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wanting to make a monitor to detect low voltage I'm very new at this. But what I'm needing is something that would light an LED at 7.4v to 8.5v and either change color or just turn off at 3.5v if any one is able to formulate a schematic would be a complete Hero in my eyes. thank you folks for your time

TI 431 figure 9-12 gives an example of a LED turning ON when the voltage is between two given voltages (a window for the acceptable voltages).

Oh very sorry didn't think that one threw to well . I'll try again. Using a charged 7.4v battery LED would be on then go out when voltage drops to 3.5v Basically a low voltage warning as not to deplete batt. below the 3.5v mark

We can't light a LED if the voltage is less than its minimum required voltage, which is generally more than one volt.

Using R1B = 0 ohm, and R1A/R2A = 0.4 (using Vref = 2.5 volt) should turn the LED on for a voltage less than 3.5V (but larger than the LED_on_voltage, or 2.5V, whatever is the maximum) since that will define a voltage window from 2.5V to 3.5V ( or V_LED_ON to 3.5V).

Sure, if you have another voltage source (other than the one tested), you can use an OP-AMP as a comparator, which will be able to return a voltage even if the tested battery is dead. We can also think about using a PMOS where the tested battery will be at its gate, and the second voltage source (around 5V) at its source, so that when the tested battery voltage will drop below Vs - Vgs(th), the PMOS will start turning on, with a voltage high enough to turn on the LED at its drain (if its V_LED << Vs- Vgs(th) ). Vs = voltage of the second source, at the source pin of the PMOS, and Vgs(th) , here assumed a positive value, dependent of the PMOS (the datasheet may indicate a Vgs(th) negative, use the absolute value in the computation). You have to protect the green LED (so it sees less than 20 mA, or whatever is suggested in its datasheet, if any is available), and maybe the second source against the 7.4V of the fully charged tested battery (a Schottky diode may be considered).

I thank you very much for the info much apricated.