# Detection of low dc voltage to power a led

#### Gsalet

Joined Aug 8, 2017
21
Hello

I am working on a project that I need to detect when the circuit is energized, the circuit feeds a dc motor 4.5 volts operates forward and reverse by changing polarity. I want to light a led when in forward and another led when in reverse. I put two leds in parallel with the motor each led polarized opposite of each other, Each led lights depending on rotation of the motor or current flow, only problem with this is the motor speed is reduced due to the reduction in current caused by the parallel led.

What i need is a circuit using a transistor to sense the voltage then acts as a switch to turn the led on via separate circuit. I assume I will need 2 transistors to accomplish forward and reverse.

Thanks for the help
George

#### ericgibbs

Joined Jan 29, 2010
16,388
What i need is a circuit using a transistor to sense the voltage then acts as a switch to turn the led on via separate circuit. I assume I will need 2 transistors to accomplish forward and reverse.
Hi George,
Adding two extra transistors is one option.
Post a circuit sketch, we can check it out.
E

#### crutschow

Joined Mar 14, 2008
30,801
Must be a pretty small motor if it's affected by the small current from an LED (?).

#### LesJones

Joined Jan 8, 2017
3,813
Hi crutschow, I was thinking along the same lines at first but he did not say that he put a current limiting resistor in series with the LEDs or how feeble the 4.5 volt power supply is.

Les.

#### Dodgydave

Joined Jun 22, 2012
10,506
Use a Bi Coloured led and 100 ohms series resistor across the motor, Red one direction. Green the other....

#### Audioguru again

Joined Oct 21, 2019
5,179
I agree that the 4.5V is an almost dead battery that drops its voltage when it is loaded.

#### Gsalet

Joined Aug 8, 2017
21
Yep your right pretty small motor used to rotate a fireplace gas valve,
I did not use a resistor
4.5 volts = 3 - AA batteries
Attached is a sketch (pretty basic) that i used for testing in design stage.(almost sounds like i know what i am doing <GRIN>)
Also note that there is a solenoid in a separate circuit that I was considering putting a led in series with but figure it would drop the voltage

Probably should let everyone know what the entire project is
This is for testing gas fireplaces. My intent is to place my testing device in series with the communication wire between the ignition control module and the gas control valve. My Idea is that if I can verify that the ignition control module is sending power to the gas control valve at the appropriate time, I can determine if the ignition control module is bad and or the gas control valve is bad.
The second step is to send power to the gas control valve to manually put it through its steps to verify operation

There are 3 circuits (1) feeds the motor to turn the gas valve to pilot mode where it (2) actuates a micro-switch that send power back to the ignition control module which (3)energizes a gas pilot solenoid which allows gas to flow to the pilot and then starts the spark ignitor, when the flame lights a detector rod senses the flame holds for 30 seconds then rotates the motor to the low operating position which breaks the micro-switch and we are now in operation.

Sorry for being so long winded
George

#### Audioguru again

Joined Oct 21, 2019
5,179
You show two LEDs connected directly to 4.5V. But a red LED conducts when the voltage is higher than about 1.8V and other colors conduct when the voltage is about 3V, so the LED current will be extremely high with 4.5V. The very high current will kill the LED, the battery or both.
An LED must have a resistor in series with it to limit its current to its recommended current.

If you connect an LED in series with the 4.5V and in series with the solenoid then the LED gets 1.8V to 3V and the solenoid gets only 4.5V - 1.8V= 2.7V or gets 4.5V - 3V= 1.5V. Then the solenoid will not work at such a low voltage.

#### Dodgydave

Joined Jun 22, 2012
10,506

Make R 100 to 220 ohms..

#### Alec_t

Joined Sep 17, 2013
13,153
I hope you've thought through the necessary safety precautions and circuit interlocks to prevent the main gas valve being opened (or kept open) if the igniter fails, the pilot flame goes out, the microswitch contacts weld closed, the motor drive signal fails or any other error occurs?

#### MisterBill2

Joined Jan 23, 2018
13,139
View attachment 196945

Make R 100 to 220 ohms..
This is the first comment that makes sense. Indeed there MUST be a resistor in series with the LED, and as the drawing shows, only one resistor is needed.

#### MisterBill2

Joined Jan 23, 2018
13,139
I hope you've thought through the necessary safety precautions and circuit interlocks to prevent the main gas valve being opened (or kept open) if the igniter fails, the pilot flame goes out, the microswitch contacts weld closed, the motor drive signal fails or any other error occurs?
READ POST #7, where it describes the system and the way it verifies that the pilot light is burning. That same controller also shuts off the gas if the pilot fails.

#### Gsalet

Joined Aug 8, 2017
21
I hope you've thought through the necessary safety precautions and circuit interlocks to prevent the main gas valve being opened (or kept open) if the igniter fails, the pilot flame goes out, the microswitch contacts weld closed, the motor drive signal fails or any other error occurs?
This is just a testing device used when diagnosing the system....
But you brought up a good question. The answer is fire places use a thermocouple to power the pilot gas and a Thermopile to control the main burner creating a double safety so if ether one does not produce enough current at the pilot the whole unit shuts down and cannot restart without going through the whole start up procedure.

Thanks for the help
George

#### RIKRIK

Joined Oct 11, 2019
146
The motor i would imagine, no expert by the way. wouldn't the motor have stop spinning after a certain point, making all leds turn off?. maybe a latch circuit would work?.