Need help determining resistor values for a voltage divider

Thread Starter

Quahetus

Joined Mar 5, 2018
5
My Daughter and I are building a "Battery Buggy" for Science Olympiad. We have the power source (12v battery pack), switches, and motor all wired up and working. Now we want to add a laser pointer to help with alignment. The rules state the laser pointer must be powered from the same 12v battery pack as the motor. Rules also limit electronic components to batteries, switches, relays, potentiometers, resistors, and motors. The laser pointer requires 4.5vdc and pulls 50mA. I need to get the 12vdc battery pack to 4.5vdc. I figured the only way to do it with the component limitation is to build a voltage divider with two resistors. Problem is, I don't remember how to determine the resistor values. Can anyone help with this? I am attaching the circuit diagram.
 

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dl324

Joined Mar 30, 2015
8,913
Welcome to AAC!

If you use a voltage divider, the current in the divider needs to be 10 times the load current. That's half an amp.
 

ericgibbs

Joined Jan 29, 2010
8,733
hi Q,
Do you have a type number or datasheet for the laser pointer that you could post.?
E
 

Thread Starter

Quahetus

Joined Mar 5, 2018
5
Hi E,
Sorry, I do not. I can tell you it was designed to be mounted on a handgun and operate on 3 LR44 batteries. I removed the batteries and connected it to my variable power supply. I set the power supply to 4.5v and the ammeter reads 50mA while it operates.
 

ericgibbs

Joined Jan 29, 2010
8,733
hi Q,
I would suggest a 7805 [ 5V regulator IC] and a series diode say a 1N4001, to drop the Voltage to ~4.3V for the pointer.
Are you able to get those components.?
E
 

DickCappels

Joined Aug 21, 2008
5,882
I endorse the the use of a 5V regulator (LM7805) -put a diode in series with the output to drop it down to about 4.4V if you are concerned about the voltage being too high (It won't be).

If you have to do it resistively you can put a 1.5K 1/2 watt resistor in series with the battery if that is permitted. Measure the voltage across the laser pointer and if necessary, change the value of the resistor until it is close enough to 4.5 volts. I

If you must use a voltage divider, you can add a 1 meg resistor across the laser diode. It doesn't do anything but increase the component count ;-)
 

Thread Starter

Quahetus

Joined Mar 5, 2018
5
hi Q,
I would suggest a 7805 [ 5V regulator IC] and a series diode say a 1N4001, to drop the Voltage to ~4.3V for the pointer.
Are you able to get those components.?
E
Good suggestion, but the rules prohibit us from using semiconductors.
 

Thread Starter

Quahetus

Joined Mar 5, 2018
5
I endorse the the use of a 5V regulator (LM7805) -put a diode in series with the output to drop it down to about 4.4V if you are concerned about the voltage being too high (It won't be).

If you have to do it resistively you can put a 1.5K 1/2 watt resistor in series with the battery if that is permitted. Measure the voltage across the laser pointer and if necessary, change the value of the resistor until it is close enough to 4.5 volts. I

If you must use a voltage divider, you can add a 1 meg resistor across the laser diode. It doesn't do anything but increase the component count ;-)
Good info. The use of the divider is not required. Doing it that way was what I assumed was the "correct" way. If a single series resistor accomplishes the voltage drop, that seems to be the easiest solution.
 

DickCappels

Joined Aug 21, 2008
5,882
Expect to see a little voltage drift as the laser diode warms up. It probably be small but if you consider it a problem a 5V regulator is a good fall-back solution.
 

MrChips

Joined Oct 2, 2009
19,275
Use Ohms Law for calculating the series resistor required.

Resistance
R = V / I = (12V - 4.5V) / 50 mA = 150Ω

Power dissipated is
P = I x V = 50mA x (12V - 4.5V) = 0.375W

Use a 150Ω ½W resistor.
 

Ylli

Joined Nov 13, 2015
776
LR44 batteries have a significant internal resistance. Depending on who's data you use, the internal resistance is probably somewhere around 10 ohms for each battery. So the design power source for this LED is 4.5 volts with a 30 ohm series resistance. IOW, 50 mA may be too much current for this LED.

I'd suggest a more conservative 20 mA, in which case you will need a series resistor of about 400 ohms. I'd go with a 390 ohm 1/2W.
 

Thread Starter

Quahetus

Joined Mar 5, 2018
5
Got it working! Ended up with a 250 ohm in series and a 1M ohm across the laser pointer. Works well even as the battery pack gets low. Thank you all for your help!
 
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