Diode Logic Gates circuit analysis?

[radiodude]

I need to understand how diode-resistor gates work. I've read a lot about what's supposed to happen based on inputs, but I need to see some numbers & analysis to understand what's going on. Where do you start in terms of trying to analyze the following AND gate?

http://img441.imageshack.us/img441/8827/andns2.gif

How I would attempt it is by assuming all diodes are off and representing each diode's voltage in terms of Va, Vb, and Vr. That'll tell me what is required to make the diode turn on (forward-biased). But then I get mixed up because if I tried to do a KVL starting @ Vdd and downwards through R2, R1, and Va, I need to know what the current is through R2 and R1, which leaves me stuck. If they're all off I know i would be 0, but what if they weren't?

Any suggestions on how to analyze this circuit? Thanks.

PS: I suppose this should be moved to Homework help, sorry for posting in the wrong forum.

What I've come up with so far, and I doubt it's correct:

0) If all diodes off, Vout = 1k/(10k+1k) * 5 = 0.45V
1) Vr < 5.7 for D3 to be off [Vd3 = -5 + Vr]
2) Va < 4.75 for D1 to be off [Vd1 = 0.45 - Va + 5]
3) Vb < 4.75 for D2 to be off [Vd2 = 0.45 - Vb + 5]

Note: Von for forward-biased diode = 0.7

 

[chroot]

If you turn on either D1 or D2, by applying a logic-low signal to Va or Vb, you get a current path through the resistor divider formed by R1 and R2. The division is between 5V and 0.7V, the forward-biased voltage across the conducting diode. Since the resistor divider has a ratio of 10:1, the output voltage will be 1/11th of (5V - 0.7V), or about 0.4V.

If both D1 and D2 are off, because you've applied a logic-high signal to both Va and Vb, then there is no current path, and R1 and R2 no longer drop any voltage. The voltage at the output will then drift back up to Vdd.

- Warren

 

[oswaler]

V

Dear student,

Thank you for your question about diode logic gates circuit analysis. As you mentioned, there are many resources available that explain the theory and concept behind diode-resistor gates. However, to truly understand how these gates work, it is important to see some numbers and analysis.

First, let's start by understanding the basic principle behind diode-resistor gates. These gates use diodes as switches to control the flow of current through a resistor. When the diode is forward-biased, it acts as a closed switch, allowing current to flow through the resistor. When the diode is reverse-biased, it acts as an open switch, blocking the flow of current.

Now, let's take a closer look at the AND gate you provided. This gate consists of two inputs (Va and Vb), a resistor (R1), and two diodes (D1 and D2). The output (Vout) is taken across the resistor R2.

To analyze this circuit, we can start by assuming that all the diodes are off. This means that they are reverse-biased and act as open switches. In this case, the current through R1 and R2 would be zero, as you correctly mentioned.

Next, we can consider different scenarios where one or both of the inputs are high (5V). Let's take the case where both inputs are high. In this case, D1 and D2 would be forward-biased, acting as closed switches. This would allow current to flow through R1 and R2. By applying Kirchhoff's voltage law (KVL), we can determine that the voltage drop across R1 would be 5V, and the voltage drop across R2 would be 5*1k/(10k+1k) = 0.45V.

Similarly, if only one of the inputs is high, only one of the diodes would be forward-biased, and the output voltage would be 0.45V. If both inputs are low (0V), then both diodes would be off, and the output voltage would be 0V.

To summarize, the output of this AND gate is 0V when both inputs are low and 0.45V when one or both inputs are high. This is because the diodes act as switches, allowing current to flow through the resistor only when the inputs are high.

I hope this explanation helps you