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enter image description hereHello, please shed some light if my analysis is correct regarding the circuit. It is a DC circuit with an LED, LED driver and MOSFET.

I'm initially getting 24V power into my PCB from a AC/DC wall mount barrel jack. So Point A on the circuit will be 24V.

The LED driver I'm using is a constant-current driver outputting 700mA current. As this circuit is a serial circuit with no parallel paths for current, I will assume that this 700mA is constant throughout the circuit.

Now according to the LED driver data sheet (https://www.meanwellusa.com/upload/pdf/LDD-L/LDD-L-spec.pdf), the output voltage of the LED driver will always step down by 3V from input voltage.

So at Point B in the schematic, the voltage (when MOSFET is ON) will be 24V - 3V = 20 V.

So ultimately my LED will receive 20V at 0.7A current. The LED's forward voltage is 17.2V, so at point C in my circuit, the voltage will be 20 - 17.2 = 2.8 V. So in effect there will be 2.8 V left over or unused voltage so to speak?

Now, an analysis of power and heat dissipation.

The MOSFET is dissipating power, P = VI = (0.0175)(0.7) = 0.01225 W, which seems negligible.

I'm confused on how to determine the power/heat dissipated by the LED driver. If the voltage drop across the LED driver is 3V and 0.7A current if flowing through it, then P = P = VI = (3)(.7) = 2.1 W.

But the date sheet also states the LED driver's efficiency is 95%. Does that mean 95% of the total load power? In this case, the load (LED) is consuming P = VI = (17.2)(0.7) = 12 W, so 5% of this will be dissipated by the LED driver as heat which would be 12 * 0.05 = 0.602 W.

I'm struggling to determine the heat/power dissipation of the LED driver and if it will require a heat sink.

As for the LED, as I mentioned it has a forward voltage of 17.2V and a current at 700 mA at a test temperature of 85 C. As I understand it, if I use sufficient heat sinking to keep the LED temperature at or below 85 C, then the above analysis will hold true and the forward voltage will be held at around 17.2V.

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  • \$\begingroup\$ You're using an N-channel MOSFET as a high-side switch; you realize that means you'll need to provide Vcc+5 V to turn it on, not just 5 V, right? Here, that would be 29 volts. \$\endgroup\$
    – Hearth
    Oct 17, 2021 at 1:02
  • \$\begingroup\$ @Hearth I see thank you I was not aware of that. To convert to low-side switch, would the proper placement of the MOSFET be between the Vin- of the LED driver and ground or should it be between the cathode LED side and the Vout- of the LED driver? \$\endgroup\$
    – Jai
    Oct 17, 2021 at 1:06
  • \$\begingroup\$ I've edited the picture and updated the question to account for low-side switching now I believe. \$\endgroup\$
    – Jai
    Oct 17, 2021 at 1:16
  • \$\begingroup\$ Yes, that would be correct, however you should now note that your driver is no longer ground-referenced, which may be important. Make sure you consider what could possibly happen when the FET is turned off--the driver's outputs will all float up to 24 volts. As will its inputs, if it has any. Be absolutely certain this won't damage anything before doing it. Or better, just use a high-side switch with a P-channel device, along with a level-shifting N-channel one. \$\endgroup\$
    – Hearth
    Oct 17, 2021 at 2:49

2 Answers 2

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Here's what we know about the constant current buck converter:

  • It provides a current of exactly 700mA from its output side, through whatever load is connected between \$^+V_{OUT}\$ and \$^-V_{OUT}\$.

  • It is supplied with a constant voltage of 24V.

  • It is 95% efficient.

  • Its output voltage \$^+V_{OUT} - ^-V_{OUT}\$ can be anything between 0V and \$V_{IN} - 3V\$, or 0V to 21V.

Since efficiency E is the ratio of input power to output power, we can represent this relationship as:

$$ P_{OUT} = E \times P_{IN} $$

Using the power law \$ P = I \times V \$, and plugging in what we know, this is the formula we get:

$$ \begin{aligned} I_{OUT} \times (^+V_{OUT} - ^-V_{OUT}) &= E \times I_{IN} \times V_{IN} \\ \\ 700mA \times (^+V_{OUT} - ^-V_{OUT}) &= \frac{95}{100} \times I_{IN} \times 24V \end{aligned} $$

Notice that \$+V_{OUT}\$ and \$-V_{OUT}\$ are unknown at this point, since we haven't yet considered the load:

  1. We don't yet know what values they have with respect to each other (the potential difference between them), and

  2. importantly, we don't even know what values they have with respect to 0V (ground) on the input side. They may be completely isolated from the input, information which you must obtain from the driver's data sheet.

Point (1) is a result of the fact that any constant current source adjusts its output voltage to maintain the required current (700mA in this case), and this will be whatever voltage the load develops when passing that current.

For your LED, that seems to be 17V, which will rise as the LED's temperature falls, and vice versa. How much that voltage rises and falls with temperature is a parameter you obtain from the LED's datasheet. As long as the forward voltage of your LED never exceeds 21V, this unit will work fine, and the condition \$I_{OUT}=700mA\$ will be maintained.

From now on, instead of writing \$(^+V_{OUT} - ^-V_{OUT})\$, I will refer to this difference simply as \$V_{OUT}\$

With a value of \$V_{OUT}=17V\$, we can return to the \$P_{IN}\$ vs. \$P_{OUT}\$ relationship, to find the last unknown \$I_{IN}\$:

$$ \begin{aligned} I_{OUT} \times V_{OUT} &= E \times I_{IN} \times V_{IN} \\ \\ 700mA \times 17V &= \frac{95}{100} \times I_{IN} \times 24V \\ \\ I_{IN} &= \frac{\frac{100}{95} \times 700mA \times 17V}{24V} \\ \\ &= 0.52A \end{aligned} $$

Now we have all the information necessary to calculate input power \$P_{IN}\$ and output power \$P_{OUT}\$ (power delivered to the LED).

$$ \begin{aligned} P_{IN} &= I_{IN} \times V_{IN} \\ \\ &= 0.52A \times 24V = \\ \\ &= 12.5W \end{aligned} $$

$$ \begin{aligned} P_{OUT} &= I_{OUT} \times V_{OUT} \\ \\ &= 700mA \times 17V = \\ \\ &= 11.9W \end{aligned} $$

Their difference will be the power \$P_{DRV}\$ is wasted in the driver unit:

$$ \begin{aligned} P_{DRV} &= P_{IN} - P_{OUT} \\ \\ &= 12.5W - 11.9W \\ \\ &= 0.6W \end{aligned} $$

All the above serves to illustrate the relationships, and how they fit in with the behaviour of a DC-DC converter. It's long-winded though, and could have been cut down to a very simple:

$$ \begin{aligned} P_{DRV} &= P_{IN} - P_{OUT} \\ \\ &= \frac{1}{95\%} \times P_{OUT} - P_{OUT} \\ \\ &= (\frac{1}{95\%} - 1) \times P_{OUT} \\ \\ &= (\frac{1}{95\%} - 1) \times I_{OUT} \times V_{OUT} \\ \\ &= (\frac{1}{95\%} - 1) \times 700mA \times 17V \\ \\ &= 0.6W \end{aligned} $$

I'll rewrite the LED and driver power equations here, to illustrate my next point:

$$ \begin{aligned} P_{LED} &= I_{OUT} \times V_{OUT} \\ \\ P_{DRV} &= (\frac{1}{E} - 1) \times I_{OUT} \times V_{OUT} \\ \\ \end{aligned} $$

Since LED current \$I_{OUT}\$ is constant, and assuming effiency also remains constant (negligible change over the entire range of operating conditions), you can see that power disspipated in each element is a function of \$V_{OUT}\$ only. Therefore, to find the worst case power dissipation in the driver and LED, you need to find the worst case value of \$V_{OUT}\$.

In both cases, maximum power is when \$V_{OUT}\$ is at a maximum, and as I said before, you'll need to refer to your LED's datasheet to find out what this will be. It will occur at the lowest temperature that you expect your LED to experience.

I do not think you will have to worry about driver power increasing very much over the entire LED operating temperature range. However, in the terrible case where total LED forward voltage rises to 21V, the maximum that your converter can provide, power dissipated in the driver and LED would be:

$$ \begin{aligned} P_{LED} &= I_{OUT} \times V_{OUT} \\ \\ &= 700mA \times 21V \\ \\ &= 14.7W \end{aligned} $$

and

$$ \begin{aligned} P_{DRV} &= (\frac{1}{E} - 1) \times I_{OUT} \times V_{OUT} \\ \\ &= (\frac{1}{0.95} - 1) \times 700mA \times 21V \\ \\ &= 0.77W \end{aligned} $$

As for your reference to a MOSFET, I can't see from anything in your question which MOSFET you are talking about. I assume you mean that you wish to switch the LED on or off using an exernal transistor. Here are my thoughts on that:

  • Why do that when you have a digital on/off input on the driver?

  • Since I don't know if \$^+V_{OUT}\$ and \$^-V_{OUT}\$ are isolated, or in other words, if there's a common ground and which side it's on, the practicality is questionable.

  • You had better be very sure of how your driver will behave when the load is connected and disconnected in this way. It might even be discouraged. Again, the datasheet is your friend.

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  • \$\begingroup\$ Thank you very much for the explanation. I could not find any mention on the datasheet if it is isolated or not. For Vin-, it simply says don't connect to Vout-. I assume I would connect Vin- to GND. \$\endgroup\$
    – Jai
    Oct 17, 2021 at 15:10
  • \$\begingroup\$ Yes I removed the MOSFET from circuit as it was not necessary. \$\endgroup\$
    – Jai
    Oct 17, 2021 at 17:39
  • \$\begingroup\$ Would one driver be sufficient for a more than one LED lights connected in series? For example if I want to power 4 LED's each with a forward voltage of 15V at 700mA, I would need to supply input power around 63V to the driver and it will ensure a 700mA series current through all the LEDs? \$\endgroup\$
    – Jai
    Oct 18, 2021 at 15:46
  • \$\begingroup\$ The datasheet says no. 36V maximum in. \$\endgroup\$ Oct 18, 2021 at 16:34
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When I apply a 5V to the gate of the MOSFET, it has an RDSon of 25 mOhms (https://assets.nexperia.com/documents/data-sheet/BUK9Y29-40E.pdf). Given that the voltage across the MOSFET is V = IR, V = (.7A)(.025 Ohm) = 0.0175 V drop across the MOSFET.

Do you even need a MOSFET? The driver has an on/off control. Rather than cut the power to the driver, you could just turn it off.

So at Point B in the schematic, the voltage (when MOSFET is ON) will be 24V - 3V = 20 V.

The datasheet doesn't say, but likely this is an isolated driver, so no DC path exists between the supply/ground and the current driver. Therefore you cannot say what the voltage at the driver is relative to the input. Since the voltage is floating, it could be 0, -24, +50 or any other value relative to ground.

So ultimately my LED will receive 20V at 0.7A current. The LED's forward voltage is 17.2V, so at point C in my circuit, the voltage will be 20 - 17.2 = 2.8 V. So in effect there will be 2.8 V left over or unused voltage so to speak?

I think you are misunderstanding the meaning of constant current. Constant current sources like LED drivers are designed to sense the current through the diode junction and adjust the voltage to drive the set current. In this case if your LED has a Vf of 17.2V at 700 mA, and you set the driver to 700 mA, then the voltage across the diode (not relative to ground) will be very nearly 17.2V. As the diode heats up and the forward voltage drops, the driver will lower the voltage to compensate such that the current is held constant.

I'm confused on how to determine the power/heat dissipated by the LED driver. If the voltage drop across the LED driver is 3V and 0.7A current if flowing through it, then P = P = VI = (3)(.7) = 2.1 W.

You're analyzing this as though it were a linear regulator (basically, a resistor), but this is a Buck converter so the input voltage/current is being transformed. The voltage drop across the driver is whatever you are supplying (24V). The current through it will depend on the load.

But the date sheet also states the LED driver's efficiency is 95%. Does that mean 95% of the total load power? In this case, the load (LED) is consuming P = VI = (17.2)(0.7) = 12 W, so 5% of this will be dissipated by the LED driver as heat which would be 12 * 0.05 = 0.602 W.

It means that 95% of the power consumed by the driver is supplied to the load. So if the load is supplied 10W, then the driver must be supplied 10/.95 = 10.5W. Hence, 0.5 is dissipated. It is unlikely you would need to cool an LED driver at such low power levels.

As for the LED, as I mentioned it has a forward voltage of 17.2V and a current at 700 mA at a test temperature of 85 C. As I understand it, if I use sufficient heat sinking to keep the LED temperature at or below 85 C, then the above analysis will hold true and the forward voltage will be held at around 17.2V.

If the forward voltage is specified at 85C, and you intend to run it cooler, then you will actually have to supply more voltage. Fortunately the maximum you driver can supply is 21V, so you have some voltage to spare.

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  • \$\begingroup\$ I've updated the schematic picture. I'm turning the led driver on/off directly from my microcontroller now. If I understand you correctly, we cannot say for sure what the voltage at point B is, this will depend on how much voltage the driver is dropping. \$\endgroup\$
    – Jai
    Oct 17, 2021 at 3:17
  • \$\begingroup\$ At a maximum it will be 21 V, as it must drop 3 V regardless. But it will lower the voltage from 21V if need be to ensure the current is 700mA. Say when you first turn the LED on, the forward voltage will be 17.2V and current 700mA. But as the LED heats up, the forward voltage will drop. Is this equivalent to saying that the LED internal resistance lowers? So for a given voltage across the LED, if its internal resistance drops, the current will spike up. To avoid this, the LED driver will lower its output voltage? \$\endgroup\$
    – Jai
    Oct 17, 2021 at 3:20
  • \$\begingroup\$ At 17.2 Vf and 700mA, the LED will consume around 12 W power, so the LED driver will consume around ~0.6W power as heat. Can we say this 12 W consumed by LED and 0.6W consumed by driver will be constant? If the Vf of the LED drops, it would be consuming less power right? P=VI. Would the LED driver in turn be consuming more power/heat to compensate? \$\endgroup\$
    – Jai
    Oct 17, 2021 at 3:28
  • \$\begingroup\$ @Jai If the driver is actually isolated then you cannot say what the voltage is relative to ground (due to isolation). Since the circuit would be floating, voltages much greater than 21V are possible. If it isn't isolated, then -Vout is at GND and point B is at whatever the forward voltage is relative to ground. \$\endgroup\$ Oct 17, 2021 at 3:29
  • \$\begingroup\$ As the LED heats up, the power consumption of both the diode and driver will decrease slightly. \$\endgroup\$ Oct 17, 2021 at 3:30

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