Electrical – Some questions about a transistor latch


Below is a latch made up of two transistors:

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When the base current of Q2 is increased this also increases the base current of Q1 which in turn increases the base current of Q2 more. By such positive feedback, both transistors go into saturation and establish an ON switch.

I have two questions regarding this latch:

  1. When this latch is saturated by applying voltage at B1, what happens if B1 is left floating just after that. Will the switch remain saturated?
  2. If the latch is ON i.e saturated, how can it be turned OFF indefinitely besides turning off the Vcc? Would grounding B1 work for that?

Best Answer

1: Yes, it will remain on.

2: Short answer: You can't. Long answer: You can, but it's hard.

This is roughly equivalent to a thyristor, or SCR. There are ways to turn off a thyristor by drawing away current that would otherwise go to the base of Q2 and keep it on, but this is not exactly easy control circuitry to make because you need to route all of the load current through your control circuitry for a short time. And it needs to be all of it, or else the thyristor won't turn off. Shorting the base to ground is one way to do this, but whatever switch is used for that short needs to be rated for the appropriate amount of current, and have a low enough impedance to drop the base voltage almost all the way to zero. It's easier if you just apply a negative voltage to the base of Q2, but you still need high-current switches. Even higher current, in fact.

Note that I may have some details wrong here; it's been a while since I worked with thyristors that were designed to be turned off like this.

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