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Lets say I need an LED to indicate when a DC voltage reaches 5 volts or more. I don't want the LED to be dim when it is at 4.5 or anything. I just want it to be full bright at 5, 0 bright at anything below 5. Is there a way I can do this with discrete components (no IC's, just resistors, capacitors, transistors etc.)? I do not require the input has the same power source as the meter.

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    \$\begingroup\$ Why the discrete component requirement? I know you want to understand all of this stuff at the lowest level, but this can literally be done with a single op amp, a pot, a resistor and LED. \$\endgroup\$
    – Matt Young
    Commented May 11, 2013 at 20:47
  • \$\begingroup\$ Maybe an op-amp would work, I only have 2 though and I want to build a bunch with different voltages to trigger the LED. I guess I could buy some if an OP-amp has many advantages over other components \$\endgroup\$
    – skyler
    Commented May 11, 2013 at 21:53

3 Answers 3

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To narrow the voltage range over which the LED turns on, you can use a transistor in order to create voltage gain, to drive a second transistor which drives the current into the LED.

schematic

simulate this circuit – Schematic created using CircuitLab

Results from simulation, where the input ramps from 0V to 8V over the period of a second:

Input Voltage:

Input Voltage

LED Current:

LED CUrrent

The LED goes from zero to fully on over a range of something like 80 mV.

Zener diode D1 biases the input so that the turn on takes place at around 5V.

Note that only about 150 microamps of current are required across R1 to develop 0.7V across Q2's EB junction. That small amount of current does not require the BE junction of Q1 to be fully turned on! This is why the Zener value isn't based on 5V - 0.7V. From the simulation, the transition occurs between 5.12V and 5.2V, and not starting at 5.3V as the approximate junction model would suggest. Higher values of R1 do not significantly improve the upper 5.2V voltage, but degrade the soft turn on voltage. The next available lower Zener model in the simulator is a 4.3V unit, which leads to a turn on that falls short of 5V. Raising that toward 5V requires lower values of R1, which cause Q1 to pull as much current as Q2.

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  • \$\begingroup\$ You can decrease the 80mV further if you change Q2 for a Darlington. \$\endgroup\$
    – jippie
    Commented May 12, 2013 at 9:05
  • \$\begingroup\$ @jippie Good one. I played with different transistor models. Using a power one like 2N3055 also decreased it quite a bit. But that is silly in real life. :) \$\endgroup\$
    – Kaz
    Commented May 12, 2013 at 9:17
  • \$\begingroup\$ Remember to add a heat sink if you go for the 2N3055, it looks cool ;o) \$\endgroup\$
    – jippie
    Commented May 12, 2013 at 9:18
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Didn't test it, but it should work. The SCR latches when its gate current exceeds a certain value. Notice that the LED won't turn off unless V(much) is interrupted.

schematic

simulate this circuit – Schematic created using CircuitLab

Sizing is very depending on the thyristor, leaving that to the reader.

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I would use a comparator, which of course is an op amp. You can get a quad package like the LM339 to do 4 at a time. Resitors or zener diodes can set different on voltages.

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  • \$\begingroup\$ So how Would I do this with a tlo82cp op amp 5v detection \$\endgroup\$
    – skyler
    Commented May 12, 2013 at 2:47
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    \$\begingroup\$ -1: OP asks for a solution without ICs. \$\endgroup\$
    – user17592
    Commented May 12, 2013 at 6:46
  • \$\begingroup\$ Sorry about the IC answer. Got off track when someone mentioned it. You can get close with a zener diode in series with an led, but it will be very dim as the voltage nears the switch on voltage. To use an op amp, you would have the reference voltage set on the negative input and the voltage to measure on the positive input. The refernce voltage can be created with a zener diode. When the positive input goes higher than the negative, the op amp will turn on. Sorry I don't have schematic drawing software installed right now. \$\endgroup\$
    – BrianK
    Commented May 13, 2013 at 13:19

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