Calculating Current in a Diode Circuit with Given Voltages

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In the discussed diode circuit, with VA at 5V and VB at 3V, the current through the load resistor (RL) can be calculated assuming a barrier voltage of 0.7V for each diode. The voltage across the resistor is determined to be 4.3V, resulting in a current of 4.3 mA flowing through RL, with only diode A conducting while diode B remains reverse biased. The circuit's behavior is also likened to that of an OR gate, where a logic '1' is defined as any voltage greater than 4.3V. The conversation emphasizes the importance of understanding diode characteristics in circuit analysis. Overall, participants found the discussion helpful for clarifying the circuit's operation.
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http://img545.imageshack.us/img545/7113/diodeq.jpg
In the circuit shown in the figure (c) above if VA=5V and VB=3V, calculate the current through RL.

Please explain.
Thank you
 
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I just finished a unit on circuits in class so I wanted to try this out, but the picture you have provided isn't working.
 
hope the image works now :)
 
Here you go:

yhowKZh.png
 
Thank you but pardon me. I cannot grasp what is written on the image. It's not clear. And what if the diodes have a Barrier voltage of 0.7V each.
Thank you.
 
I'm sorry, but I thought you preferred really really poor picture quality. Anyway, assuming a diode voltage drop of 0.7V is a good start because by doing that you'll soon enough run into a contradiction (of sorts) and see that one of the diodes is reversed biased. Think about it some more and you'll see it, I'm sure.

EDIT: SchemeIt is a good alternative for drawing circuits. It's free and very easy to use. Check it out!
 
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I got the idea thank you. If VA=VB=5V and if the diodes have a barrier voltage of 0.7V each then voltage across the resistor will be 4.3V. Am I right?
 
Assuming you're using standard silicon diodes (voltage drop about 0.7 V), the current that flows through RL is 4.3 mA, only diode A conducts and diode B is reverse biased.
 
Proff Young said:
If VA= VB= 5V and if the diodes have a barrier voltage of 0.7V each then voltage across the resistor will be 4.3V. Am I right?
Yes. By the way, your circuit is also functionally equal to a _____ gate. Do you know which one?

Logic_gate.gif


P.S Let a logic '1' be any voltage > 4.3V and a logic '0' be any voltage < 4.3V. The gate output 'F' from the figure above would be the node between the diodes and load resistor.
 
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  • #10
Gordianus said:
Assuming you're using standard silicon diodes (voltage drop about 0.7 V), the current that flows through RL is 4.3 mA, only diode A conducts and diode B is reverse biased.

Thank you.
 
  • #11
gnurf said:
Yes. By the way, your circuit is also functionally equal to a _____ gate. Do you know which one?

Logic_gate.gif


P.S Let a logic '1' be any voltage > 4.3V and a logic '0' be any voltage < 4.3V. The gate output 'F' from the figure above would be the node between the diodes and load resistor.

I hope it is the OR gate. Thank you for helping, I have a better idea now.
 
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