# I know basic gates but not their structure

I know everything except the structure of logic gates. Recently a HighOrderThinkingSkill question asked the gate from structure where I was totally stumped please help me:

simulate this circuit – Schematic created using CircuitLab

where A and B are input and : $$1\to>5V\\0\to<1V$$

How to find and be prepared for future questions?

The answer key says it's an AND gate, but I don't understand why.

• Usually I solve these by trying combinations of 0V and 5V on A and B - if you solve for Vout in each case, you can get a good idea of what's happening. Mar 13, 2015 at 13:38
• @Gregd'Eon if anyone of the input is 0(<1V) both diodes are forward biased and so current will flow , so i think it's OR Mar 13, 2015 at 13:39
• Also: this circuit looks really strange, because Vout is tied to Vcc. Are you sure this schematic is accurate? Mar 13, 2015 at 13:41
• @Gregd'Eon yes. Mar 13, 2015 at 13:41
• ok - looks like @Camil's answer covers everything here then. Mar 13, 2015 at 13:42

Analyse the working of the circuit for all possible values of A and B. Write down a truth table of A, B and the output. Then you should recognise a pattern.

In this case, note that the output is directly connected to 6V. What does that tell you?

• w8 and see my edit Mar 13, 2015 at 13:37
• i suppose it's OR because if any of the input are low then diode is forward biased and current flows? Mar 13, 2015 at 13:43
• since node vcc is 6v so output is always 6v as it is connected by non resistive wires throught the branch. Mar 13, 2015 at 13:50
• @ADG yes please. I'm pretty sure this circuit is not AND. Perhaps the 10K resistor should be on the Vcc side.
– user17592
Mar 13, 2015 at 13:55
• @ADG I'm sorry, your teacher is wrong. Tell him that since the output is tied to Vcc, the output will always have a potential of 6V with respect to ground, hence the output is always a logical 1.
– user17592
Mar 13, 2015 at 14:01

Where a logical 1 is equal to +V minus a diode drop, and a logical $0$ is equal to $0$ V plus a diode drop, using resistors and diodes it's only possible to construct AND and OR gates like this: Inversion isn't possible and Y must either be pulled up - through a resistor - to +V for the AND function, or the polarities of the diodes reversed and Y pulled to GND through a resistor for the OR function, so none of the answers to problem 11 in your workbook is correct.