Today, in class, we were to analyze this circuit:


simulate this circuit – Schematic created using CircuitLab

And calculate \$\alpha \in [0, 1]\$ associated with R2 a potentiometer. I'm confused with the explanation of the answer given in class. We assume a perfect LED.

What happens when the LED goes from open to close

To my understanding, a potentiometer is similar to this:


simulate this circuit

And therefore, I may see the circuit like so:


simulate this circuit

And therefore, by voltage divider, the LED closes at \$\alpha = 1 - 2\frac{U_{led}}{U}\$.

The part I'm confused about

Here is the part that confuses me. Once the LED is opened, by virtue of being a perfect LED, \$R_B\$ should be short circuited. With only \$R_A\$ the only resistive component of the circuit, I believe I can lower \$\alpha\$ down to \$0\$ because \$U_{R_A}\$ is no longer dependent on the value of \$R_A\$, which is nonsense, because I know for a fact to be able to shut down a LED. What am I missing?

  • \$\begingroup\$ What is your LED model? Zero current to Vf, infinite current after? Infinite current for any positive voltage? Or something else? \$\endgroup\$
    – Reinderien
    May 5, 2022 at 22:54
  • 5
    \$\begingroup\$ An LED is not a switch - it is not correct to talk about the LED 'opening" or "closing". The LED does not become a short circuit when it lights. \$\endgroup\$ May 5, 2022 at 22:54
  • \$\begingroup\$ You cite a U_LED. Does that mean that you have a fixed forward voltage? If so, what is it? \$\endgroup\$
    – Reinderien
    May 5, 2022 at 22:59
  • 1
    \$\begingroup\$ Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. \$\endgroup\$
    – Voltage Spike
    May 5, 2022 at 23:15
  • \$\begingroup\$ RB is not short-circuited. It still participates in the circuit in parallel with the LED. However the current through RA and RB is not the same when the LED is on. \$\endgroup\$
    – user253751
    May 6, 2022 at 11:39

3 Answers 3


Your circuit is:


simulate this circuit – Schematic created using CircuitLab

with example values filled in.

Recognise that D1 is piecewise: conducting or not conducting (not open or closed). While it conducts it has a fixed forward voltage \$V_f\$ and some non-zero current, and while it does not conduct it has zero current and some \$V < V_f\$. Since D1 is piecewise the whole circuit analysis is piecewise.

Designate \$R_3\$ as the bottom half of potentiometer \$R_2\$.

For the non-conducting case, diode current \$I_L = 0\$, and \$I_{CC} = I_3 = \frac {V_{CC}} {2R}\$.

For the conducting case,

$$ I_{CC} = \frac {V_{CC} - V_f} {R ( 1 + \alpha )} $$

$$ I_3 = \frac {V_f} {R (1 - \alpha)} $$

$$ I_L = \frac 1 {R(1 + \alpha)} \left( V_{CC} - \frac {2 V_f} {1 - \alpha} \right) $$

The next question is: which expressions do you choose for which piece, and when? Sketch this out in a very simple spreadsheet to visualise your choices:


If the right half of the current curves were to be still in the conducting state, the diode current would go negative and the bottom half of the potentiometer would experience current above what the supply can possibly push through the circuit's resistors. This does not make physical sense, so in the right half, the other curves (non-conducting) must be true.

Verify with a DC sweep of alpha (here called K) in simulation:


With these values, the diode conducts for \$0 \le \alpha < 0.2 \$ and does not conduct after. An important part of this analysis is calculating the value of alpha for which the diode stops conducting; this is the point at which the conducting and non-conducting supply currents are equal:

$$ \alpha = 1 - \frac {2 V_f} {V_{CC}} $$

You had already come to this conclusion numerically, but it seems you're missing the intuition as to why. You say

Once the LED is opened, by virtue of being a perfect LED, \$R_B\$ should be short circuited

which is incorrect. The LED does not operate as an open-or-short switch; it operates as a conducting or non-conducting diode. So in no case can you disregard \$R_B\$.

  • \$\begingroup\$ okay, I think I get it now. Please apologize my layman speak, I'm trying to build intuition. Because the led has that voltage drop in parallele to the resistor, it acts as an emf to the resistor, forcing current through it ? I cannot replace the perfect led with a 2V voltage source and \$0\Omega\$ resistor because it is in parallele to \$R_3\$, is that it ? \$\endgroup\$
    – NRagot
    May 7, 2022 at 12:33
  • \$\begingroup\$ @NRagot As a different answer shows, you actually can replace the LED with a voltage source for the purposes of analysis, so long as you also have an ideal diode that has zero drop. The LED, while conducting, sets the voltage across R3 but does not provide any current to R3; all of the current must always come from the main voltage source. \$\endgroup\$
    – Reinderien
    May 7, 2022 at 12:51

Your Prof. possibly described the ideal diode as a current switch at a certain threshold voltage.

We usually call it a Current Rectifier and in the old days the schematic would define them as CR1, CR2 but now you see D1,D2 etc. It does act like a soft switch with its quadratic I vs V until it reaches an asymptote towards a linear fixed bulk resistance. That resistance at rated current is inverse to the device power rating which sizes the bulk resistance.

Here is a simulation comparing a power Silicon diode to a low power 20 mA LED. The incremental resistance is the slope of the V-I or x-y curve.

enter image description here

Shown above is a 0 to 5V linear triangle test generator with 0 Ohms, not a 50 ohm sig gen , a weak dimmer solution using a 1k pot as the diode resistance is much lower than the Pot when it gets bright. So this is not a good solution and needs a current buffer. You could use a 50 Ohm power Pot but that could get hot or burn out.

  • \$\begingroup\$ Here without the current limiting 10 ohm resistor I can burn out the LED with a 100 ohm pot. tinyurl.com/yxpwamdo. Change the simulation speed or stop to your convenience. Everything can be changed with a mouse. \$\endgroup\$ May 5, 2022 at 23:21

In comment, you cleared up what "perfect LED" meant. If it has a threshold voltage Vth, then your circuit can be replaced by


simulate this circuit – Schematic created using CircuitLab

This has two distinct parts. For low a, such that V(aR/2R) < Vth, D1 is reverse-biased, and you can calculate the current "normally".

For higher values of a, you can start by calculating the current through R1 and R2 as being (V - Vth) / (R1 + R2), since the voltage at the wiper is clamped at Vth.

The rest is up to you.


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