We have the diode circuit in the figure. The diodes are real and Vd=0.7 Volt I have to find the currents in the diodes and in the resistances R1 and R2 enter image description here

So I apply KVL in loop 1 and I have

-20+ 0.7+0.7 +I2*5.6=0

Here I find that I2=3.3 mA.This means that the current through D1 is 3.3 mAmper.

I apply KVL to the second loop. -0.7 +3.3*I1=0 so I1=0.212

The current through D2 is I2=I1+Id2 so Id2= I2-I1 =3.088 mA

So the operating point of D1 is (3.3 ; 0.7) and of D2 is (3.088; 0.7)..Is this correct?

  • \$\begingroup\$ Seems you left out the K ohm, (use 5600 ohm and 3300 ohm). Also notice that without doing KVL the voltage across R1 is only 0.7v. \$\endgroup\$ – Nedd Feb 26 '15 at 10:30
  • \$\begingroup\$ I have made a mistake and written the current as Amper instead of mA . \$\endgroup\$ – Notyourthing Feb 26 '15 at 13:35

There is problem in your calculations:

-20+ 0.7+0.7 +I2*5.6=0

Should be:

-20+ 0.7+0.7 +I2*5600=0

Same with 3,3kOhm resistor - you should use this kind of calculation:

-0.7 +3300*I1=0

Always remember in all of your calculations to use proper unit.

| improve this answer | |
  • \$\begingroup\$ Not everything else is OK. The part of "3.3*I1" also needs to change to "3300*I1". \$\endgroup\$ – Michael Karas Feb 26 '15 at 11:08
  • \$\begingroup\$ You have right, but i think author will be smart enough. ;) \$\endgroup\$ – Sławomir Kozok Feb 26 '15 at 11:15
  • \$\begingroup\$ Maybe the original poster would be smart enough. But as stated your answer is wrong. I commented simply to give you the opportunity to correct your answer and make it correct. \$\endgroup\$ – Michael Karas Feb 26 '15 at 11:18
  • \$\begingroup\$ I actually thought about writing the answer in mA,thats why I have written it as 3.3 kohm \$\endgroup\$ – Notyourthing Feb 26 '15 at 13:34
  • \$\begingroup\$ Are the operating points correct? \$\endgroup\$ – Notyourthing Feb 26 '15 at 13:43

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