Consider following circuit

enter image description here

At t=0 I can wrap my mind around the fact that Vout equals whatever value the highest phase has since the others are negative or zero. But after that only one of the three phases has a zero voltage value. Blue is rising while red is falling. Why isnt the output here blue + red? Since purple is negative, the diode should block any current from flowing but the others are positive.

How can I mathematically prove that Vout is always equal to whatever the highest value of the voltage is? Whenever I watch a video about it, they always say "highest voltage wins" with no explanation.

Maybe its too trivial but could someone help me anyways?

  • \$\begingroup\$ If the blue voltage is the highest then t1 will on, and t2 & t3 will be off, since they have a higher voltage (the blue voltage) on their cathodes than on their anodes. etc... \$\endgroup\$ – Chu Jun 19 '18 at 23:47
  • \$\begingroup\$ If blue is at 100 volts (say) then Vout will be at about 99V, and red is only 50V, are you sure t3 won't block any current? \$\endgroup\$ – immibis Jun 20 '18 at 4:31
  • \$\begingroup\$ Diodes don't "block negative voltages" exactly. They allow current to flow one way. A voltage is only negative compared to another voltage we define as having a value of 0V, and call ground--that's a way of simplifying our math, but the diodes don't care which node we call ground. If the cathode is more negative than the anode, they conduct a negligible amount, and if it's the other way around they conduct prodigiously. Never get fooled into thinking circuit ground is a real thing...it's just a node we picked so we could describe the other nodes. \$\endgroup\$ – Cristobol Polychronopolis Jun 20 '18 at 12:21

The diodes perform a MAX function, not a sum of positive values.

  • \$\begingroup\$ Yes , diodes in logic are OR gate switches but actual voltage depends on impedance ratios and tends to be average of ripple. \$\endgroup\$ – Sunnyskyguy EE75 Jun 19 '18 at 19:54

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