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I am trying to understand how voltage drop happens. From what I know, there should be about 0.7-0.6V drop across a silicon diode. This is exactly what happens when I remove the LED from this circuit. However, when LED is connected, each diode has about 0.2V drop. And the current across diodes is 10.7nA. I did this to see what the voltage would be on the node between R1 and LED. Let's call this NODE-1.

I considered 2 possibilities, first was having 1.6V on NODE-1 (which is what happened) and no voltage drop across each diode, meaning no current flow on them since 1.6V is not enough to forward bias every single diode. Second possibility was having about 6.3V-5.4V on NODE-1, enough voltage to supply 0.7-0.6 voltage drop for every single diode. In this case, since the drop across R1 is lower, meaning less current, the LED would be dimmer.

So my question is, why does the power supply only "consider" the 1.6V drop across the LED and sends current accordingly, instead of "considering" the 6.3V-5.4V drop across diodes? (which is what happens when you take the LED out of the circuit the circuit)

  • 2
    \$\begingroup\$ The forward voltage drop across the diodes vary based on forward current. The lower the current, the lower the voltage drop. \$\endgroup\$
    – Passerby
    May 31, 2016 at 21:24
  • \$\begingroup\$ Even when a diode is not 'properly' forward-biased, some small amount of current (you measured 10.7nA) will flow. This is why you're seeing a small voltage drop across each diode. \$\endgroup\$
    – brhans
    May 31, 2016 at 21:29
  • \$\begingroup\$ I cleaned up your language here. This is Electrical Engineering and as such there is a certain level of professionalism, which includes spelling, capitalization ... i.e suitability of expression. From the way you wrote the rest of your sentences it's clear you are more than capable of doing it properly. \$\endgroup\$ May 31, 2016 at 21:30
  • \$\begingroup\$ Also, we wouldn't think you were showing off if you used paragraphs to break the text into logically related sections. You have no question mark in your question. It all helps legibility. \$\endgroup\$
    – Transistor
    May 31, 2016 at 21:35
  • 2
    \$\begingroup\$ Never put diodes in parallel without a current limiting resistor or equivalent in series with each diode branch. Diode mismatch will cause you a lot of grief when debugging larger circuits. \$\endgroup\$
    – Cloud
    Jun 1, 2016 at 2:46

4 Answers 4


why does the power supply only "consider" the 1.6V drop across the LED and sends current accordingly

Note that power supplies don't "send current," instead they send voltage. The load resistor then "draws current" based on Ohm's law (or for diodes, based on the V-I curve.)

I think your confusion is caused by the concept "nonlinear resistance." Diodes don't actually turn on and off, instead they have nonlinear voltage/current behavior. Diodes don't behave as resistors, instead their current is determined by the applied voltage, and described by (oh no!) an exponential function. Because of the LED nonlinear resistance, even a simple LED with series-resistor isn't perfectly easy to understand.

Your circuit will be doubly-confusing because you're fighting two "nonlinear resistors" against each other: the LED's nonlinear curve, versus the nonlinear curve for the whole diode-chain. Nasty!


Here's one way to look at it. Suppose we slow things down by adding a large capacitor from NODE1 to GND, like 3,300uF. Next, when we suddenly connect the battery, the voltage on the capacitor starts rising. The capacitor voltage is also across the LED and the diodes. Eventually the voltage will arrive at the "fast rising" part of one of the diode graphs. In this case the LED arrives first (it's around 1.0V for red-color LEDs, higher for other colors.) The fast-rise part of the diode chain's voltage is around .4V for each diode, times nine, so roughly 3.6V, much larger than the LED volts. As the capacitor voltage rises, the LED "wins." The rising voltage will level out as soon as the resistor's Ohm's law behavior gives the same current as the V-I equation for the LED.

In other words, the diode-chain cannot draw significant current until your LED voltage goes above 3.6V!! This won't happen with a red LED and a 2.7K resistor.

However, if you'd used a white LED and a 100-ohm resistor, the diode-chain WILL draw significant current. If a white LED draws 30, 40, 50mA, the voltage can climb well above the usual 3V seen on white LEDs.

So, the answer to your question is different for different color LEDs!

See? Nasty.

In cases like these, the only way to make completely accurate predictions is unfortunately to abandon simplified mental models. Instead, write and solve equations. (This one has two exponential equations, one for the LED and another for the diode-chain.) Or, use a circuit simulator or Spice program which is invisibly solving equations for you in the background. Adding a capacitor and imagining slowly-changing conditions can take you far in grasping nonlinear electronics. But sometimes it's not obvious where that capacitor should be placed, or which nonlinear component will dominate.

  • \$\begingroup\$ I don't think it can be correct to say that a power supply 'sends voltage'; a voltage is a value measured at a point in a circuit (relative to another point), and the voltage at that point doesn't 'go' or 'flow' anywhere. I would say a standard voltage-regulated power supply sends current to its output so as to try and maintain a target voltage there. \$\endgroup\$
    – nekomatic
    Jun 1, 2016 at 8:00
  • \$\begingroup\$ > and the voltage at that point doesn't 'go' or 'flow' anywhere. WRONG. Voltages propagate throughout circuitry. Example: wide capacitor (such as a 2-wire transmission line,) charged by discharging an adjacent capacitor to one end. Voltage propagates across its plates, leaving a constant voltage on the entire plate-pair. (Yes I'm ignoring boundary reflections.) A similar analogy exists for "charging" a shorted coil by using a constant-voltage supply. \$\endgroup\$
    – wbeaty
    Jun 1, 2016 at 18:53
  • \$\begingroup\$ > "charging" a shorted coil by using a constant-voltage supply. WRONG. :) Oops, can't edit. (I meant constant-I supply, above.) Voltage propagates as e-fields, and of course requires a conductor-pair, usually a circuit node plus a ground plane or common point. Currents propagate too, as charge-flows and associated b-fields. From the energy-flow viewpoint, everything is a transmission line, including all loops and nodes. A constant-V supply, spreads the supply-voltage to the circuit rails, and then the circuit responds in the opposite direction of propagation, "drawing" a current. \$\endgroup\$
    – wbeaty
    Jun 1, 2016 at 19:07

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Figure 1. Current versus voltage for various coloured LEDs with curve for nine series diodes shown in black.

The chart of Figure 1 is a little rough but it should get the idea across. With your 2.7k series resistor max current will be about 3 mA. We won't be able to see this on the graph so I'm going to discuss the case when the LED is lit with 20 mA.

  • At 20 mA we can see that a red LED will have a forward voltage drop of about 1.8 V. This means that the voltage at the top of the diode chain will drop to 1.8 V.
  • If we look at 1.8 V on the diode curve we can see that the current through the nine diodes is going to be very small. I had a look for low-current/voltage graphs for the 1N4001 diodes but none had graphs for less than 10 mA.

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Figure 2. The 1N4148 small signal diode datasheet goes down to 0.1 mA.

  • Even the 1N4148 datasheet only goes down to 0.1 mA. If you extrapolate the graph you can see that for every 0.2 V or so the current will go down by a factor of 10.

So my question is, why does the power supply only "consider" the 1.6V drop across the LED and sends current accordingly, instead of "considering" the 6.3V-5.4V drop across diodes (which is what happens when you take the LED out of the circuit the circuit)?

It's not the power supply considering anything. The LED hogs the current and as can be seen from the graph will cause the voltage to drop to 1.8 V. That's all the nine diodes have to forward bias them and, as the 1N4148 chart shows, you won't get much current through them at that voltage.


Because the LED is also a diode. And it behaves as such. When you put it in parallel to your long diode string, the current finds it simpler to go through the single LED (because the voltage drop is lower this way) and the other diodes see (almost) no current anymore. Hence there is (almost) no more voltage drop across them.

By construction, the voltage across the single LED equals the voltage across the long string, right ? So, because the voltage across the single LED can't be greater than its voltage drop (current being limited by the resistor), the voltage across each silicon diode ends up being the LED voltage divided by the number of silicon diodes.

  • \$\begingroup\$ I see! So this is why when i remove all diodes except for one, the voltage drops to 600mV. Can i simply say smaller voltage drop will always be the determining factor in these sort of diode related problems? \$\endgroup\$
    – Eclipse
    May 31, 2016 at 21:46
  • \$\begingroup\$ In a parallel configuration, each branch will balance the current flow to end up at the same voltage. If you use only one diode, the lowest resistance branch will be the diode and you will most likely not illuminate the LED (depending on voltage levels for the LED). To figure these out, look at Kirchhoff's circuit laws: en.wikipedia.org/wiki/Kirchhoff's_circuit_laws You will need I/V curves for Diode and LED. \$\endgroup\$
    – Joe
    May 31, 2016 at 21:56
  • \$\begingroup\$ @Eclipse Yes, in practice, that's a good way to analyze a circuit like this. Note, however, that diodes are not very often put in parallel, because it makes them redundant. You could do it to share high currents and spread the power dissipation, but it usually requires some precaution to guarantee good current balancing between the diodes. \$\endgroup\$
    – dim
    Jun 1, 2016 at 6:55

You can certainly get less than 0.7V across a silicon diode. If you simply apply 0.1V to the diode, then that is your drop. You can even apply a negative voltage, up to the reverse breakdown. If you apply -0.3V, then that's the drop. Both of these are less than 0.7.

Of course, very little current will flow if the voltage is between reverse breakdown and forward bias. That's what's happening in your circuit simulation; not enough forward voltage is applied to the diodes to produce forward bias.

Diodes aren't voltage sources. They don't magically produce a constant 0.7V when plugged into a circuit. They limit at approximately 0.7V; it's hard to coax them into having a greater forward drop: so much current is required in order to sustain a significantly larger forward drop than 0.7V that it will fry the diode.


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