 50W 34V LED is being used to light a small box. To reduce heat generated by the LED, we are switching it on and off at 50Hz.

But I was wondering will this actually help? In the case of motors, transistors, and MOSFETS we have seen that it works the opposite way. Motors need an initial high current to start up. Transistors and MOSFETS operating at high frequencies heat up. So, does the same happen with LEDs?

Side question: What other components are affected by the switching the way motors and transistors are?

Thank you.

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    \$\begingroup\$ It would probably be better to drive a constant, lower current through the LED than to switch it, but I couldn't tell you by how much. \$\endgroup\$ – Hearth Mar 13 '19 at 19:53
  • \$\begingroup\$ Thank you Hearth for your comment! I was thinking the same thing! But my curious mind still asks which is better (at a constant current and voltage): switching it on and off or keeping it on. \$\endgroup\$ – Abdulla Masud Mar 13 '19 at 19:57
  • \$\begingroup\$ I know that blinking LEDs is used to digitally adjust their brightness and conserve power. I think that blinking does actually reduce the heat because the difference between motor drives is that the transistor is interrupting the current across it and entering the linear region. In an LED blinker, a MOSFET elsewhere is still doing that job, not the LED. But I think it would still be more cooler if you ran it at a lower constant current because it makes it not as peaky and because blinking it at a higher intensity still makes it appear dimmer than it would otherwise. \$\endgroup\$ – DKNguyen Mar 13 '19 at 20:01
  • \$\begingroup\$ Do you know how to compute Ohm's law motor DCR , PWM circuits? with duty cycle? Do you know the effective DCR in the motor , LED and transistor? Then Ohm's Law applies for power. We don't know your skill level is \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Mar 14 '19 at 0:19
  • \$\begingroup\$ @SunnyskyguyEE75 Yes to PWM with duty cycle and effective DCR of LED and transistor. Not sure about motors though maybe just need refreshing. Not sure what my skill level has to do here though... maybe I am misunderstanding something. (oh and thanks for the EE portal with E-books on your profile! Wish I had found them years back) \$\endgroup\$ – Abdulla Masud Mar 14 '19 at 19:34
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If you are talking about apparent (visual) brightness of an LED that appears to be continuously "on", there is little difference between the consumption of a pulsed LED vs. a steady LED for the same visual brightness.

At lower duty cycles, the \$I^2 R\$ losses in the resistive component of the LED forward drop mount and it becomes less efficient. There may be some curvature to the brightness-current curve that will cause it to be slightly more efficient at higher currents.

Of course if you reduce the average current, thus making the LED appear dimmer, you'll reduce the power consumption and the heating.

It's easy to adjust the apparent brightness by changing the PWM duty cycle, of course. The actual (not the perceived) brightness will be almost a square wave at just about any reasonable PWM frequency- the response of an LED to current is sub-microsecond and there is no significant persistence of typical phosphors used to make white LEDs). I would guess that at some very low duty cycle you could cause eye damage while not appearing to be too bright- our eyes are not adapted to protect themselves against things like that- but it would likely be at pathologically high peak power levels and short duty cycles.

50Hz is low enough to cause a bit of an annoying flicker, especially in your peripheral vision.

Switching losses are likely not going to be a factor at frequencies many orders of magnitude higher than your 50Hz- there's a bit of capacitance which will cause losses, but not much. Conduction losses in a MOSFET are part of the \$I^2 R\$ loss (but may not be a large part, depending).

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  • \$\begingroup\$ Thanks for the explanation though! sorry that it took me a while to comment but had to process and take your explanation in and read other materials. A new thing learned. \$\endgroup\$ – Abdulla Masud Mar 14 '19 at 19:39
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Depends.

The losses of a power device is split between conduction and switching losses. Your two scenario's are

  1. LED 100% conduction -> 0% switching losses
  2. LED x % conduction -> 100 - x % switching losses.

Depending on how you set the switching frequency and duty, you can easily achieve 50% conduction, 50% switching losses in the LED and thus there will be a configuration where the overall losses in the LED are lower than 100% conduction: 50Hz @ 50% duty could easily produce 49.xxx% reduction in LED losses (assuming the switching time significantly less than the duty period).

HOWEVER, you cannot just consider the LED in this situation, you must consider what else is present in your circuit to control the current. You will need to have a V-I converter in your circuit and this now needs to become more complex as you add a switching device (a FET), which will come with its own conduction and switching losses.

Without know your present scheme or how you plan to PWM the LED, it is not possible to guarantee you will result in a loss saving. You however would need to do something significantly wrong to result in a larger power loss with a PWM at 50Hz.

NOTE: consider the "annoyance" factor of using a switching frequency of 50Hz. if a simple chopper circuit is used, that 50% will be easily visible and will be extremely annoying. Increasing the switching frequency to 200Hz or using an inductive freewheel path will improve this.

In short, yes it is possible to realise a system with a lower loss.

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  • \$\begingroup\$ Thank you for the suggestions! It does make a lot of sense that the way I control the current will also add to the heat loss (why didn't I think of that). Depends does sound like a good answer. (also, does inductive freewheel mean the freewheel diode connection?) \$\endgroup\$ – Abdulla Masud Mar 14 '19 at 5:51
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Motors need an initial high current to start up.

Correct. The starting current is high but once running with a PWM (pulse width modulated) current the startup current condition doesn't occur again.

Transistors and MOSFETS operating at high frequencies heat up.

Transistors and MOSFETs behave, in the applications you are discussing, as switches. The power dissipated in a switch is given by \$ P = VI \$ where P is power (W) dissipated in the switch, V is the voltage across the switch (not the load) and I is the current.

  • When the switch is open I = 0 so P in the switch is zero.
  • When the switch is closed there is a small voltage across the switch so P > 0 but usually not too high.
  • The problem occurs when switching from one condition to the other the switch goes through a transition region where the voltage is part way between off and fully on and the current is part way between zero and maximum. During this transition the power dissipated can be very high. With a high slew-rate (the speed with which the circuit swings from off to on and vice versa) the power dissipation can be minimised but with frequent (high frequency) switching the power adds up and can become significant.

So, does the same happen with LEDs?

No. The LEDs consume power when on and don't when off. The average power dissipated will vary in direct proportion to the switching duty-cycle (% on-time).

enter image description here

Figure 1. PWM signal transitioning from high pulse width (75%) to low (25%) and back again. Note amplitude remains constant. Source: LEDnique - Dimmable mains PSU control.

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To add to the answers you've already received, Many LEDs are quite tolerant to voltage overdrive as long as their overall power dissipation remains within ratings. I have some 1-3W white and RGB LEDs rated for ~3V that I can drive by pulse width modulation of 12V at about 200hz without burning them out. Doing this has a significant negative effect on their efficiency and longevity due to I\$^2\$R losses, so if PWM of an overdrive voltage is used for simplicity, it is best to keep the overdrive voltage close to the rated voltage.

For best efficiency at a given brightness, LEDs should be driven at a constant voltage, with the caveat being that they have a non-linear brightness when driven this way and cannot be driven below their minimum turn on voltage. There are generally two approaches taken as a result:

If accurate 0-100% dimming is required, like for RGB LEDs incorporated into a display, the LEDs are typically PWM driven with a 0-100% duty cycle and pulses of a constant voltage chosen so that at 100% duty cycle the LED will have the desired brightness.

If accuracy is less important, the LED can be driven with constant voltage above its minimum turn on voltage and switch to PWM where dimming below that point is required. This is more efficient than PWM dimming across the entire range, but providing accurate 0-100% output control with this method requires calibration and a more complicated/intelligent control circuit.

Note that because of thermal runaway, it is best to use current rather than voltage regulation for power LEDs. Your goal is simply to "reduce heating" of your (COB?) LED, so another way to state it would be that you wish to "increase efficiency", so you should use a constant current supply. Note that the switching converter you choose will have losses and generate heat as well, so find efficiency graphs for it to ensure it will have good/acceptable efficiency at your intended load current particularly, and also over your potential load current range.

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