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Quiz Show answer box --- Why won't second bulb come on?

Alektron

Jan 14, 2018
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Jan 14, 2018
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This is a schematic from a book called " Electricity and basic Electronics by Stephen R. Matt".

I'd like to build this Quiz Show answer box, but I'd like to understand how this thing works first.

This is the schematic from the book.
1.JPG

Book says that only one light should lit up, even if the second switch is pushed in too.

What I don't understand --- How come only one light, lights up, even when the second switch is pushed in? That's where I'm confused.

I understand what happens if S2 is pushed in, Red current goes thru bulb I1, R2, S2, turns on Q2 at the Base, then the Collector and the Emiter is connected, then Q2 starts to draw the yellow current path, and I2 bulb lights up.

I1 bulb does not light up because R2 is limiting the red current passing thru it which is not enough to make the bulb glow, although it must go thru the bulb otherwise Q2 would not function as a switch.

OK, so far so good; however. ( next photo)


Bulb 2 lighting up.JPG


So let's say S1 is pushed in too, but just a fraction of a second after S2. You can see that S1 draws the green current across R1 to the Base of Q1, then on Q1 the Collector and the Emiter are connected and the blue current starts to flow from bulb I1.

This is what I don't understand. Why wouldn't Q1 (the blue current path) draw enough current to light up bulb I1.

Is it because R2, which is in parallel with the blue current path is limiting the current flow thru bulb I1?

Bulb 3 lighting up.JPG
 

AnalogKid

Jun 10, 2015
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When S2 is pressed, Q2 saturates. This makes the voltage at the Q2 collector approx. 0.1 V. The Q2 collector also is the voltage source to turn on Q1 when S1 is pressed. BUT, if S1 is pressed second, there is only 0.1 V available to turn on a transistor that needs 0.6 V at the base. Thus, second transistor's base is "shorted to ground" by the first transistor's being saturated.

Note that while the circuit looks like a 2-trahsistor flipflop, it is not. The first response is not latched; it is shown only as long as the first switch remains closed. If S1 (the second switch) still is closed when S2 opens, I2 goes out and I1 comes on, and now I2 is locked out.

ak
 

Alektron

Jan 14, 2018
111
Joined
Jan 14, 2018
Messages
111
When S2 is pressed, Q2 saturates. This makes the voltage at the Q2 collector approx. 0.1 V. The Q2 collector also is the voltage source to turn on Q1 when S1 is pressed. BUT, if S1 is pressed second, there is only 0.1 V available to turn on a transistor that needs 0.6 V at the base. Thus, second transistor's base is "shorted to ground" by the first transistor's being saturated.

Note that while the circuit looks like a 2-trahsistor flipflop, it is not. The first response is not latched; it is shown only as long as the first switch remains closed. If S1 (the second switch) still is closed when S2 opens, I2 goes out and I1 comes on, and now I2 is locked out.

ak

Thank you AnalogKid
 
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