0
\$\begingroup\$

I have a 3x1.5V battery pack (AA alkaline). It's fresh and with no load sits at 4.8/4.9V.

In parallel, I have 3 LEDs, each with their own resistor. That works fine..

I can also switch in a 5V laser diode module. That also works fine..

But if I attach everything at the same time, the laser diode module is lit, but not the LEDs.

I suspect/hope this is some fairly straightforward concept - but if the battery is sitting as it is, everything works well and these are fairly low-power devices, all connected in parallel - I don't understand it.

Can anyone guide me in the right direction?

\$\endgroup\$
4
  • 1
    \$\begingroup\$ We need links to the manufacturer's datasheets (not the vendor webpage) for the LEDs and the laser diode module. Have you measured the voltage available from the batteries with just the LEDs, just the laser module, and with both? \$\endgroup\$
    – Elliot Alderson
    Commented Dec 19, 2019 at 19:46
  • 1
    \$\begingroup\$ How much current is the laser module drawing and what voltage do you have when it is connected? I guess the current draw is to much foor the poor little AA and voltage drops so far that the LED won't light up anymore. \$\endgroup\$
    – jusaca
    Commented Dec 19, 2019 at 19:46
  • \$\begingroup\$ I don't have a datasheet for either component, they have been in my component drawer for about 12 years unfortunately! Edit: but, I got the laser diode module as a batch of 10, which was reasonably cheap I remember. I'd be surprised if it's asking for >30mA... but will do some tests/checks. \$\endgroup\$
    – Kieren Johnstone
    Commented Dec 19, 2019 at 21:09
  • \$\begingroup\$ I've taken some fresh measurements. No load bat voltage: 4.63v. Laser (found one of the batch was marked as "<5mW") connected: drops to 2.2v. Just LEDs connected: drops to 3.2v. I am pretty surprised. What would be the typical solution here, in terms of simplicity to get this xmas gift finished? :) Add another 1.5v cell? \$\endgroup\$
    – Kieren Johnstone
    Commented Dec 20, 2019 at 20:23

1 Answer 1

Reset to default
1
\$\begingroup\$

When analysing non-linear impedance loads there are simple ways of expressing it with superposition to compute the net effects.

To simplify that, I propose all diodes have threshold voltage at say 10% of their current where LED's might be 10% bright and we say their forward voltage and current drawn are related to some threshold voltage Vt to get their rated forward voltage Vf.

That is Vf= Vt + If*Rs where Rs is the bulk series resistance where you can include your fixed R and equate to your battery voltage.

However your battery has a similar series resistance or ESR like Capacitors.

Vbat = Voc-I*ESR for open cct Voc

Once all passive diode loads are defined , a set of linear equations can be created and the net shared current can be computed. Keep in mind all diodes have a huge tolerance (>25%) on Rs, but that has improved a lot in LEDs with recent years with equipment process controls. It used to be like hFE tolerances.

But in your case lets just assume your Laser Vt is less than your Diode Vt and that the combination of your total load current and battery Rs resulted in the Vbat dropping below the threshold of your LEDs which I would guess are not Red.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.