# Push Button Switch Selector 'One hot' (No ICs)

Discussion in 'General Electronics Discussion' started by iimagine, Jul 28, 2019.

1. ### iimagine

55
8
Oct 23, 2013
Only one switch can be on; push button style; extendable.
I was bored. I had to design something! What do you think? Could it get any simpler?

Inside the yellow box is a 2 transistors latch on push button switch circuit. This yellow box can be copied and paste...i mean extended to as many as you like.

2. ### AnalogKid

2,393
665
Jun 10, 2015
Reference Designators ! ! !

When a switch is closed, the collector of the bipolar transistor is a little below Vbat/2. Depending on the (secret) battery voltage, this might not be low enough to turn off any latched circuits.

Also, the base current is (Vbat - 0.6) / 2 million. That might not be enough current to wake up the transistor.

ak

3. ### Alec_t

2,892
776
Jul 7, 2015
Is the circuit supposed to have an any advantage over the more conventional simple two-transistor latch?

4. ### iimagine

55
8
Oct 23, 2013
If you'd use ohm law:
Vbatt - Vf= IR (Vf = Forward voltage of the led)
Vbatt - Vf = 0.0251 * 110 (I = 25.1mA; R = 110 as shown in the pic)
Vbatt - Vf = 2.761
Common LED have a typical Vf of 2 to 3V at ~20mA to 25mA
So Vbatt must be 5 to 6V and since typical supply voltage is usually 3, 5, 9,12.... It is safe to assume Vbatt is 5V in this case right?

You forgot the 1nF bypass cap.
Ib = (Vbatt - 0.6) / 1000000
Ib = (5 - 0.6) / 1000000 = 4.4uA

Last edited: Jul 29, 2019
5. ### iimagine

55
8
Oct 23, 2013
It is a common two-transistors latch. There are 2 switches, 2 transistors per switch and can be extended for more. The whole purpose is that if you turn on 1 switch, the rest will be turn off. So if you turn on switch 2, switch 1 will be off

6. ### AnalogKid

2,393
665
Jun 10, 2015
Your calculations are based on a series of assumptions that are not valid individually, let alone in combination. Since you are the one asking for free help, you could just supply the information.
No, not ever. First, there is no "standard" battery with a 5 V output. Plus, I don't think anyone here is going to reverse-engineer your circuit just to determine what the operating voltage *might* be. It is entirely possible that your battery symbol is for a single AA battery, in which case the circuit will not work no matter what the component values are. Keep in mind that with only 3 posts, there is no history to examine to determine your experience and skill levels.
No, I didn't. It does not affect the steady-state base current in any way. Your schematic clearly shows two 1 M resistors in series with the base-emitter junction.

Disconnect the capacitor.
Calculate the base current with one switch closed.
Reconnect the capacitor.
Wait 0.01 seconds (5 time constants).
Recalculate the base current.

ak

7. ### iimagine

55
8
Oct 23, 2013
I'm sorry, did I ask for help somewhere?
Wow, ok a single AA battery lights up a LED with a 110ohm resistor in series. (This is not a joule thief!)
Do you even know what you are talking about? or how the circuit is operated? why should I wait? the purpose of that cap is to make sure that the bjt only need to pull down all N-mosfets with a single burst
Right, I agree, its just my way of drawing circuit connecting it like that was meant to be for power supply. I apologize for confusing you