# How does voltage drop work?

I've been working with electronics for a while now, and I understand how to work with/account for voltage drop. But I'm still mystified as to how it works, particularly with components in series. Say I have two small incandescent lamps in series hooked up to a power supply. Through what physical mechanism is the voltage spread evenly over both lamps, resulting in both lamps running dim, instead of the first lamp consuming all the power and running at full brightness and the second running dark?

EDIT: It seems that people were getting hung up on the fact that I was using LEDs in my example and were focusing on whether the LEDs would run at all with lower-than-rated voltage. This wasn't my intention - I am focused on the phenomenon that both drop an equal amount of voltage instead of the lead LED dropping more voltage than the second LED. As a result I changed the example to incandescent lamps which show a closer-to-linear relation between voltage and brightness.

• Seems to me you understand it, partly anyway. LEDs have roughly the same current to voltage ratio, so if you hook them up in series the same current must run through them all. Because of the current to voltage ratio the voltages across them distributes evenly (almost). – itsproject Aug 5 '14 at 18:22
• If the lamps are connected in series, the voltage doesn't necessarily "spread evenly," the current does. If the bulbs are the same (have about the same resistance) then they'll each have about the same voltage drop. If the bulbs were very different in resistance (filament size etc), they might have the same current flowing through each and yet have different voltage drop across each, still adding up to the supply voltage of course. With bulbs in parallel they'd have identical voltage and different currents. – Matt B. Aug 5 '14 at 19:16
• What do you mean "...voltage spread evenly over both volts?" – JYelton Aug 5 '14 at 20:38
• @JYelton - perhaps OP meant "lamps": "...voltage spread evenly over both lamps?" – Kevin Fegan Aug 6 '14 at 2:37
• @Kevin That's what I assumed, and edited it as such; but in a subsequent edit by OP, it was changed back. – JYelton Aug 6 '14 at 2:38

Through what physical mechanism is the voltage spread evenly over both lamps, resulting in both lamps running dim, instead of the first lamp consuming all the power and running at full brightness and the second running dark?

Ohm's law. Both lamps, presumably having been manufactured similarly, have similar resistances. Given that both have the same amount of current running through them (KCL), they will have similar voltage drops across them.

• So what is the physical basis upon which Ohm's law is founded? What are the physical properties of materials that manifest Kirchoff's and Ohm's law in the real world? – TSL Aug 5 '14 at 18:31
• It takes a certain amount of energy to move electricity through a material, whether due to bouncing electrons from one atom to the next or to keep them moving when they get too close to a nucleus. We call the resultant absorption of energy "resistance". – Ignacio Vazquez-Abrams Aug 5 '14 at 18:45
• I think I get it (at least a little more than I started). Thanks! – TSL Aug 5 '14 at 19:14
• You can use the water analogy. Though like all analogies it breaks down in certain circumstances. Basically imagine the light bulbs as balloons with two necks, and imagine wires as water hoses. The balloons have a certain natural tendancy to draw in water (current) as it flows through them. Resistance is like the strength of the rubber that the balloons are made of. If the rubber (resistance) is the same for both bulbs, they'll have the same amount of pressure (voltage) relative to the source of the water. – Calphool Aug 5 '14 at 20:54

Think of each component as a two terminals black box.

This component will have a current that flows from one terminal to the other, and a voltage that you can measure across these terminals. What characterizes the component is the V-I characteristic, i.e. the function that given a certain voltage across the component allows you to calculate the current that flows into it, and vice versa. For example, for a resistor the V-I function is the known Ohm's law: $$V=I\cdot R$$

The physical mechanism that leads to a particular V-I characteristics is not a single mechanism but varies wildly between components, and would be quite a huge topic to be treated in a single answer. Let's take your LEDs: they obey an exponential law, meaning that: $$I_{LED}=I_0(e^{\frac{V_{LED}}{V_T}}-1)$$ where $V_T\approx 26$mV is the thermal voltage and $I_0$ is a parameter of your LED. Since you say that your LEDs require 3.3V, assuming that forward current would be 20mA you can calculate $I_0\approx 1.5\cdot10^{-57}$A from the above expression.

Now you connect two LEDs in series, each of them will have its own current and voltage, so how can you solve the situation? You throw in the equations coming from the circuit: $$V_{LED1}+V_{LED2}=3.3V\\ I_{LED1}=I_{LED2}\\ I_{LED1}=f_1(V_{LED1})\\ I_{LED2}=f_2(V_{LED2})$$

Now you can happily solve this system (note that it is not linear). It's quite easy because: $$f_1(V_x)=f_2(V_x)=f(V_x)==I_0(e^{\frac{V_x}{V_T}}-1)$$

So you can write almost immediatly $V_{LED1}=V_{LED2}$ that leads to $V_{LED1}=V_{LED2}=\frac{3.3V}{2}=1.65V$.

Please note that all of the above holds if the IV characteristic is true. There are some components that have "piece defined" characteristics, so that after finding a result you have to check that the characteristic you've chosen at the beginning is the right one.

If the LEDs are rated at 3.3 volts then more likely neither will be seen to glow at all - LEDs need to have a "certain amount" of voltage across them before they appear to be glowing.

On the other hand if you were "forcing" the rated current through them both you would have to be developing about 6.6 volts across the two in series - one LED couldn't arbitrarily decide to be conducting full current with very little volt drop allowing the other one to be at full brightness.

So, turn it round - for one LED to be fully on it has to be taking the rated current which has to mean that rated current is flowing thru the other hence it must also be at full brightness.

Your all LEDs are will not work. Because your LED needs 3.3V. If you are connected two LEDs in series. Current remain constant. But Voltage will change (Divide - if you are use same LEDs) .

• I left an edit which further clarifies what I'm asking here. I'm not asking about the brightness, I'm asking about why the voltage drop is equal over both LEDs. – TSL Aug 5 '14 at 18:26
• The resistance of same LEDs is equal. – Arshid KV Aug 5 '14 at 18:37

In series circuit - voltage splits proportionally to resistance.

In addition - bulb resistance is variable - depends on filament temperature. Higher temperature is higher resistance.

When you have two bulbs with diffrent power - one with smaller power will get more voltage initially and heat up fast. Resistance will rise. Bulb with higher power will heat up to lower temperature and resistance will be much smaller than in normal conditions.

You have to use identical or very similar bulbs with same power to split voltage evenly.