Latch circuit with transistors which is initially on? [closed]

Question

I just learned about latch circuits and built the one displayed here in Fig 75. Latch Circuit (B):

The LED is initially off. How can this be modified to be initially on?

Answer 1

The two transistor together form something like a SCR, and are either on or off. There must be a wide enough hysteresis threshold between on and off so that the circuit keeps its state when both switches are open.

Since the initial state with no power applied is no current, it is easier for the circuit to start up with no current, meaning the off state. It will be difficult to have the circuit reliably start in the on state, but otherwise have the condition of both switches open preserve the state. Since the circuit powers up with both switches open, and must then turn on, how does it not turn on when both switches are open after power up?

There are some alternatives:

1. Use a capacitor to provide initial charge to turn on one of the transistors (which will turn on the other) when the power supply first comes up. This will rely on some minimum dV/dt of the supply. Put another way, this won't work if you bring up the supply slowly.

2. Use another transistor to invert the output of this circuit.

3. Use a different two-transistor bi-stable topology where one transistor is on and the other off. Now you can connect the LED to whichever one you favor a little so that it powers up on.

   Here is a example of such a bi-stable circuit:

   

   Let's say Q2 is on and saturated, so has 200 mV C-E. Assuming green LEDs that drop 2.1 V, that leaves 2.7 V across R1, which means the current thru it will be 5.7 mA. The collector of Q2 is too low for Q1 to turn on, which would require about 2.1 V for the LED and abou 700 mV for the B-E drop. Since Q1 is off, the base of Q2 is connected to 5V thru R2 and R3. Again assuming 2.1 V LED and 700 mV B-E drop, the base of Q2 is at 2.8 V, leaving 2.2 V across R2+R3. The Q2 base current is therefore (2.2 V)/(R2 + R3) = 210 µA. That means Q2 must have a minimum gain of (5.7 mA)/(210 µA) = 27, which many small signal transistors can achieve at this voltage and current.

   To allow forcing each of the two possible states, add momentary switches between the transistor bases and ground.

   To have the circuit power up in a predictable state, add a little power supply bleed to one side. But make sure this is not so much that the circuit flips back to that state after having been deliberately switched to the other state by momentarily shorting the base of a transistor to ground.