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I have a voltage regulator outputting 5v to an LED grow light that draws 1A. Can I increase the output to 6.5v, and add an analogue quarts clock in series, intended to run off a 1.5v battery?

My concern is that the current draw from the clock is not constant, and this will create a drop in the LED light that could damage it.

From reading online, the clock draws 0.1mA constantly and and then draws 5mA for 0.02 seconds per second when the solenoid fires.

themartin
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  • are you asking if the quartz clock can withstand a 6.5 V power supply? – jsotola Aug 26 '22 at 03:38
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    In fact, the current used by the clock will be miniscule, far too little to light up your LED. As a result, most of the 6.5 V will appear across the clock, not the LED, and will probably destroy it. – Dave Tweed Aug 26 '22 at 03:39
  • The easiest way would be dropping voltage with a few small diodes. A small-signal diode such as 1N4148 will drop 0.7V. I'd keep your output at 5V, put the clock in series with (5-1.5)/0.7 = 5 diodes , and put that combination in parallel with the LED lamp. You could use a 1.5V linear regulator if that's available to you (most folks have a bunch of diodes just laying around...) – Kyle B Aug 26 '22 at 03:54
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    This will not work at all. Remember that when you connect components in series, the same current must flow through all of them. – Drew Aug 26 '22 at 03:54
  • You could perhaps use two forward biased 2A or greater power diodes which will give about 1.4 to 1.6 volts drop to power the analog clock. They will get fairly hot with almost 1 watt each, so provide sufficient ventilation and maybe a heat sink. It would be more efficient to use a separate circuit with about 470 ohms in seris with the diodes and about 10 uF capacitor to handle the 5 mA pulse. – PStechPaul Aug 26 '22 at 03:58
  • @jsotola I thought if the clock was added after the light, there would be only 1.5v remaining. – themartin Aug 26 '22 at 04:01
  • @DaveTweed I thought that adding them series would allow the LED to use 5v, leaving 1.5v available in the circuit for the clock to use. I also thought that the clock would draw a small amount of current through the LED, so the current required would just add up? Battery -> LED -> Clock. Obviously, I'm a beginner. – themartin Aug 26 '22 at 04:02
  • @themartin two series connected components (LED and clock) affect each other, so there is no `only 1.5v remaining` – jsotola Aug 26 '22 at 04:14
  • I didn't know that @jsotola. Will the 1A LED force 1A through the clock because they are wired in series, even when the clock is after the LED? – themartin Aug 26 '22 at 04:17
  • @KyleB Your answer makes sense to me. Since the LED and Clock+Diodes are wired in parallel, they will draw different amounts of current (good) but the same voltage (dropped by the diodes for the clock)? – themartin Aug 26 '22 at 04:18
  • the LED is not a power supply .. it does not force current through any component – jsotola Aug 26 '22 at 04:18
  • the LED rating is the maximum current that should be allowed to flow through the LED ... higher current may cause the LED to burn up – jsotola Aug 26 '22 at 04:21
  • I will need to further my understanding of current @jsotola – themartin Aug 26 '22 at 04:21
  • @themartin Right. U understand the idea. I promise what I propose would work. I are a engineer ;) Seriously though, the only way it wouldn't is if that 0.1mA figure you quoted is something way way less. If it doesn't, source yourself a proper linear voltage regulator. – Kyle B Aug 26 '22 at 07:53

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The series operation could only work if you waste the same series LED current across some 1.5V limiter. This would be unwise and inefficient.

Can you think of a better solution?

Tony Stewart EE75
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  • It seems from other comments that it would be better to connect them in parallel and drop the voltage with diodes on the side of the clock? – themartin Aug 26 '22 at 04:19
  • Yes, there are many solutions to a 1.5V regulator that will waste little power. A series R and shunt diode or LDO – Tony Stewart EE75 Aug 26 '22 at 04:43
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No. Either the clock burns out from the higher current draw or the leds won't turn on from a reduced voltage drop. As the current of the clock will vary greatly between the motor moving, the motor not moving, or the motor moving a load, the leds will not see a steady draw if any.

If you really need to do this, put a 1.5V regulator in parallel with the led light. Even an non-low drop out (LDO) linear regulator will work. The parallel draw will not affect the other side and you get both powered by a single source.

Hell, a few silicone diodes in series would be enough. 5 diodes in series would be just about 3.5V give or take a few tenth of a volt and give you 1.5V. But again since the motor which is the highest draw in a mechanical clock, will vary, an actual linear regulator would be better. Lm317 would fit nicely.

Passerby
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  • Thank you for your reply. I see that the LM317 is adjustable. Am I correct in assuming I can set the voltage by attaching a potentiometer and dialling it in? – themartin Aug 26 '22 at 05:18
  • Yes, just follow its required parts (capacitors), meet the required resistor ratio, and ensure the minimum load is met (10mA iirc). There's better adjustable regulators but the 317 is simple enough. – Passerby Aug 26 '22 at 05:20
  • I realise that the holding current is related to the 1.25 volts across the resister, so I'll need 240 Ohm and 47 Ohm. But could I replace the 240 Ohm resister with a potentiometer, and dial it in until the voltage reads 1.5v? – themartin Aug 26 '22 at 05:55
  • Yes. A 1k pot. Or 5k pot and 470 ohm resistor. Multiple options available. – Passerby Aug 26 '22 at 06:04
  • Cheers! Do you intuitively understand the relationship between these elements or do you tend to have to refer to formulas a lot? Will using larger resisters as per your comment lead to better head dissipation or something like that? – themartin Aug 26 '22 at 12:17
  • @themartin experience mostly. The formulas are on the datasheet and there are multiple calculators online for adjustable regulators that will calculate it for you. Like https://circuitdigest.com/calculators/lm317-resistor-voltage-calculator The larger resistors would change the power dissipated in the resistors, but using a different value is mostly for part selection. Sometimes you don't have a 10k pot that's sensitive enough for what you want but you do a 1k. It gives you flexibility. More than one way to regulate a cat. – Passerby Aug 26 '22 at 13:54