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I am trying to create an SR Latch with two NOR logic gates that lights an LED. I am implementing the gates using 2 transistors and 2 resistors each.

Here is the schematic:

schematic

simulate this circuit – Schematic created using CircuitLab

The goal is for Switch 1 to be the Reset button and for Switch 2 to be the Set button. My concern is if there will be enough current to light up the LED (I am hoping to get 5mA-20mA flowing through it).

I am confused on how to find the resistance and current because different paths split into multiple other parallel paths. (i. e. the path that \$R_2\$ is on splits into two, and one of those paths splits into two more)

I know that \$\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + ...\$ and that \$I = V(\frac{1}{R_1} + \frac{1}{R_2} + ...)\$ But how do I apply this to the different paths of the circuit?

I would like to know how to find how much current is flowing through a component and how much resistance is on a path in schematics similar to this one (meaning they use logic gates implemented similarly).

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    \$\begingroup\$ The forward biased base-emitter voltage of an NPN silicon transistor will be about 0.7 volts (same as a silicon diode), so the LED can never light - it would require about 2 volts, depending on colour. \$\endgroup\$ Commented Dec 10, 2017 at 22:50

2 Answers 2

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Move the LED. Add a current-limiting resistor in series with the LED, and place that series R_LED combo in parallel with R2. Try 100 ohms in series with LED.

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Trying to reinvent the SR latch wheel?

There are a few problems not worth solving with your design.

  • Vbe4 < Vf of D1 so D1 never turns on. Even if resolved that there is no reason to use so many discrete parts when 2 NAND or 2 NOR gates will work as an SR latch.

in any case assuming LED was on If= (5V-Vf)/3.3k


schematic

simulate this circuit – Schematic created using CircuitLab

If using standard 74HCxxx logic, I might derate 20mA LED to 12 mA and calculate R for 12 mA using datasheet for 74HC00 NAND with Vol at 4.5V for V+.

This is the classic SR latch using 2 NAND gates with active logic low inputs. (!)

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  • \$\begingroup\$ Thanks for your answer. I just wanted to make it for learning purposes. I guess I just don't understand how to find Vbe4. \$\endgroup\$
    – alexm
    Commented Dec 10, 2017 at 21:58
  • \$\begingroup\$ Vbe4 is about 0.7V when biased on and 0.2V when off from Q1 or Q2 s o the LED Vf of ~2V for R/Y or ~3V for G/B/W would never work. \$\endgroup\$
    – D.A.S.
    Commented Dec 11, 2017 at 2:51
  • \$\begingroup\$ You can examine 1970 datasheets for discrete TTL NOR gates to look at datasheets or web images for SR FF , here dual gate SR latch and 2 more gates turns into JK FF electronics-tutorials.ws/sequential/seq13.gif It is better to learn how it was done than guess. \$\endgroup\$
    – D.A.S.
    Commented Dec 11, 2017 at 2:53

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