I've got 3 sets of solar powered LED lights, 20 warm white LEDs each. Water ingress into the circuit boxes is proving a pain so I'd like to remove the 3 solar circuits and replace with a single mains PSU in a waterproof housing.

Each string is currently powered by a single 1.4V 1400mAh NiMh battery and I suspect the LEDs are wired in parallel, I don't see any limiting resistors in the string. I'd like to use a USB 5V ~1A phone charger.

I have tried various LED calculators but my knowledge of electronics is limited and the LEDs are fixed in place which makes experimentation difficult (up a ladder).

I'd be most grateful for your thoughts.

These are the lights:
http://www.amazon.co.uk/Innoo-Tech-Powered-Outdoor-Christmas/dp/B008O18EIC

Cheers.

 

You have to measure the current, and then to decide the resistors.

The normal 1.4V battery may not light up the LED, does the LED string has any other power device?

The USB 2.0 only can be offer 500mA for each port, USB 3.0 can be offer 2A 10W for each port.

 

Thanks for the response.

You have to measure the current, and then to decide the resistors.

This could be tricky as none of the solar panels are working so none of the lights come on.

The normal 1.4V battery may not light up the LED, does the LED string has any other power device?

There is a small circuit board with an 8 pin IC and a few discrete components. I'm guessing this both charges the battery from the solar panel and adapts the output from the battery to drive the LEDs. There isn't any power other than the rechargeable battery. You are also correct that connecting a battery to the led light string doesn't make work.

The USB 2.0 only can be offer 500mA for each port, USB 3.0 can be offer 2A 10W for each port.

It's a USB charger (rather than an actual USB port) so the 1A is the output rating given on the power supply.

 

Could you attach the pictures of the LEDs string and the circuit board, and draw a block diagram to describe and shows that how is the circuit connection?

 

Can you identify the polarity of the LED string? Assume the string draws 400 mA, LED about 3V with 5V supply need to drop 2V = 5Ω, 2W. If strings are not in parallel, they will not light, if connected in reverse, they will not light, if all of the guesses are right, should have 20 bright LED's.

 

It is highly unlikely that they are drawing 400mA at 3V. If they do, they would last at most 1 hour on a full charge. I would assume considerably less current than that. Todays high brightness LEDs can be quite bright at as little as 1mA.

Bob

 

1400/ 400= 3.5 hr. WE could also assume 15 mA for 4.7 hr.but with line operated power supply they will stay on untill manually shut off unless OP wants to add a dusk to dawn or timer.

 

ΩThanks for the help. All of the parallel LED circuits I've found online have a current limiting resistor for each LED but I'm not going to be able to do this as the LEDs are already in strings with no resistors. Is it possible to use a single high wattage resistor or could it be that the LEDs have some form of current limiting built in?

Manufacturers claim >6 hours of light from full charge so I guess that gives a draw of about 200mA for the entire circuit.

I've got one of the boxes down but the IC has had it's numbering removed. However I can see a 0.5W 33Ω resistor and a couple of 0.25W res. and a small cap.

 

There are lots of parallel LED ckts out there in the early LED flashlights. I have several that also operate on 3 AA's, 4.5V on 3V LEDs & not a failure yet in over 20 years. Decide on how bright you want them & use resistor from 5Ω to 10Ω, 2 to 3 W.

 

1400/ 400= 3.5 hr. WE could also assume 15 mA for 4.7 hr.but with line operated power supply they will stay on untill manually shut off unless OP wants to add a dusk to dawn or timer.

Not when you are running >3V LEDs from a 1.2V battery. There is a boost converter in between and you need to draw 3X the current from the battery as the LEDs will need. Hence my rough calculation of 1 hour if they were drawing 400mA. If the manufacture says they will last 6 hours, then you can only draw about 233mA from the battery and only deliver about 70 mA to the LEDs. My point was that you assumption that 400mA was going to the LEDs might lead to a nice short flash.

Bob

 

Do not believe OP is using any part of old ckt.

 

It doesn't matter. The point is that the old circuit was putting at most 70mA through the LEDs, and you are suggesting a resistor that would put 400mA. This might very well work, and the LEDs would be much brighter than the were originally, but it might also blow them all out.

IXP