Voltage Divider question - 4.25v 0.4a required

[Tintin]

Update

So I've had a play today, and as there is nowhere I can find within a 2 hour drive to buy components, and also because I am impatient, I took on board what Uber said about running them in series.

I looked for a ready made solution and found out that the cheap (£8 from Asda) 3 in one charger from Nexxus had a car charger to USB output (that you plug in various tips for different mobiles) that says it gives 5vdc <500ma.

It uses the MC34063a IC from TI (datasheet here) which is a DC to DC chip giving up to 1.5a.

I've roughly wired it into both strings in parallel and the voltage is 4.82 and the combined current is 0.54a.


The lights are (unsurprisingly) a touch brighter.

I'm gonna monitor the heat of the IC, but I think it will cut out if it gets too hot and that hasn't happened yet.

Taking on board the comments about LEDs neededing current limiting, not voltage, is it worthwhile limiting the current below what I am getting?

I've attached a few pics of the lights.

View attachment 11317

View attachment 11318

 

[Tintin]

That sounds like a good idea but wasn't it suggested earlier along with some calcs on resistor values.

 

[Bodach na mara]

I am a bit concerned at the replies that suggest rather low series resistor values based on a current demand of 0.4 A. From the first post, it appears that the spec for the power supply is a capability of 4.5 V amd a power of 0.8 VA, which suggests that at 4.5 volts the unit can deliver just under 0.2 amps. It is not the delivery of the power supply that is important in specifying a series resistor, but the maximum current that the LED string can accept. My experience of LEDs has been with devices with maximum currents in the 20 mA region, about 1/10th of the currents considered in other posts. Now undoubtably there will be a combination of series and parallel arrangement within each string, but you need to find out the limiting current for the string. Once you know that, you can start finding out how to achieve it. You also need to know the forward voltage drop when the string is conducting. Again, you cannot simply use the value for the power supply output.

The quick-and-dirty method is a series resistor with the strings themselves in series. Assuming the current limit is I(lim) and the voltage drop is Vf for each string, and the maximum system voltage is V(max), the safe resistor value is R = (V(max)_ Vf) / I(lim)

In a warning light circuit which I installed recently, the values were:- I(lim) = 20 mA, V(max) = 13.5 Voltd (from regulated DC/DC convertor operating on 24 V system) and Vf = 2.2 V Using these values gave a value for R = (13.5 - 2.2)/ 0.02 = 565 Ohms. Power disipation is just under 1/4 watts so a stock value resistor of 560 Ohms was found to be OK. I would usually go up to the next preferred value (690 ohms) but the 20 mA was a target value, not a limit in this case.

The more elegant solution is a current regulator set to deliver 90% of the limiting current.

 

[VicS]

I am a bit concerned at the replies that suggest rather low series resistor values based on a current demand of 0.4 A. From the first post, it appears that the spec for the power supply is a capability of 4.5 V amd a power of 0.8 VA, which suggests that at 4.5 volts the unit can deliver just under 0.2 amps. It is not the delivery of the power supply that is important in specifying a series resistor, but the maximum current that the LED string can accept. My experience of LEDs has been with devices with maximum currents in the 20 mA region, about 1/10th of the currents considered in other posts. Now undoubtably there will be a combination of series and parallel arrangement within each string, but you need to find out the limiting current for the string. Once you know that, you can start finding out how to achieve it. You also need to know the forward voltage drop when the string is conducting. Again, you cannot simply use the value for the power supply output.

The quick-and-dirty method is a series resistor with the strings themselves in series. Assuming the current limit is I(lim) and the voltage drop is Vf for each string, and the maximum system voltage is V(max), the safe resistor value is R = (V(max)_ Vf) / I(lim)

In a warning light circuit which I installed recently, the values were:- I(lim) = 20 mA, V(max) = 13.5 Voltd (from regulated DC/DC convertor operating on 24 V system) and Vf = 2.2 V Using these values gave a value for R = (13.5 - 2.2)/ 0.02 = 565 Ohms. Power disipation is just under 1/4 watts so a stock value resistor of 560 Ohms was found to be OK. I would usually go up to the next preferred value (690 ohms) but the 20 mA was a target value, not a limit in this case.

The more elegant solution is a current regulator set to deliver 90% of the limiting current.

Sorry but the OP says he has measured one string as drawing 0.2 amps at 4.25 volts

These are the figures used for my calculations. Two strings in series being 8.5 volts, the current still being 0.2 amps

You just need a resistor that will drop the difference between 8.5 and the supply volts at 0.2 amps.

Simple! and you'll find a similar example of calculating series resistors for single LEDs in the Maplin catalogue.

If one had some more detailed specs max voltage perhaps and current at Vmax one would be able to work out a minimum value for the series resistance to run the strings at their max output if required.

There's a sporting chance you could run the two in series without a resistor but without the specs one cannot be sure.. Put 3 in series then you could!

 

[pteron]

So I've had a play today, and as there is nowhere I can find within a 2 hour drive to buy components, and also because I am impatient, I took on board what Uber said about running them in series.

I looked for a ready made solution and found out that the cheap (£8 from Asda) 3 in one charger from Nexxus had a car charger to USB output (that you plug in various tips for different mobiles) that says it gives 5vdc <500ma.

It uses the MC34063a IC from TI (datasheet here) which is a DC to DC chip giving up to 1.5a.

I've roughly wired it into both strings in parallel and the voltage is 4.82 and the combined current is 0.54a.


The lights are (unsurprisingly) a touch brighter.

I'm gonna monitor the heat of the IC, but I think it will cut out if it gets too hot and that hasn't happened yet.

Taking on board the comments about LEDs neededing current limiting, not voltage, is it worthwhile limiting the current below what I am getting?

I've attached a few pics of the lights.

If you feel like a bit of hackery, you could modify the buck converter to work in current mode. Fig 8 on page 8 is the mode it will be working in at the moment. To convert it to current mode, you remove R2, and move R1 to be in the ground of the resistor string with the output of the buck converter connected to the positive of the resistor string. The feedback to pin 5 comes from the connection of R1/diode string ground. To set the resistor value, you use the formula

R = 1.25/desired current in amps

You may need a bit of capacitance on this node to stabilise the loop.

The buck converter is working to make the voltage at pin 5 equal to the internal reference voltage (1.25V). In voltage mode, this voltage is derived from the output through a potential divider made up of R1 and R2. In current mode this voltage is generated across R1 as the current that flows through the string passes through it. Forcing a constant voltage across a resistor turns it into a current source - just what we want.

Edited to add: if you hack it like this, you must permanently connect the LED string - without a load the buck converter will keep on increasing the output voltage to attempt to get 1.25V on pin 5, and will almost certainly try to destroy its output device through overvoltage.

 

[mikefleetwood]

I rather suspect that would be a step or two beyond what the OP would feel confident with. (plus he hasn't got access to a components shop "within a 2 hour drive")