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Could you explain to me why U_out is -4,4V and only 3rd diode is open in the following example? Given that R is 10kOhm.

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My physics resistant brain tell me that voltage at each anode is is 10,15 and 20(I know it's now just teasing you) shouldn't all diodes be open if cut-in voltage is 0.6V? Why correct answer is that U_out is -4,4 and only 3rd diode is open? I think I don't really understand how voltage works. Given voltage is similar to water pressure and current can't never flow in stop direction through diode. Then only "pressure" maker is +15V and maybe U_out. How I get to know what is U_out and which diodes allow current to flow if I need knowledge of which diode is open to calculate U_out from Kirchoff's 2nd law and I need to know what is U_out to calculate which diodes are open. I am really confused about this example. I think I am missing knowledge of some crucial law or something, but you can't learn something if you dont' know name of it. And I don't have the time to study subject thoroughly now. Please tell me what I am missing here.

EDIT: U_out was confusing me(I thought it's some SEM not just voltage between wire and "ground"). It's really easy after this realization.

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3 Answers 3

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Just as "water seeks its own level" (-E. Norton), DC current flows the easiest towards the lowest point. In this case the -5v level is the lowest point so this dominates the circuit. Resulting in the voltage to the right of D3 being at about (-5 +0.6) or -4.4 volts, which in turn keeps the other diodes off.

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  • \$\begingroup\$ Much thanks. I get it now. So like D3 is "relaxing" pressure for U_out and 2 other anodes to -4,4V. \$\endgroup\$
    – Damaon
    Feb 16, 2015 at 18:56
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The third diode pulls the common voltage down to -4.4V. This is not high enough to forward bias the other two diodes, so they remain off.

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  • \$\begingroup\$ How you calculate this common voltage? \$\endgroup\$
    – Damaon
    Feb 16, 2015 at 18:29
  • \$\begingroup\$ With math and logic. 5V + 0.6V = 5.6V. 0V + 0.6V = 0.6V. -5V + 0.6V = -4.4V. Since it can't be all three values, it must be the one that turns the other two off. \$\endgroup\$ Feb 16, 2015 at 18:40
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Temporarily forget about VF and think of the diodes as of "polarity-dependent switches" that commutate (connect or disconnect) the input voltage sources to the common output. Then the source with the lowest voltage will be connected to the output... and the others will be disconnected... Then, if you are pedantic enough, remember VF and add it to the voltage of the connected voltage source...

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