voltage stepdown chip and diode location

would it be a better idea to put a rheostat or potentiometer between the 24V source and voltage stepdown chip, and then adjust it until I get 20ma draw through the LED strip to make it be the same as it was with the battery?

That way I protect the voltage step down chip from potential overload with too many resistors... Am I thinking along the right line?

That would not work; a switch mode regulator acts like a "negative resistance" as the input voltage changes within its working range, as it maintains the same load voltage (and current, with a fixed load).

That's a constant power, so eg. if you half the input voltage it will double the input current to maintain the same power throughput.

rjenkinsgb said:

That would not work; a switch mode regulator acts like a "negative resistance" as the input voltage changes within its working range, as it maintains the same load voltage (and current, with a fixed load).

That's a constant power, so eg. if you half the input voltage it will double the input current to maintain the same power throughput.

Click to expand...

ok... I think I understand...
I was trying to put a resistor before the volt stepdown IC because I dont know how much current the 24V system will give, and if its alot, and I put a high amount of resistance after the chip, I could run the possibility of exceeding its max amp rating. (If i understand it correctly, the resistors just convert the excess "amps" into heat, and I would not be limiting how much amps are going through the voltage step down chip, just reducing the amount of "amps" available after the resistors).

rjenkinsgb said:

I was going to suggest a 100 Ohm resistor to start with.

If the LEDs are white they will probably drop around 3V, so 100 Ohms from a 5V supply (2V across the resistor) will give a current of around 20mA as given in the OPs first post.

If the supply is actually 6V, then 150 Ohms for ~20mA

Click to expand...

I get 6.5v from the stepdown chip.

Diver300 said:

What had been missed in this thread is that the CR2032 batteries will have a resistance of maybe 40 Ohms each. Two on series will have a resistance of somewhere around 80 Ohms.

If you have constant current supply, set it to 10 mA. If the LEDs are too dim, increase it a bit.

If you have a constant voltage supply, set it to 6 V and put 200 Ohms in series. What is probably better to do is to get a bunch of 1 kOhm resistors. Put about 5 in parallel and use them in series with the LED strip. The more resistors you put in parallel, the brighter the LEDs will be.

Click to expand...

would putting resistors in parallel produce less heat than resistors in series?
(I am slightly concerned I may have too much heat and will need to add ventilation)

Dave001 said:

would putting resistors in parallel produce less heat than resistors in series?
(I am slightly concerned I may have too much heat and will need to add ventilation)

Click to expand...

No, it's exactly the same either way - ohms law applies.

However, dropping 3V across 200 ohms is only 45mW dissipation.

It might be helpful to look at a tutorial on powering LEDs using current-limiting series resistors. I don't see that anybody has explained the basic fundamentals here and the questions illustrate that lack of knowledge. I'd explain more but it's nearly 1 am here, and there are plenty of good tutorials on the web.. I believe Sparkfun has a good one.

OTOH letting the magic smoke out a few times has a way of clarifying one's mind.

Dave001 said:

I was trying to put a resistor before the volt stepdown IC because I dont know how much current the 24V system will give, and if its alot, and I put a high amount of resistance after the chip, I could run the possibility of exceeding its max amp rating. (If i understand it correctly, the resistors just convert the excess "amps" into heat, and I would not be limiting how much amps are going through the voltage step down chip, just reducing the amount of "amps" available after the resistors).

Click to expand...

The "stepdown chip" - a miniature switched-mode PSU - will only draw as much current as it needs for whatever its supplying, plus a trivial amount to power itself.

The current at 24V will be little over 1/4 the output current at 6.5V

Resistors in series with the output to the LEDs will reduce the load current - and the input current, as less power is needed.

The power dissipated by a 150 Ohm resistor from 6.5V, feeding an LED with a forward voltage of 3V:

6.5 - 3 = 3.5V across the resistor.
3.5V / 150 Ohms = 0.023 amps; 23mA

The resistor power dissipation would be 3.5V * 0.023A = 0.082W, less than 1/10th of a watt.

Power dissipation is not a problem with a normal 1/2W or 1/4W resistor.

Unless I missed something, it looks like the load current still is 20 mA. In that case, there is a real possibility that the static current to run a switching regulator is greater than the load current. Plus, there still is power loss in the switching element, magnetics, etc. Per the datasheet, the worst case no-load input power is over twice as much as the LEDs being driven. This is a good indicator of how efficient the part is at very low load currents.

Consider a *linear* <gasp> regulator.

Delivering 6 V / 20 mA from a 24 V source, the power dissipation in the regulator is only 0.36 W. Yes, that also is greater than the load power, but I think there are several advantages to going a.n.a.l.o.g.

First, lets address the instant assumption that a linear regulator is inefficient. In this application, so what? Or, more accurately, yes but not enough to matter and not as much as you think. A 7806 in a TO-220 package would get warm to the touch, but not require any heatsink. It needs small input and output capacitors, but so does that switching part, so that's a wash. And don't let the "switching" buzzword fool you. It also will get warm. That part is 96% efficient only at full load current with one particular input voltage. As above, in this application the "96% efficient" part is dissipating more power than the load. Two switching regulators is a lot of grief go through for less than 1/8th W.

Speaking of EMI, check out the recommended EMI filter for that little switching puppy. If that is what they put on the datasheet, it is a clear indication of how noisy the part is. And to be clear for the less experienced, the linear regulator has no EMI.

The linear regulator type is a much more common part number - better availability and price. Way better. Like 8:1 better.
https://www.digikey.co.uk/en/products/detail/onsemi/MC7805ACTG/1481210

But wait - there's more. Now that it has been determined that current limiting is needed, let's change the linear regulator to a 7812, followed by a 300 ohm, 1/4 W resistor. Total regulator power dissipation is shared between two devices for lower operating temperature - chip power is less than 1/4 W and resistor power is less than 1/8 W.

And finally - constant current. Analog with a great big capital A.

Using the same combination of one regulator chip and one resistor, you can configure a 7805 or 7812 into a constant current regulator. The LM317 is famous for this application, but the larger the regulator headroom, the less heat is pushed into the resistor. With the number of LEDs you have now, a 7812 would have approx. the same power distribution as above.

This will deliver 20 mA to the LEDs even if the number of LEDs doubles (with a 7812) or triples (with a 7805). A nice advantage of this solution is that the static current in the regulator chip is delivered to the load rather than to GND. This does not reduce the heat in the component, but it gets double duty out of the current causing that heat. AND - if you stick with the two-regulator configuration, the unused regulator is drawing zero power, because if it is not driving a load, there is no static current. Now *that* is efficient.

ak

ps. I also recommend Ron's single-regulator system configuration in post #3. It eliminates the need for Shottkey diodes, costs a lot less, and is just plain less work.

Because of omitted information in the original question, the KEY piece has been missed here.

The load being powered is a fairy string of LEDs, originally powered by a pair of 2032 coin cells. This diagram shows the fairy string – a number of LEDs in parallel, possibly a current-limiting resistor and a battery. The resistor may not be fitted, depending on the high internal resistance of 2032 cells to limit current.



For calculations, this can be reduced to a basic LED circuit as shown. The necessary parameters are:

Vs – source voltage

Vf – LED forward voltage

If – LED current

R – current limiting series resistor



The source voltage can be 24 volts, no need to regulate it down.

Vf depends on the color of the LEDs. Red is around 2 volts, white around 3.7 volts.

For parallel LEDs, the current is calculated from the desired current per LED × the number of LEDs. But we can take a shortcut here. 2032 batteries supply at most 20mA for any length of time. Start with a resistor calculated to deliver 10mA. If that's not bright enough, calculate for 20mA.

Using Ohm's Law,

R = (Vs -Vf)/I

To supply 10mA to white LEDs from a 24 volt supply:

R = (24 – 3.7) / 0.010 AMPS

R ~ 2k ohms


Power dissipated in the resistor:

Pd = I^2 R = 0.01^2 × 2000 = 0.200 watts.

Use a ½ watt resistor, near 2k ohms.

Use two diodes as discussed to isolate the supplies. Save the voltage regulator for another project.

Here is a schematic of a constant-current regulator. Current-setting resistor R1 is larger than what might be expected, because the static current through the 7812 contributes approx. 5 mA to the output current.

ak

And for completeness, here is the more common constant-voltage regulator, followed by a current-limiting resistor. Note that except for the resistor value, all of the components are the same.

Popcorn makes a good point. If the number of LEDs is not going to change and the 24 V sources are stable, then things get very simple. Note that now all of the heat is in the single resistor, so it should be at least a 1 W part.

ak

Geez. It's a meter of tiny "fairy light" LEDs designed to run from two 2032 cells. Fixed load of at most 20mA. Why complicate the circuit?

Two diodes, which could be as small as 1N4148s or any of the 1N400x series. One resistor. Worst case dissipation of the resistor is less than ½ watt at 20mA, ¼ watt at 10mA. Three parts. Three simple connections and done.

Thanks to all the responses and informative feedback For The Popcorn rjenkinsgb AnalogKid

From what you all are saying, I can use resistors and not need the voltage stepdown chip?

So I looked into it, and the power source is 24V 2.5A (60W)

I already have the voltage step down chips on hand.
1N400x diodes on hand

Under normal operation (how the lights came supplied) they are 0.02A @ 6V = 0.12w
I played around with it using my regulated DC power supply and they work just fine with lower V higher current, but begins to get hot at above 1.15watts
(at around 1.85w sparkles and smokes Papabravo )

Using the voltage step down chip (or without it if it's not needed), what resistors do I need to bring it down to about 0.5w consumption by the LED?

Dave001 said:

From what you all are saying, I can use resistors and not need the voltage stepdown chip?

Click to expand...

Yes. This is what I attempted to explain (poorly I guess) in post #31. You must drive LEDs through a series current resistor. They fairy string light could only draw around 20mA maximum from CR2032 batteries. Doing what you're doing is ROOSTING the fairy light string. Please try to understand my explanation in post #31.

The circuit you need is exactly shown here. Between your power supplies and fairy lights, you need:

● Two diodes. Rated for 50+ PIV. Whatever you have should be fine. 1N4148 would be a compact choice.

● One 1 watt resistor – try around 2k to start to supply the fairy light string about 10mA. If that's too dim, reduce to 1k.


With the circuit shown (3 components total) and a resistor value of 1k – 2k, the fairy light string will be operating just as it did originally with the battery supply.

Read post #31. This is a basic LED circuit.

For The Popcorn said:

Yes. This is what I attempted to explain (poorly I guess) in post #31. You must drive LEDs through a series current resistor. They fairy string light could only draw around 20mA maximum from CR2032 batteries. Doing what you're doing is ROOSTING the fairy light string. Please try to understand my explanation in post #31.

The circuit you need is exactly shown here. Between your power supplies and fairy lights, you need:

● Two diodes. Rated for 50+ PIV. Whatever you have should be fine. 1N4148 would be a compact choice.

● One 1 watt resistor – try around 2k to start to supply the fairy light string about 10mA. If that's too dim, reduce to 1k.

View attachment 142969
With the circuit shown (3 components total) and a resistor value of 1k – 2k, the fairy light string will be operating just as it did originally with the battery supply.

Read post #31. This is a basic LED circuit.

Click to expand...

Huge thanks! ordering resistors now.

Got a 1w resistor kit, heres the results
At 1.5K Ω it was drawing 0.02A, resistor got hot, LED lights were a bit dim (same as with the battery).

At 220Ω it was drawing 0.10A, LED lights nice and bright and remained cool but resistor got very hot and discolored/burned on the outside.

I would like to run the LED lights at around 0.10A, and because the circuit assembly will be in a confined space with very little air flow /Cooling, what can be done to reduce how hot the resistor gets?
I would assume a different wattage resistor is what I need, correct? What wattage should I get?