# electronic circuit

Discussion in 'General Electronics Chat' started by superway, Nov 13, 2009.

1. ### superway Thread Starter Active Member

Oct 19, 2009
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Hello,

Please explain this circuit for me, I still don't know how this circuit works.
Thanks. I enclosed an attatchment below

kN

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2. ### BillB3857 AAC Fanatic!

Feb 28, 2009
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Do you not understand the symbols or how the individual devices work? What does the A represent as far as an input? What does it come from?

3. ### superway Thread Starter Active Member

Oct 19, 2009
128
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A can be a signal voltage. If Tamper SW is active, A is short it to Gnd.

4. ### BillB3857 AAC Fanatic!

Feb 28, 2009
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No, A would not short to ground if SW is closed. The XOR you have drawn isolates SW input from the A input line. If you are wanting M to be disabled when SW is closed, you are using the wrong type gate. In the circuit you have drawn, if SW is open, M will be an inversion of A. If SW is closed, M will be equal to A The diodes on the input side provide level protection to the input of the gate.

On edit: I am assuming that the signal is a digital signal type signal such as a switch closing and opening, rather than an analog such as audio.

5. ### SgtWookie Expert

Jul 17, 2007
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Re-drew the schematic so that it's easier to read:

S2 just represents a digital input to the circuit.

BillB3857 already gave the correct description of the circuit. The diodes just protect against voltage inputs outside of TTL/CMOS ranges.

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6. ### superway Thread Starter Active Member

Oct 19, 2009
128
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If S1 is open, A is at Vcc of S2, is M low state?
If s1 is still open, A is at GND of S2, is M high state?
Thanks

7. ### BillB3857 AAC Fanatic!

Feb 28, 2009
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You've got it! Definition: XOR is a binary operation like AND and OR. If two bits are XORed together the results are as follows.
A B --- A XOR B
0 0 ----- 0
0 1 ----- 1
1 0 ----- 1
1 1 ----- 0
Another way to express this is that the result is true if they are different or false if they are not the same..

8. ### SgtWookie Expert

Jul 17, 2007
22,198
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Oops!
Minor correction:
the result is true if they are different or false if they are the same.

9. ### John P AAC Fanatic!

Oct 14, 2008
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It looks as though grounding point A controls the operation of the high-power gate at the right. The interesting thing is that switch SW1 makes the output change sense versus the state of the input. So for one setting of SW1 a ground at point A causes the output to go on, and for the other setting of SW1, the same input causes the output to turn off. It seems as if the designer was told to make it flexible in operation.

10. ### BillB3857 AAC Fanatic!

Feb 28, 2009
2,473
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I missed the point that the buffer/driver was an inverter. Thanks for catching that, Sgt.

On Edit.... It sure would be nice to know the application of the circuit shown, or the desired application. I'm not sure the OP is getting the desired results with his circuit.