OR Gate Voltage Drop: A=5V, B=4.6V

[dobry_den]

Hi! I've got another diode OR gate problem. When the voltage at the input A is 5V and it's 4.6V at the input B, does the output voltage still remain at 4.4V? Does it mean that the D₂ diode is forward biased and its voltage drop* is smaller than the voltage drop of the D₁ diode?

*the constant-voltage-drop is about 0.6V, thus the voltage drop across the D₂ diode should be just about 0.2V

The attached image is a scheme of the OR Gate when A is 5V and B is 0V

 

[berkeman]

Yeah, just the A input is enough to hold up the output. Just get familiar with the V-I plot for diodes, and their behavior will be a lot more intuitive.

 

[dobry_den]

Ok. So due to the 5V A input, the voltage-drop across the diode 2 will be quite small (just about 0.2). It's apparent from the V-I plot that the current going through this diode will be quite small. But what makes the A input to "dictate" the value of the output voltage? What prevents the output from being equal to 4.6 - voltage_drop_across_D2, which would increase the potential difference across the diode 1 and the current goint through this diode?

 

[berkeman]

You basically can still write the KCL for this circuit, you just need to use the diode equation to express the current and voltage for the diode legs. Write the KCL, and then solve for the currents and voltages of the diodes. You can assume some pretty generic Is saturation current -- just work backwards from the basic diode equation, assume room temperature (in Kelvin) and a 0.6V forward diode drop at some reasonable current like 10mA. That will give you an Is that you can use in your KCL.

Diode Equation: $$I_d = I_s ( e^{\frac{q V}{k T}} - 1 )$$

 

[dobry_den]

Thanks very much for your replies...! Diodes will certainly require more interest and time from my side...:)