Transistor as invertor switch

Discussion in 'The Projects Forum' started by abuhafss, May 6, 2013.

1. abuhafss Thread Starter Active Member

Aug 17, 2010
165
2
In a 555 astable circuit, I would like to use a transistor as an inverter switch between points A and B such that when the base is high, A and B remain open and when the base is low, A and B get closed. Can somebody help me how to do it.

Thanks for the help.

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2. aws505 Member

Mar 11, 2013
59
7
So you would like to have a transistor that acts to short the points A and B in your schematic? How is the base of this transistor driven? By another 555 perhaps? Also, your logic for the base of the transistor is backwards, when the base of the transistor is held high, it will short A to B and when the base of the transistor is low, there will be an open-circuit from A to B.

3. abuhafss Thread Starter Active Member

Aug 17, 2010
165
2
Yes the logic is backwards.
When there is signal at the base, A-B open circuit.
When no signal at the base, A-B short

4. Bernard AAC Fanatic!

Aug 7, 2008
4,233
414
I would use a 4066, analog-digital switch if the R values are in 10k range, or maybe a transistor photo coupler, collector to highest +V.

5. aws505 Member

Mar 11, 2013
59
7
You can always go the lazy route and make an inverter on the gate of the transistor. It only requires an additional transistor and a resistor. Very cheap'o way to implement this behavior.

6. abuhafss Thread Starter Active Member

Aug 17, 2010
165
2
Thanks for your advice, I have been considering to use RTL inverter but I am confused how to manage the connections between point A and B. Could you please show me your example on my provided picture?

7. Bernard AAC Fanatic!

Aug 7, 2008
4,233
414
I believe that you will have trouble using a transistor as a switch as there is no solid reference for the base. I am using something similar, basic period is 20 msec, a short + pulse,.7 ms, or long, 1.8 ms to drive a servo. If input is LO , pulse is long, if input is HI, U1 & U2 conduct, shorting R6 & R7, giving shorter pulse. Leaving out U1, using X as input ,operation is reversed, but X must go to top rail-U3-8.
Can you explain the timing that you wish to achieve?

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8. tubeguy Well-Known Member

Nov 3, 2012
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It might be good to let us know what your final goal is with circuit. Could you post a more complete schematic?