Electronics- 9v battery to 5v output

Hi guys,

I’m new to electronics and would appreciate some guidance.

I’m trying to ‘hack’ a LED Fire-effect light bulb to work on a 9v battery to make it portable.

Using a multimeter I can see that the bulb converts mains voltage to 5v.

I’ve tried the following:

  • I’ve used a 5v LM7805 voltage regulator, this works well until the battery gets to around 6v which is below the regulator’s limit - causing the Circuit to stop with more than half the battery charge left.

  • I’ve used a MP1584EN buck converter, but have found that the voltage output doesn’t say a constant 5v as the battery drains - it looks like the converter fractions up the outgoing voltage (so if 5v is 55% of 9v, when the battery is at 6v the converter is delivering 3v)

Is there a way to get a constant 5v output from a 9v battery which uses up as much of the battery charge as possible? (I don’t want to be throwing away half-full batteries thinking they are empty)

Thanks for your help guys, and look forward to meeting you at your next open evening.

Martyn

With linear regulators, you want to look at data sheets to find one with the smallest ‘dropout voltage’. Dropout voltage is the minimum voltage difference between the input and output. There are better performing, but less common, types than the LM7805. If you do a search on a component supplier like Farnell or RS you should find some.

A buck converter steps down voltage. You want a buck-boost converter if you want a constant output with the input voltage ranging above and below the output voltage. This module purports to do what you want it to, although I do not have any experience with it:

https://www.proto-pic.co.uk/pololu-5v-step-upstep-down-voltage-regulator-s7v7f5.html

may sound daft but to keep the Buck happy, 2 x 9v batteries delivering 18v to it may actuality be better, especially at the lower end of the curve.

The Buck is defo the best route BTW, Liniers give off a good % of the energy as heat

Courty

Go with the buck converter if you want to get more out of your batteries. What sort of battery are you using? The higher the internal resistance the lower the voltage you’ll get from them under load, particularly when they’re at the point of near depletion. 9v Zinc carbon batteries can have an internal resistance as high as 35 ohms, whereas 9V alkaline batteries as low as 1 ohm. I’d definitely try what Paul suggested and try two or even three in series in any case since that chip can take up to 28v.

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Hi Guys
I’ve been doing some experimenting with a few difference regulators the S7V7F5, S9V11F5 & S7V8F5 and i come across the same issue.

When it’s a ‘fresh’ 9v battery the bulb works fine, however when the battery drops to around 7v (the output from the regulator remains at 5v) the bulb dims to more than half brightness.

What am I missing/should be looking out for?

Thanks for you help guys, what i hoped would be an easy starter ‘hack’ is still puzzling me.

Martyn

9v -

7v -

(The photos doesn’t really show up the brightness difference, but you can see if on the reflection on the battery)

You are missing the chemistry of the battery.

Batteries are weird beasts , you can workout how much charge is left in a battery by the voltage their output.

A battery with 7v doesn’t mean is charged , your led are facing because there is not much capacity left in to the battery.

Eevblog have a series of videos about batteries and why the presence of voltage doesn’t mean a battery is charged

Worth also noting that 9V batteries have relatively low power density compared to other batteries, they are kind a con battery.

So, did you try alkaline batteries in parallel? Also, are you measuring 5v from the regulator without a load on it? You should try taking a measurement with the batteries connected to the LEDs, what voltage do you get then? You should also find out how much current the strip needs when powering all those LEDS at the preferred brightness (use a bench power supply), I reckon it’ll be too much for any 9v battery.