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V = 0,7 VI = 198,7 µA, VTH = 6V e RTH = 26,67 kΩ; How do I do it?

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  • \$\begingroup\$ I do discuss some of the details about Thevenizing a resistor divider pair here. No diodes. But it covers some of what you need. Does it help? But you can also easily see that there are two very simple resistor dividers in circuit (b). If you removed the diode for a moment, would you be able to work out the voltages that exist between each remaining pair of resistors? \$\endgroup\$
    – jonk
    Commented Jun 1, 2020 at 0:30
  • \$\begingroup\$ If you follow jonk's link, you'll see how to replace a voltage source with a voltage divider after it into another, lower, voltage source with a simple output resistance. Once you've done that to all the dividers, you'll have simple series components that are easier to solve. \$\endgroup\$
    – Neil_UK
    Commented Jun 1, 2020 at 5:38

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