Voltage Divider question - 4.25v 0.4a required

Tintin

Well-known member
Joined
21 Mar 2009
Messages
4,754
Location
Kernow
Visit site
I've got two nice strings of coloured LED lights on board that the kids like that power from two 240v wall warts rated at 4.5vdc 0.8va (one for each string).

Its ages since I've done any component electronics (o-level physics) so I could do with some help.

I've measured the actual current and voltage, these being 0.2a and 4.25vdc.

I want to knock up a voltage divider but need help with the resistor values for R1 and R2, giving Vout of 4.25vdc for power out of, say, 0.5a. This will enable me to wire them in parallel.

Or should I go with Vout of 8.5vdc and wire in series? Is there any benefit to this that anyone can think of?

Final question - should I base the Vin on 12.6vdc (standing voltage of batteries) or 13.2vdc (the max charging voltage I seem to get) ?

Thanks for any help,

chris
 

VicS

Well-known member
Joined
13 Jul 2002
Messages
48,334
Visit site
I'd wire them in series with each other and with a 18 ohm resistor (1 watt minimum) to cater for 12.6 volts
22 ohms for 13.2volts

But 13.2 volts isn't a charging voltage!

so 27 ohms for 14.4 volts
 

Tintin

Well-known member
Joined
21 Mar 2009
Messages
4,754
Location
Kernow
Visit site
Hi Vic, Thanks. Do you think 27 ohm will be enough to limit the current? I ask because there doesn't seem to be anything other than 12 LED's in each string,

thx
chris
 

VicS

Well-known member
Joined
13 Jul 2002
Messages
48,334
Visit site
i took each string as 4.25 volts at .2 amps so 8.5 volts for two in series
We need to drop 14.4 - 8.5 = 5.9 volts

5.9 volts at 0.2 amps = 5.9/0.2 = 29.5 ohms. Nearest preferred value = 27 ohms. Take it up to 33 ohms if you like, but you are not expecting 14.4 volts.

Power dissipated is I²R = 0.04 x 27 = 1 watt
 

mikefleetwood

Well-known member
Joined
19 Dec 2005
Messages
3,670
Location
In my shed
Visit site
Just to second Vic's input - 27 Ohm should be fine, it will limit the current to the same value as previous (I havn't a calculator to hand to check Vic's maths, but I'm sure he's right). You might find the LEDs a bit dimmer if the battery's not charging. If you're more likely to use the LEDs when the engine's not running, then might be better to opt for one of the lower values Vic suggests.

(I typed this before I saw Vic's post #4)
 

Ubergeekian

New member
Joined
23 Jun 2004
Messages
9,904
Location
Me: Castle Douglas, SW Scotland. Boats: Kirkcudbri
www.drmegaphone.com
I've measured the actual current and voltage, these being 0.2a and 4.25vdc.

I want to knock up a voltage divider but need help with the resistor values for R1 and R2, giving Vout of 4.25vdc for power out of, say, 0.5a. This will enable me to wire them in parallel.

Or should I go with Vout of 8.5vdc and wire in series? Is there any benefit to this that anyone can think of?

If you use them in parallel you need to drop 13.8 - 4.25 = 9.55V with 0.4A, which means a resistor of 9.55/0.4 = 23.875 ohms. However, if one of the strings then blew, or was disconnected, you'd have a much lower voltage drop over the resistor. It's impossible to say what the current and the voltage would be through the remaining LEDs without a full spec, as they are non-linear, but it will certainly be more than it's supposed to be.

Wiring them in series is fail-safe. Wiring them in parallel ain't.
 

Martin_J

Well-known member
Joined
19 Apr 2006
Messages
4,285
Location
Portsmouth, UK
Visit site
What would be wrong (or right) in using a voltage regulator... They're only 99p and with a voltage input from 12v up to 35v they still output 5v across their outputs.. No more components required although would 5v be too much across the string of leds.

http://www.maplin.co.uk/1a-positive-fixed-voltage-regulators-46475

Or you could buy one that has 9v across it's output terminals and then use the leds in series.
 

mikefleetwood

Well-known member
Joined
19 Dec 2005
Messages
3,670
Location
In my shed
Visit site
Because of the way LEDs "work", some sort of series resistance is needed to regulate the current. Because of this, I would not recommend direct connection to a regulated supply. I imagine the original mains supply worked by limiting the current available to a value safe for the string of LEDs. The OP doesn't say how many LEDs there are in the string, but I would expect them to be wired in a series/parallel combination.

Whilst wiring the strings in parallel will result in less variation of LED current, hence brightness, with variation in battery voltage - the originally suggested series LED strings and resistor is probably best.
 

pteron

Member
Joined
1 Dec 2004
Messages
870
pteron.org
Because of the way LEDs "work", some sort of series resistance is needed to regulate the current. Because of this, I would not recommend direct connection to a regulated supply.

You need to regulate the current, not the voltage. With a 3 terminal regulator such as an LM317, you put a resistor in the output and take the feedback from after the resistor. It's better than simply using a resistor as you get the current you want at all input voltages (within reason).
 

Cymrogwyllt

Well-known member
Joined
30 Dec 2010
Messages
10,983
Visit site
Please feel free to correct me.

I was taught that a potential devider as a rule of thumb needs 10 times the pulled out current to flow through it in order to be close to the mark.

The current through the divider should then be of the order of 4A. That suggests a total resistance of approx 3.5 Ohm in the combination and a lower resistor of about 1 Ohm and 1.7W power. I assume that the connection will be taken between the 0V and the centre connection.
 

pteron

Member
Joined
1 Dec 2004
Messages
870
pteron.org
Please feel free to correct me.

I was taught that a potential devider as a rule of thumb needs 10 times the pulled out current to flow through it in order to be close to the mark.

The current through the divider should then be of the order of 4A. That suggests a total resistance of approx 3.5 Ohm in the combination and a lower resistor of about 1 Ohm and 1.7W power. I assume that the connection will be taken between the 0V and the centre connection.

You'd be mad to do it as a potential divider for just that reason - you are throwing away power in the divider. Plus, LEDs don't want a fixed voltage, they want a fixed current, so using a PD is the wrong way to go about it.
 

pteron

Member
Joined
1 Dec 2004
Messages
870
pteron.org
The parallel option also has an issue that the two strings may not share current equally and one string would end up brighter than the other.

The way to do it is on p18 of this data sheet http://www.national.com/ds/LM/LM117.pdf

The 1A current regulator. Replace R1 with a 6 ohm (0.25W) resistor and you've got your 0.2A.

Use this with the two strings in series and you lose less power in the regulation vs the parallel combo and so your efficiency is as high as it can be without using a switching converter. It will also protect you from load dump and other nasties.
 
Top