Electronic – How to show that this is an OR gate

Question 1: Assuming that D1 is off and D2 is on, we have the following circuit:

We can first calculate the current throught diode 2

\$ i_{D2} = \frac{9V-(-6V)}{22k \Omega + 43k \Omega} = 0.2308 mA \$

and then the voltage across diode 1

\$ V_{D1} = 9V-0.2308mA \cdot 22 k \Omega = 3.92 V \$

As a diode may only block negative voltages, the assumption of diode 1 being off and D2 being on is invalid.

Question 2: Assuming that both diodes are turned on yields the following circuit:

Calculating the current through diode 2

\$ i_{D2} = \frac{0V-(-6V)}{43 k \Omega} = 0.1395 mA \$

and then the current denoted as ix

\$ i_x = \frac{9V - 0V}{22 k \Omega} = 0.4091 mA \$

Using Kirchoffs Current Law (KCL) stating that all currents entering a node should be equal to the current leaving the node we get

\$ i_x = i_{D1} + i_{D2} \$

Which may be rearranged to

\$ i_{D1} = i_x - i_{D2} = 0.4091 mA - 0.1395 mA = 0.2696 mA \$

With diode analysis, one thing you have to be completely aware of is that it's a guessing game. Yes, I said guessing. How well you guess will depend on your experience, so even if you are the worst guesser ever, its ok, because the math will tell you.

When Vd >= 0.7, it conducts. When Vd < 0.7, it does not conduct. Vd is the is the difference between anode and cathode.

Now to the guessing game.

There are 4 cases to test.

• Case 1: D1 = OFF | D2 = OFF
• Case 2: D1 = ON | D2 = OFF
• Case 3: D1 = OFF | D2 = ON
• Case 4: D1 = ON | D2 = ON

Your initial guess was that D2 conducts, and D1 does not. You didn't specify what your reasons were for that selection, but even if it was a blind guess, it does not matter because you'll need to prove your guess was correct. Don't ever guess, and not verify. Always verify your guess.

Case 3: D1 = OFF and D2 = ON

$$ -10V - 10k\Omega I + 0.7V - 5k\Omega I - 10V = 0 $$ $$ I = \frac{20V - 0.7V}{15k\Omega} = 1.28667mA $$

Lets look at the anode of D2 $$ 10V - (1.29mA)(10k\Omega) = -2.86667V $$

So let's look at the voltages at node between the two diodes.

$$ 5k\Omega I -10V = -3.6V$$

So now the main question, do these values hold true for our assumptions.

For D2 to be on, \$V_{D2} >=0.7\$ and \$V_{D2} = -2.9V - (-3.6) = 0.7\$

For D1, to be off \$V_{D1} < 0.7 \$ and \$V_{D2} = 0-(-3.6V) = 3.6V \$

But.. D1 is supposed to be off...

Repeat the process for all cases until you can confirm that the math holds true with your assumptions.

But your initial guess that D2 is on, and D1 is off is wrong.