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We have two circuits in the figure. For each find I and V. The answers are

  • 3 mA, 3V
  • 4mA +1 V

This isn't homework,before you tell me to add this to the homework.I just found this and I want to know how to solve this.They look similar to the logic gates,the first one to OR gate and the second to AND gate,but where do I go from there?

http://d.yimg.com/hd/answers/i/d7b3f80ff2a84f94aee9235060209fa6_A.jpeg?a=answers&mr=0&x=1384809801&s=99eb4c6770f14e7259a1448dd67f11b2

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  • \$\begingroup\$ We don't just give you answers to homework questions here, but we can help you understand the theory behind them if you get stuck. It seems these problems are intended to show you how diodes can be used to implement a min or max function. Consider what would be different in the first problem if the third input were 2 Volts instead of 1 Volt, or 0 Volts. Explain specifically what you are stuck on. \$\endgroup\$ – Olin Lathrop Nov 18 '13 at 20:00
  • \$\begingroup\$ I understand that 3 Volt is the biggest potential out of three diodes but why is that potential exactly,the potential in the main branch? \$\endgroup\$ – Notyourthing Nov 18 '13 at 20:09
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It's digital logic, 3 being High and 1 being Low, anything in between won't have much effect.

The left side is an OR gate. So its kinda like High or low = 1 or 0 = High High being 3 volts, and low being 2 or 1 volt relatively speaking.

In the latter, right side, is like an AND gate, So it's kinda like High and Low = 1 and 0 = 0 Again, High being 3 volts and Low being 1 volt.

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  • \$\begingroup\$ @Notyourthing no problem, wondering what college you go to, we dealt with a similar scenario in mine. \$\endgroup\$ – Iancovici Nov 18 '13 at 20:21
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The provided answers (3 mA, 3V and 4mA +1 V) assume ideal diodes -- no voltage drop when forward biased and no current when reverse biased.

In the left diagram, the diodes connected to 1 & 2 volts are reverse-biased and so the only current flowing is from the 3 volt source.

In the right diagram, the diodes connected to 2 & 3 volts are reverse-biased and so the only current flowing is into the 1 volt source.

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