Reversed and Forward Biased Diodes in Parallel

Question

^{simulate this circuit – Schematic created using CircuitLab}

We were tasked to find the node voltage V. My analysys:

• The way I approached this problem is to first make the assumptions of the state of the diodes. My guess is that D1 and D6 will be OFF while D5 will be on. My reasoning for D1 and D6 being OFF is because they have the opposite polarity from the voltage source that are parallel with them. D5 on the other hand i think will be ON because of a positive potential difference between the anode and cathode. But i'm having trouble with the state of D2 and D3.
• My guess is that given the direction current source and the fact that current moves from higher to lower potential, D2 will be ON since D2 has the +- potential.
• If D2 is ON, will V just be the forward voltage of the diode? In this case, 0.7 V since these are given to be silicon diodes.

Answer 1

Your intuition is correct, however you assumed that the voltage drop on the current source I1 is zero, which is incorrect. It is like to assume that a current flowing through voltage souce is zero. On the contrary - you can imagine that a current source generates whatever voltage is needed in order to cause the current to reach the specified value.

Lets consider resistor R1. Its left side is being held on voltage 1V by V1. Its right side has the voltage V, which we are looking for. The current through R1 is IR1=(1-V)/100R, flowing right.

Similarly, the current through R2 is IR2=(V-(-1V))/100R, also flowing right.

The current through both diodes D2 and D3 is I1=1A, it does not matter which diode is conducting.

Sum of the currents in point V is zero: -IR1+I1+IR2=0. Therefore:

-(1-V)/100+1+(V+1)/100=0

(V-1)+100+(V+1)=0

2V+100=0

V=-50V

Answer 2

The way I would do this is to observe that there is 100\$\Omega\$ to -1V and 100\$\Omega\$ to +1V, which means the Thevenin equivalent at the Vo point is 50\$\Omega\$ to GND.

There is 1A flowing through D2 because of the current source so the voltage at Vo will be -50V and the voltage at the current source will be -50.7V. Anything (except an open circuit) in series with an ideal current source does not affect the current.

You can look at the circuit and see that D1 and D6 are reverse biased. D2 is forward biased (arrow goes in the direction of the current source) and D3 is reverse biased. D5 is forward biased because of the polarity of the voltages.

The current through D5 is (2V - 0.7V)/100\$\Omega\$ or 13mA. The voltage at the D5-resistor junction is -1V+0.7V = -0.3V or +1V - (0.013A * 100\$\Omega\$) = -0.3V.