Solving a circuit with a directional loop

Question

I came up with the following circuit:

Please disregard the labeled values in the diagram - the resistances are identical for the sake of simplicity, and of value \$R\$, and the voltage has value \$V\$. Assume ideal circumstances regarding the diode, wires, etc.

I'm not able to solve it because I can't seem to find the total resistance.

As stated in the diagram, there is a parallel branch which splits into a diode facing against the positive end of the circuit and a resistor, in series with an identical resistance.

In my attempt at solving this, I imagined that once the positive current reaches the bottom node, it splits into current going through the resistor towards the negative terminal and current that re-loops through the resistor that was just passed.

Given that assumption, the circuit acts as a series parallel circuit whose net resistance can be found by $$ \Sigma R = R + x $$ where \$x\$ is the value of the resistance in the infinite parallel branches that result from the current re-looping. Then, \$x\$ would seem to be given by $$ x = (\frac{1}{R} + (\frac{1}{R} + (\frac{1}{R} + ...)^{-1})^{-1})^{-1} $$ $$ x = (\frac{1}{R} + x)^{-1} $$ $$ x^2 + x/R - 1 = 0 $$ $$ x = \frac{-1/R + \sqrt{1/R^2+4}}{2} $$ But this value of \$x\$ isn't even dimensionally correct!

Is there another approach I can use to find the resistance?

Answer 1

In my attempt at solving this, I imagined that once the positive current reaches the bottom node, it splits into current going through the resistor towards the negative terminal and current that re-loops through the resistor that was just passed.

It might do that. But you know that both the resistor and the diode are passive devices. And since they're connected in parallel, the voltage across them is the same. Since they're passive devices (meaning, they're not adding energy to the circuit like a source can), current can only flow through them from high voltage to low voltage. This means you know that current through the diode, if it goes at all, has to go in the same direction as it goes through the resistor.

Answer 2

The diode is reverse biased, so no current will flow thru it (other than leagkage, but you said ideal components and that would be much much smaller than the resistor currents anyway). You therefore are simply left with two 4.7 kΩ resistor in series with a 5 V supply. The resistances add, for a total of 9.4 kΩ. From Ohm's law, the current is 5V / 94.kΩ = 532µA. Since the resistors are equal, the node between them will be at half the supply voltage, which is 2.5 V. Yes, it really is that simple.

Answer 3

If you want to solve this problem by hand, and not for example via the use of numerical methods on a computer, you will need to make some assumptions on the diode. Specifically, you will need to decide how sophisticated the model of the diode will be.

For example, two common models are:

1: The ideal diode:

Which has infinite resistance when OFF (open circuit), and null resistance when ON (short circuit).

2: Voltage source and limiting resistor model (more accurate): Where the diode on the model is ideal.

Solution
Since you are considering the diode ideal, then you have to solve this in two steps.

• Suppose that the diode is ON
  Then, you have a short circuit and therefore the parallel resistor is totally ignored. In that case the the resistance is simply 4.7 kΩ. But look at the graph on the first link. The current has to be POSITIVE, but in this case it's NEGATIVE.

This supposition is false, the diode isn't ON.

• Suppose that the diode is OFF
  Then, you have an open circuit and therefore two resistors in series. In that case the resistance is 9.4 kΩ. But, does this actually make sense? Note that since we have two equivalent resistors, then the voltage across the parallel diode is -2.5V (notice negative sign), which satisfies the graph on the first link.

This supposition is true.

Therefore the answer (finally) to your question, using the ideal diode model, is 9.4 kΩ.

EDIT -
I noticed the poster has been having doubts about the possibility of having a looped current.
Well let's consider that possibility. We can at least agree that the current is in the direction to where the diode is pointing. In that case, if the current \$i_D\$ loops to the parallel resistor there will automatically be a voltage across that same resistor according to: $$V = R i_D$$

Since the diode is parallel to that resistor, then it also is subjected to that voltage. But that means it would have a negative voltage, making it impossible to be conducting current.

Answer 4

In this case the diode is open, since the 5V supply is generating a current in the direction of the diode's cathode. Since the diode is open, the resistance are connected in series and you can add the values.

In general, as a first approximation (ideal diodes), you can assume that diodes are either open or conducting. You can then assume one of these two states and solve the circuit. Then you test your assumption. If you assumed that the diode was open, then you need to check that the voltage between the anode and the cathode is negative, if it is not, then your assumption is wrong. If you assumed that the diode is conducting, then you need to check that the current (defined as positive when entering by the anode) is positive. Again, if it is not, then your assumption is wrong.

In this example, if you assume that the diode is conducting, then the current will be negative, implying that the assumption is incorrect. If you assume that the diode is open, then the voltage between the anode and the cathode is -2.5V, implying that the assumption is correct.