Help with resistors

[starscream]

Hi all. I'm planning on adding an LED indicator to the circuit of my receiver so i can tell when it is on. I want to run it off 2 9v lithium batteries because the size of them will fit well in the enclosure. Firstly the LED indicator runs on 2v dc and draws 20mA so i need to reduce the 18v coming from the batteries down to 2v. Secondly i need to reduce the remaining voltage leading to the receiver down to 9v.

So my question is 1. will my circuit work?
2.what type of resistors will i need?

Sorry if my diagram doesn't make sense, i am not very familiar with electronic circuits.
I really appreciate any help. Thanks!

 

[Martaine2005]

Hi starscream,
My question would be why two batteries in series if you only need 9 volts?
Just wire them in parallel and you'll have 9 volts.. Plus twice the run time.
Then just use an automotive illuminated rocker switch.. Job done..
Unless there are reasons you can't do this?

Martin

 

[starscream]

I thought that the LED would draw away some voltage from the 9 v battery and only supplying 7 v to the receiver.

 

[Anon_LG]

Just wire them in parallel and you'll have 9 volts.. Plus twice the run time.

You can usually do this, only under certain conditions. Either the batteries must be exactly matched, as in same chemistry, make, model, charge level and at least approximately the same counts of charge-discharge cycles. This is even more so crucial when working with lithium's, as a reverse current caused by voltage imbalance may cause a catastrophic failure, resulting in heating, burning and possibly explosion. Or a diode to ensure one way current flow must be included, the diode should prevent one battery from running current through another. It is possible however that your lithium's have inbuilt blocking diodes to ensure this occurs anyway.

Nonetheless, the batteries do not need to be in series.

 

[Anon_LG]

I thought that the LED would draw away some voltage from the 9 v battery and only supplying 7 v to the receiver.

Your LED is in parallel, parallel components always have equal voltage across the parallel branches of the circuit. Current is instead divided between the branches, the proportions of which depend on the sum of resistances of each branch. In series, you statement would be true however, voltage is divided between the components and the semiconductor junction of the LED would cause a voltage drop, anywhere from about 0.8 volts up to 2 volts.

 

[starscream]

Oh i see! So if i used just one 9v battery and wired an LED rated at 9 v parallel in the circuit then i shouldn't need any resistors?

Problem is i only have LED's rated at 2v, so in that case, will i need a resistor only for the LED?

Thanks.

 

[cjdelphi]

The LED is current driven, you can use any voltage over 2v however, the voltage drop across the resistor is then 7v, lower the voltage the less waste, the primary reason for the resistor is to limit the current to stop it exceeding 20ma

 

[BobK]

Lavaguava's info about paralleling batteries is not correct.

Once batteries of the same chemistry are paralleled they cannot have different voltages (since the terminals are tied together) so most of what he said cannot happen. If one discharges faster than the other it will stop discharging until the other catches up.

The only issue is that they must be at the same state of charge and voltage when you connect then together. Otherwise, the higher voltage one will charge the lower voltage one at perhaps too high a current.

Bob

 

[starscream]

Thanks for the input guys, got my circuit working perfectly so that's good. Plus i know a bit more about circuits now. Really appreciate it.

 

[Herschel Peeler]

Just because the LED is rated at 20 mA does not mean you MUST draw 20 mA through it. How much brightness do you really need? It may work just fine for you at 5 mA. Since this is battery powered you will save a bit of battery life at 5 mA.
9 V - 2 V = 7 V. At 5 mA means about a 1500 ohm resistor, 1/4 watt is fine.