can I use resistors in a DC circuit to reduce voltage?

Thread Starter

pnolans

Joined Oct 18, 2019
9
Please be gentle. It's been 50 years since I had electronics shop. I DO intend to take some sort of "Electronics 101" class/instruction.

I live in a trailer with 120V and 12V circuitry. I have a few LED devices that nice mood lighting.
They run off of 3 AA batteries. I would like to integrate them into my 12V system, in order to use reusable power (solar panels). I know I can use a transformer to lower the voltage to 4.5V. Can I use a resistor circuit of some kind to reduce voltage?

Thanks in advance for any help,

Pat
 

ericgibbs

Joined Jan 29, 2010
8,722
Hi Pat,
Welcome to AAC.
You can add a series resistor, but we need to know the current thru the LED's , so that we can calculate the resistor value.
E
 

dendad

Joined Feb 20, 2016
2,986
Get hold of some of those 12V to 5V cigarette lighter USB power regulators and your 4.5V LEDs will probably run ok on the 5V.
If you want, add a series 1N4004 diode to the 5V to drop it down to around 4.3V.
Better than resistors as it will waste less power.
 

dl324

Joined Mar 30, 2015
8,870
Welcome to AAC!
I know I can use a transformer to lower the voltage to 4.5V. Can I use a resistor circuit of some kind to reduce voltage?
You can't use a transformer unless you have an AC voltage.

You can use a resistor if you determine the amount of current they draw from the batteries. Then use Ohm's Law to calculate the resistor value.

Another option is to use a zener diode to drop the voltage; assuming these are the typical 20mA maximum current type of LED.

You could add a 5V regulator or a automotive USB port. This option will waste less power if they're switching regulators.
 

Thread Starter

pnolans

Joined Oct 18, 2019
9
Thanks for the replies! I forgot about the transformer and AC current. DUH.
 

dl324

Joined Mar 30, 2015
8,870
I forgot about the transformer and AC current.
In your defense, it has been 50 years.

I was surprised and disappointed how quickly my coding skills got rusty after not using them for a couple years after I retired. I knew what I wanted to do, but had to keep looking up syntax.
 

Thread Starter

pnolans

Joined Oct 18, 2019
9
Welcome to AAC!You can't use a transformer unless you have an AC voltage.

You can use a resistor if you determine the amount of current they draw from the batteries. Then use Ohm's Law to calculate the resistor value.
So, the LED's are rated at 20mA. If I were to use a 12V battery, I would want a 600 ohm resistor?

Or am I messing something up?
 
Single LEDs have a voltage drop depending on color. The variation in voltage drop shows up as an intensity variation. You can place the LEDS in series as long there is enough voltage to power them. R<=(12-2.4)/20e-3 and size the wattage accordingly.
There are LEDS that are designed to operate at 5, 12 and 24 V which have an internal resistor. They get tricky. You might want to use 13.8V, but there is lots of parameter variation.

PWM is an easy way to dim LEDs. You don;t want to put LEDS in parallel.
 

Thread Starter

pnolans

Joined Oct 18, 2019
9
Single LEDs have a voltage drop depending on color. The variation in voltage drop shows up as an intensity variation. You can place the LEDS in series as long there is enough voltage to power them. R<=(12-2.4)/20e-3 and size the wattage accordingly.
There are LEDS that are designed to operate at 5, 12 and 24 V which have an internal resistor. They get tricky. You might want to use 13.8V, but there is lots of parameter variation.

PWM is an easy way to dim LEDs. You don;t want to put LEDS in parallel.
Uh... I don't feel bad that an awful lot of what you "Kept Simple" ... :) in the above reply went zoooooommmm over my head. I searched the forum for PWM and read a few bits of the articles. Pulse Width Modulation, huh? I need to go back to the 8th grade. Transistors were a big thing back then. Maybe I'll make re-learning electronics my retirement hobby. o_O

Thanks for the reply!

Pat
 

Papabravo

Joined Feb 24, 2006
12,389
Are you familiar with how to compute the average value of a non-constant waveform?
For example let us say I have a pulse generator that I can program to be on (+5Volts) for 20 milliseconds and off (0 Volts) for 80 milliseconds?
Answer:
\[V_{avg}\;=\; (5)(.2) + (0)(.8)\;=\;1\;\text Volt\]

That is all well and good, but as we know LED's are controlled by current. So let us assume that we have set up our LED to consume 18 mA when +5 is applied to the LED and a current control resistor (more on the resistor later). For one cycle of the waveform we have:
\[I_{avg}\;=\;(.018)(.2) + (0)(.8)\;=\;3.6\;\text mA.\]

What the LED actually "sees" is an average current of 3.6 mA, and depending on the persistence of the LED material may or may not blink, and it will be dimmer than if it was powered at 18 mA. continuously.

PWM is just building a pulse generator that allows changing the pulse width of the high part of the cycle. The length of the pulse relative to the period is called the duty cycle.

Does that help?
 
Last edited:
@pnolans PWM is also a TLA (Three Letter Acronym) <G>

R<=(12-2.4)/20e-3

R (Resistance in ohms Ω)
12 - Nominal system voltage (Volts)
2.4 V - Typical LED drop (Red LED)
20e-3 = 2-x10 superscript(-3) or since calculators can't do that, E , so 20 * 0.001 = 20 mA

Datasheet: https://www.alliedelec.com/m/d/6355b8aba0b01578df0bb7b871ceefd7.pdf

This one has a typical Vf (Max) of 2 V.


Then I'll leave it as en exercise to compute the resistor wattage needed.
 

Thread Starter

pnolans

Joined Oct 18, 2019
9
@pnolans PWM is also a TLA (Three Letter Acronym) <G>

.
I used to work for a company called DEC. :) So, not only was our name a TLA, but we used TLA's everywhere. My particular area of expertise was with a SW product that DEC named Rdb. Another TLA.
 

Thread Starter

pnolans

Joined Oct 18, 2019
9
I used a variety of DEC equipment early in my career. Started with a PDP-11. Happy RSX, PIP, VAX, VMS, DCL, ... to you!
I supported PDP's some early on, but became a VAX guy soon after. But I would still run across stuff where I needed to use PIP. So, Happy MCR to you! :)
 

Thread Starter

pnolans

Joined Oct 18, 2019
9
I liked VMS - a lot. The company I worked at first dipped their toes into Unix-land with Eunice, then Ultrix. Then we moved away from DEC equipment and VMS forever.
Hmmm... shame... I was still using and supporting VMS when I retired from HP 3 years ago. Our customer was committed to keeping their system running. :)
 

Tonyr1084

Joined Sep 24, 2015
3,577
Getting a little off topic are we ? ? ?

Using resistors to drop voltages. Yes, you can, but when you do you have to consider the load resistance as well. Or it's reactance. It can get quite complicated. Using resistors as voltage dividers is most useful when a reference voltage is needed. Otherwise, the main purpose of a resistor is to resist the flow of current.

To address your question - you want to emulate a 4.5 volt circuit to drive your LED mood lighting. You have available 120 VAC and 12 VDC. You need 4.5 VDC. Using a buck converter to drop your voltage to 4.5 volts will give you sufficient power to drive your LED's and effectively reduce the voltage to 4.5 volts. It's probably going to be the most efficient way of doing so. Buck converters are cheap. You'd plug it into your 12 volt circuitry and NOT your 120 VAC circuit. You can find buck converters on Amazon, eBay and probably lots of other places. And I'd bet you can use one of those 12 volt cigarette lighter converters on the 12 volt system to drop the voltage down to 5 volts.

AA batteries are typically 1.56 volts new. Three of them will give you 4.68 volts. That's only slightly lower than 5 volts. A 1N4001 diode should drop 0.4 volts across its junction, giving you 4.6 volts. And those LED mood lights probably don't draw much current. Depending on how much current is being drawn you may need to choose a different diode. A 1N400X family should give you plenty of current capabilities.

Good luck with your project.
 

Thread Starter

pnolans

Joined Oct 18, 2019
9
Getting a little off topic are we ? ? ?

Using resistors to drop voltages. Yes, you can, but when you do you have to consider the load resistance as well. Or it's reactance. It can get quite complicated. Using resistors as voltage dividers is most useful when a reference voltage is needed. Otherwise, the main purpose of a resistor is to resist the flow of current.

To address your question - you want to emulate a 4.5 volt circuit to drive your LED mood lighting. You have available 120 VAC and 12 VDC. You need 4.5 VDC. Using a buck converter to drop your voltage to 4.5 volts will give you sufficient power to drive your LED's and effectively reduce the voltage to 4.5 volts. It's probably going to be the most efficient way of doing so. Buck converters are cheap. You'd plug it into your 12 volt circuitry and NOT your 120 VAC circuit. You can find buck converters on Amazon, eBay and probably lots of other places. And I'd bet you can use one of those 12 volt cigarette lighter converters on the 12 volt system to drop the voltage down to 5 volts.

AA batteries are typically 1.56 volts new. Three of them will give you 4.68 volts. That's only slightly lower than 5 volts. A 1N4001 diode should drop 0.4 volts across its junction, giving you 4.6 volts. And those LED mood lights probably don't draw much current. Depending on how much current is being drawn you may need to choose a different diode. A 1N400X family should give you plenty of current capabilities.

Good luck with your project.
Thank you very much... that was extremely helpful for a newbie like myself.
 

MisterBill2

Joined Jan 23, 2018
3,861
A resistor can certainly work to drop the voltage for the LEDs, BUT presently many of them run on a lot more than the 20 mA of years gone by. I have a bunch of strings of very bright white LEDs that run on about 3.2 volts and on 80 to 100mA. They will give off some light at 20mA but not nearly what they deliver with more current.
 
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