Some power supply circuit questions...

[Jess]

Hi, I have some basic questions about a circuit I want to try:

IR led and laser, 14500 batteries, drivers are the same as the DDL but use LM1086 for lower dropout.

It's a ground interrupt via rotary switch. 4 modes: led, led & laser, laser, off.

1 - is it "kosher" to use the batteries as independent power supplies but with a shared ground? From an engineering standpoint, is there a reason to not do this?

2 - the 1N4001 diodes are to prevent either load from finding a ground when they should be off. Ex: the picture shows the circuit in Laser-Only mode, and the LED remains off because the diode in the LED bus blocks the "alternate ground pathway". ...I think. Will a diode provide that function on the "back side" of a circuit?

3 - the diodes have a .7v dropout. Theoretically, there's only a 0.3v potential remaining "after" the laser. Does this mean that the diode will rob 0.4v from the laser, or will the laser still have it's full 2.0v?

Sorry if my questions are "basic", but I'm not an ee, just a hack.

Thanks!

 

[sk8er4514]

Hi, I have some basic questions about a circuit I want to try:

IR led and laser, 14500 batteries, drivers are the same as the DDL but use LM1086 for lower dropout.  

It's a ground interrupt via rotary switch.  4 modes: led, led & laser, laser, off.

1 - is it "kosher" to use the batteries as independent power supplies but with a shared ground?  From an engineering standpoint, is there a reason to not do this?  

2 - the 1N4001 diodes are to prevent either load from finding a ground when they should be off.  Ex: the picture shows the circuit in Laser-Only mode, and the LED remains off because the diode in the LED bus blocks the "alternate ground pathway".  ...I think.  Will a diode provide that function on the "back side" of a circuit?

3 - the diodes have a .7v dropout.  Theoretically, there's only a 0.3v potential remaining "after" the laser.  Does this mean that the diode will rob 0.4v from the laser, or will the laser still have it's full 2.0v?  

Sorry if my questions are "basic", but I'm not an ee, just a hack.    

Thanks!

1- should be ok.

Looking at your circuit though I would recommend having the switch on the positive side of the battery.. it just seems weird to me to have the switch on the negative side.. maybe thats just my personal preference though, I can't think of a reason why it would matter electrically but I've never done anything with supplying negative to the switch.

Also I have a feeling the LED battery will run out of juice much quicker than your laser one.. just fyi

2 (and some 3)- yeah I think so. You might want to try the laser only circuit with the diode in it, I think it may be quite dim as the driver or laser might not get enough voltage off your 3.6V batteries. Calculating out your diagram on that circuit it adds up to 4V required..... I know that my 10440 3.6v batteries when fully charged actually measure 4V, so your 14500 batts might be the same and wouldn't be a problem. just try it out.


If you want to get into even more design, check out PSpice here: http://www.electronics-lab.com/downloads/cnt/fclick.php?fid=513

you can create your circuit and simulate it and see all the electronic characteristics in action. might be a bit tricky to design ur LED and laser tho, not sure.

thanks,
Kendall

 

[Tw15t3r]

1.
Having two power supplies with a shared ground is common in engineering to save wires mainly, while still having isolated power supplies, i.e. the power supplies are independent of each other.
However, in your case, as the LED takes up more power than the laser, the battery powering the LED will definitely drain faster. If you're ok with that, then it's alright, but if not, you should connect the two batteries in parallel as one power supply.

2.
Diodes function the same at any part of the circuit, be it at the back or front. As long as there's no way to bypass the diode in your circuit, the diodes will perform as you have intended.

3.
Yes, there will be "insufficient" voltage to your laser, because the voltage available to the laser after the driver is 2.3V, hence the laser will share the voltage with the diode according to their ratio, i.e. 2:0.7. Which works out to be 1.7v to the laser, and 0.6v to the diode.
I'd suggest germanium diodes or even schottky diodes if you can get them, because their dropout voltage is much lesser. Around 0.3v.

 

[Jess]

Awesome!

Thanks for the great answers.

I will find some lower dropout diodes.

Thanks!

 

[Warske]

... in your case, as the LED takes up more power than the laser, the battery powering the LED will definitely drain faster. If you're ok with that, then it's alright, but if not, you should connect the two batteries in parallel as one power supply.

That works, but be aware that batteries wired in parallel will self-discharge faster when not being used. This is because temperature changes and gradients cause one battery to always have a slightly higher voltage than the other, causing it to charge the other battery. Energy is lost in the process because charge and discharge is not 100% efficient.

It isn't clear to me why you could not use just one battery, since the combined current is very little more than the current for just the LED.

Also, the specs on the LM1086 regulator show a typical dropout of 1.3 volts at 1.5 amps. At only 40 mA for the laser, you might find that the dropout is actually a bit less than 1.3 volts, and you might not have a problem with the voltage lost across the diode. You could wire one up and test it to find out.