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Hi all, I am new to circuit calculation and need help to guide the question above.

Much appreciate and thank you for spending your time and help!

The question is "Calculate the diode current for ID1 and ID2. "

Is it my calculations below are correct?

Since the D1 anode is connected with the ground, D1 is in reverse bias condition and has no current flow? But if ID1 has no current flow, have any calculation method to prove it?

For ID2, the D2 is in forward bias condition, the calculation is (10v - 0.7v) / (6k ohm + 3k ohm) to get the current for ID 2?

  • \$\begingroup\$ You have -10V in the circuit. You seem to assume it is 0V so you may want to fix current and voltage calculations and assumptions about D1 being forwars or reverse biased. \$\endgroup\$
    – Justme
    Commented Oct 31, 2023 at 6:11
  • \$\begingroup\$ May I know more details about how to assume D1 being forward or reverse bias ? Since D1 is connect with ground, if D1 is forward, so the calculation formula is ? \$\endgroup\$
    – Guan
    Commented Oct 31, 2023 at 6:46
  • \$\begingroup\$ Calculate the voltage at D1 cathode. That's Vb. When you have Vb, you know if Vb value is correct or incorrect as it determines if D1 is forward or reverse biased. \$\endgroup\$
    – Justme
    Commented Oct 31, 2023 at 6:58
  • \$\begingroup\$ To find the voltage of D1 Cathode, is it 10v -0.7v ? and there are 9.3v at the point B ? Sorry, I really not so understand about it \$\endgroup\$
    – Guan
    Commented Oct 31, 2023 at 7:16
  • \$\begingroup\$ Welcome! Is this homework? \$\endgroup\$
    – winny
    Commented Oct 31, 2023 at 8:02

2 Answers 2


When you are analysing the circuits with a diode, first, you need to find out which one is on and which one is off. Then, you can start calculating the currents. The starting point is very important. After a while, you know where to begin. For that specific circuit, I start with assuming the D1 is off. Then the circuit would be as this:

Diode D1 Off

Now, let's calculate the voltage on point B. Assuming the Vf of the D2 is 0.7v, it will be a simple voltage divider:

$$ V_{B}=(-10) + ((3)/(3+6) * (10-(-10)-0.7)) $$

This will lead to Vb=-3.56v.

Well, if the voltage of the point B is less than -0.7v, it will lead to D1 turning on. By D1 turning on, the voltages of points A and B would be as this:


The rest is straightforward. For ID2, you have voltages on both sides of the 6K resistor:

$$ I_{D2}=(10-0)/6=1.67mA $$

For the ID1, first, we need to calculate the current going through the 3K resistor (from top to bottom). Let's call it I3:

$$ I_{3}=(-0.7-(-10))/3=3.1mA $$

With Kirchhoff's circuit laws for currents:

$$ I_{D1}=I_{3}-I_{D2}=3.1-1.67=1.43mA $$

I hope I didn't make arithmetic mistakes. But you should get the gist of how to solve these circuits.

  • \$\begingroup\$ Hi Saadat, appreciate and thanks for your details explanation! But, I still have one question, may I know why the node A will be 0v? Is it all the voltage drop to the 6k ohm? \$\endgroup\$
    – Guan
    Commented Oct 31, 2023 at 14:30
  • \$\begingroup\$ After knowing the D1 is on, you know the voltage of the point B is -0.7v. If the D2 is on, too, the voltage on point A is 0.7v more than the voltage of B. This will lead to 0v. By reaching the 0v on point A, the current is still positive, and it proves that assuming D2 is on, is correct. \$\endgroup\$
    – Saadat
    Commented Oct 31, 2023 at 14:34
  • \$\begingroup\$ Noted, thank you again! \$\endgroup\$
    – Guan
    Commented Oct 31, 2023 at 14:36
  • \$\begingroup\$ No worries. Glad to help. \$\endgroup\$
    – Saadat
    Commented Oct 31, 2023 at 14:37


Since the D1 anode is connected with the ground, D1 is in reverse bias condition and has no current flow?

  • You haven't established that D1 isn't forward biased
  • To do this you must imagine D1 is removed
  • Then calculate what voltage is present at node B
  • If that voltage is more negative than -0.7 volts then, D1 is conducting
  • If D1 is conducting then the voltage at node B must be -0.7 volts

Can you take it from here?

  • \$\begingroup\$ Hi, if imagine D1 is removed, then is it the current calculation will be (10v - 0.7v) / (6k ohm + 3k ohm) and equal to 0.00103A. For the 6k ohm diode voltage drop which is 6.18A, so the node B, is greater than -0.7? Sorry for that If I understand wrongly as I really not understand it. Appreciate and thank you for your help. \$\endgroup\$
    – Guan
    Commented Oct 31, 2023 at 13:09
  • \$\begingroup\$ The 3 kohm resistor connects to minus 10 volts hence node B definitely is quite negative @Guan \$\endgroup\$
    – Andy aka
    Commented Oct 31, 2023 at 13:28
  • \$\begingroup\$ Can I know how to get the node B voltage? As if removed the D1, then will left D2 on the circuit, so the circuit voltage will drop to 3k, 6k ohm resistor and D2? If possible, can you provide me the formula to help me more understand about it ? Appreciate and thanks again for your time \$\endgroup\$
    – Guan
    Commented Oct 31, 2023 at 14:04
  • \$\begingroup\$ Taking into account the diode drop of 0.7 volts, there will be 19.3 volts across 9 kohm. That's a current of 2.144 mA hence, the voltage across the 3 kohm resistor is 6.433 volts and, this means that node B is at -3.567 volts. Hence, D1 (when returned) is dominant (forward biased) and node B is restricted to -0.7 volts. \$\endgroup\$
    – Andy aka
    Commented Oct 31, 2023 at 16:34

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