two transistor on off switch--what is goind on?

[chimera_786]

Hello. So, its been sometime since I messed around with transistor circuits. I figured it'd refresh my knowledge. This is what I intended ( I have attached a schematic):

I have wired up two transistors so that an led is always on when a push button is open. I have accomplished that with PNP transistor in a simple pass configuration. Now, the second transistor, an NPN, should only light up another led when the push botton is pressed and should remain lit till the time the button is pressed.

I wired it up, but the thing is that, when i press the button, the pnp led remain on but dim and the npn led lights up. When I let the button  go, the pnp leds becomes bright again and the npn led goes out (as intended).

Where am I going wrong with this basic set up? I really am amazed how easy it is to forget basic transistors circuits (for me atleast)

For image purposes, please see attached schematic. THANKS!!!!!!

[cybergibbons]

The current path for D1 is through D2 isn't it?

[chimera_786]

Nevermind, I fixed it! YAY!!! It was a stupid mistake--> I was not biasing the base of the npn at all .. silly me. I have attached the scheatic of the correct version! Feel free to use and build upon for future needs.

[jimmc]

Still not good I'm afraid.

With the switch open the lower pole will be clamped to approximately 4.3v (5v - V_BE) by the base-emitter diode of Q3.
Disconnect the base of Q1 and look at the voltage at the junction of  R3 & R5...
Switch ON = 5(200/(200+1000)) = 833mV
Switch OFF = 4.3(200/(200+1000)) = 717mV
Now 833mV is too low to guarantee turning Q1 hard OFF at low temperatures
(More correctly I should say that 833mV behind 167ohm (200x1000/(200+1000)is too low to guarantee sufficient base current to saturate Q1)
and
717mV is too high to guarantee not turning Q1 partially ON  at high temperatures.

V_BE changes by approx -2.5mV/^oC so a 50^oC step would change V_BE by more than operating the switch.

If you want to improve the margins of your circuit, try connecting R1 between the base of Q3 and the switch, change its value to 10k and also increase  R3 to 1k.

Jim

[MikeK]

Why not just remove the connection between the bottom pole of the switch and the base of Q3?  I missed that you wanted the second LED to stay lit until another press of the button.

Dave has a recent blog episode where he uses a switch as a push-on/push-off.

[chimera_786]

Still not good I'm afraid.

With the switch open the lower pole will be clamped to approximately 4.3v (5v - V_BE) by the base-emitter diode of Q3.
Disconnect the base of Q1 and look at the voltage at the junction of  R3 & R5...
Switch ON = 5(200/(200+1000)) = 833mV
Switch OFF = 4.3(200/(200+1000)) = 717mV
Now 833mV is too low to guarantee turning Q1 hard OFF at low temperatures
(More correctly I should say that 833mV behind 167ohm (200x1000/(200+1000)is too low to guarantee sufficient base current to saturate Q1)
and
717mV is too high to guarantee not turning Q1 partially ON  at high temperatures.

V_BE changes by approx -2.5mV/^oC so a 50^oC step would change V_BE by more than operating the switch.

If you want to improve the margins of your circuit, try connecting R1 between the base of Q3 and the switch, change its value to 10k and also increase  R3 to 1k.

Jim

Thanks for the heads up jim. I did tweak my values based on your suggestion and the transistors work the way they were intended.