I am a beginner in electronics. I am trying to understand AND gate on this page.

  1. I have only solved circuits that make loops and don't have a ground. How to make a positive voltage (for A=1) at a point A? Can we make this circuit in looped that is no ground or lone wire?
  2. How to find direction of a current at any point? I have read and watched the handwavy explanation and I still don't get it. Previously I used KVL/KCL to find out the direction but I don't know how to it here?

Ground, in most electronics, is just the point in the circuit we call "Zero volts", and use as a reference when measuring voltages elsewhere in the circuit. It is where we connect the black lead of our meter.

The gate circuits on the page you mention are in a sense, only partial circuits: in use, there will be a additional circuits connected to points A and B, and to "ground", to form complete ("looped") circuits to drive the gate.


simulate this circuit – Schematic created using CircuitLab

(Ignore component values - they are the schematic editor's defaults)

The node marked "Ground" is the reference point in the circuit - it is where we put the black lead of our meter when measuring voltages elsewhere in the circuit - we must always measure voltage between two points.

With both switches open, R2 and R3 will pull Input A and Input B "High" (to +5 volts), and the output of the gate will also be High (+5 volts, as measured between Output and Ground)

If either switch is closed, the associated input will be connected to Ground (zero volts), and the diode will pull the output down to about 0.7 volts from ground (a typical forward-biassed diode will have about 0.7 volts across it)

In Real Life, the output would go on to some other circuit, and the inputs could come from other gates rather than the switches.

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  • \$\begingroup\$ Can you draw the looped version for it? Thanks. \$\endgroup\$ – AirTycoon May 23 '14 at 16:40
  • \$\begingroup\$ <strike>"use as a reference when measuring voltages elsewhere in the circuit." I know how to calculate voltage between two points. Can you explain it bit more ? </strike> I understood the quoted text. But still the concept of ground is little unclear. Can you show me an example (please make it looped circuit I cant handle partial circuit just yet.) where ground is essential ? Can you show how we use KVL/KCL in the partial circuit. Thanks. \$\endgroup\$ – AirTycoon May 23 '14 at 17:02
  1. There's no need to connect a positive voltage to point A (or B) since it will go high because of the positive supply connected to the top of the resistor.

  2. Conventional current flow is defined as charge moving from a more positive to a less positive voltage.

Having said that, with a DC power supply's (a 9 volt battery, say) positive terminal connected to the top of the resistor and its negative terminal connected to ground, the output terminal will rise to 9 volts if the inputs to the diodes are allowed to float.

However, if either point A or point B (or both) are connected to ground, the diode(s) will be forward biased, current will flow through the resistor, and the output voltage will drop to one diode drop above ground - about 0.7 volts.

Since, with the diodes floating, their cathodes will be high, we can make a truth table describing the circuit's logic, with a zero indicating a low and a one indicating a high:

A   B  OUT
0   0   0
0   1   0
1   0   0
1   1   1

In the "positive true" logic convention a 1 is considered true and a 0 false so - for this circuit - the output will go true ONLY when A and B are true, satisfying the AND requirement.

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  • \$\begingroup\$ "There's no need to connect a positive voltage to point A (or B) since it will go high because of the positive supply connected to the top of the resistor." Sorry but I did not get it. What will go high ? How would one prove value and direction such points. How would I prove what you just said in last 3 paragraphs ? \$\endgroup\$ – AirTycoon May 23 '14 at 17:09

All DC circuits must have a ground. You can't define logic levels without a ground.

When one input or both are low (the same voltage as ground), the diode(s) get forward biased and enter conductive state. This pulls the voltage at output low with respect to the ground.

If both inputs are high, both diodes are reverse biased, thus the voltage at output is the voltage at V without the resistor drop (high level at low current).

If you want to "see" the current flowing through the diodes, then replace them with LEDs. Whenever they are forward biased, they will light.

I have only solved circuits that make loops

Every circuit that draws current is a closed loop. Otherwise the current wouldn't flow through it.

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  • \$\begingroup\$ what does it mean by "input is high" ? Is it possible to turn it into closed loop ? What would happen if there were no ground ? \$\endgroup\$ – AirTycoon May 22 '14 at 19:26
  • \$\begingroup\$ Input is high means that a voltage equal to V (supply voltage) is applied with respect to the ground. \$\endgroup\$ – Cornelius May 22 '14 at 19:28

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