# Where does current go when current is flowing dividing between two branches, one with a diode, one with resistor?

This question applies to both images in circuit, but for sake of clarity lets focus on image (b).

Assuming both diodes (D1, D2) are ON (forward bias) and assume both diodes are ideal diodes.

I tried two approaches:

• (i) Apply KCL at node (that connects positive terminal of D1 and D2)
• (ii) Using intuition, realize that the D1 branch has no resistance, so naturally all the current will flow through there.

These two methods give conflicting results for the current flow through the circuit.

• (i) current flow through D1 will be 0.25 mA
• (ii) current flow through D1 will be 0.5 mA

I have my suspicions that scenario (i) is the correct answer, but I am not certain.

I think there is something going on with the -3V at the bottom of the circuit that I can't explain.

Is there something special about diodes in this circumstance?

Questions:

• Q1) "What's your definition for an ideal diode, forward biased voltage is zero or something else?"
• A1) Yes, forward biased voltage is greater than or zero

I think I understand it now, this is my first post so I'm not sure what to do now, but I probably won't answer any more questions for a bit while I continue my work.

My work:

KCL method:

Current flows to zero resistance:

• Intuition plays no role. If you assume both transistors are on, then both must fulfill the equations. If the equations make no sense then the assumption that both are on must be false, and only one of them is on. What's your definition for an ideal diode, forward biased voltage is zero or something else? Commented Mar 14, 2023 at 20:18
• The only thing special about ideal diodes is that if they have current flowing through them, their voltage is zero. When you did your (i), were the diode voltages set to zero? In your (ii), in case (b), what is the voltage across D2? Is this consistent with the behavior of an ideal diode? Commented Mar 14, 2023 at 20:33

To generalize this problem:

Ideal diodes can either be a) forward biased, and thus act as short circuits, or b) reverse biased, and thus act as open circuits. Start by assuming any diodes are forward biased and then carry out your calculations, checking as you go to make sure they make sense.

In Fig. b, they do: a certain amount of current comes in, the right branch draws a smaller amount of that current (which you know by using i=V/R for the resistor), and therefore the left branch draws the rest. In Fig. a, they don't: a certain amount of current comes in, the right branch draws a larger amount of that current so the left branch would have to have current coming in from GND, which it couldn't. When a circumstance like that happens, change your assumptions. Test the circuit if D1 was reverse biased, or if D2 was. One of the configurations will work.

I also want to point out your assumption (ii) about all current flowing through the left path. That would only be the case if these two branches were in parallel and joined together at the bottom node. But the left branch connects to GND and the right branch connects to -3V, so therefore that assumption is invalid. So be careful that your intuition is actually correct before you use it to second guess yourself!

• "I also want to point out your assumption (ii) about all current flowing through the left path. That would only be the case if these two branches were in parallel and joined together at the bottom node. But the left branch connects to GND and the right branch connects to -3V, so therefore that assumption is invalid. So be careful that your intuition is actually correct before you use it to second guess yourself!" Thanks so much for that clarification about my assumption. I think that clears up this problem for me a lot.
– Nice
Commented Mar 14, 2023 at 21:10

Yes, you are correct in your first working. For fig b) consider first the current through the 6 kOhm resistor. Given the diode has zero forward voltage, 3V drops across 6 kOhm, producing 0.5 mA.

Now considering the 12 kOhm resistors. As D1 has zero forward voltage, the voltage at the positive terminal of D1 must also be 0V. Again D2 has zero forward voltage, so the voltage across the 12 kOhm resistor is 3V, which gives 0.25 mA passing through. This current passes through D2, so we can KCL for the positive diode terminals.

0.5 mA flows in through the 6 kOhm, 0.025 mA flows out through D2, ergo 0.025 mA must flow through D1.

Figure a) is actually far more interesting. You might initially expect 0.25mA through the 12 kOhm, and 0.5mA through the 6 kOhm resistors, but this would require -0.25 mA to flow through the diode, of course, this cannot happen to an ideal diode. In this case, D1 would close, with zero current flowing through it. We can then reconsider the circuit with D1 removed. Calculating the voltage at V, we can ignore D2 with its zero voltage drop. The voltage dropped across the 12 kOhm is a simple resistor divider
Voltage drop = total volts x R1/(R1 + R2)
= (3V + 3V) x 6k / (6k + 12k) = 4V If 2V is dropped, the voltage at V must be -1V

" (ii) Using intuition, realize that the D1 branch has no resistance, so naturally all the current will flow through there."
is a misunderstanding. D1 presents a zero resistance path, as long as it is forward-biased, but that does not mean all current must flow through it.

Scenario (i) is correct, because your statement

Using intuition, realize that the D1 branch has no resistance, so naturally all the current will flow through there

is wrong.

Just because there's no resistance between two points does not mean that that path will "suck" away all the current. Current will always be divided, in accordance with KCL, regardless of the resistances present in the various branches.

simulate this circuit – Schematic created using CircuitLab

• The anodes of both diodes are at 0V.

• $$\(+3V) - (0V) = 3V\$$ across the 6kΩ resistor establishes 0.5mA downwards through it.

• $$\(0V) - (-3V) = 3V\$$ establishes 0.25mA downwards through the 12kΩ resistor.

None of those facts are difficult to convince yourself of, and are not debated. At this point it's already clear, by KCL, that current through D1 must be the 0.25mA.

However, if, as per your intuition, you assume that the 0Ω resistance from D1's anode to ground would "suck up all the current", I would remind you of this: if the voltage across something (D1 in this case) is zero, and the effective resistance of that thing is zero, then using Ohm's law, or intuition, to find the current through it is ill advised:

$$I = \frac{V}{R} = \frac{0}{0}$$

Current is strictly undefined in these conditions. In other words, it can and must be defined by conditions elsewhere in the circuit, and the only tool in your toolbox for figuring it out is KCL.

KCL isn't optional, it's the law!

Update: At the risk of confusing you, it might help to examine the scenario where D1's effective resistance is not zero, but really close to it, say 1mΩ instead:

simulate this circuit

The presence of R2 causes the diode's anode to rise very slightly above 0V, but not my much. This scenario circumvents the problem we had of current being undefined, because now there's a real, non zero voltage and resistance that permits us to determine current in that path using Ohm,'s law. I don't think you would argue here that R2, being so small, would pass all the current, or even most of it. I think rather that you would trust KCL in this case.

But let me ask you this, if we make R2 smaller and smaller, can you think of any reason why suddenly the current through it would jump from 0.25mA to 0.5mA the instant it reaches zero? On the contrary, the smaller R2 gets, the closer current gets to 0.25mA, until at zero ohms, it finally reaches that target value.