# LED and lamp in series - why does the bulb not light?

My 6 year old son has just started experimenting with a Snap Circuits style kit and already we have a very basic question.

If we arrange an LED and lamp in parallel powered by batteries then both LED and lamp light up brightly.

However, if we arrange the LED and lamp in series then only the LED comes on.

Obviously current is passing through the lamp (if I unscrew the bulb the LED goes off).

So why does the bulb not light?

I'm a bit of a cheapskate so rather than buy proper Snap Circuits I bought a similar generic set from China on eBay (See: Electronic Blocks Kit W-58)

(Apologies if this is too basic for this forum but I haven't yet found the answer via Google)

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Can you give us a bit more background information? I remember that Snapcircuits are some sort of prefabricated modules that can be attached to work together and are supposed to be beginner-friendly, but they will most likely have some extra components that might be important for a good answer to this question. If you have any technical specifications of modules (at least which modules are you using exactly) or detailed, focused photos, they could be helpful. –  AndrejaKo Jan 22 at 22:03
The LED is a red led with a hidden 33 ohm resistor in series, the lamp is rated at 2.5V/300mA. Did you use a series resistor (such as 100 ohms) with the LED? –  Spehro Pefhany Jan 22 at 22:29
Not that I am aware of. –  Mark McLaren Jan 22 at 22:33
@Mark McLaren Unlike the lamp, whose brightness is controlled by power (product of voltage and current) dissipated by the wire, LED's brightness is proportional to the current going through the LED if the voltage is higher than a certain level. It's explained in Kaz's answer, but here's a bit more data: First, the voltage at the LED module is 3 V - Vf - 33* I=0, when we have just the LED in parallel with the bulb. Here, Vf is the special voltage LED needs to turn on. (cont.) –  AndrejaKo Jan 22 at 22:51
When we have both LED and the bulb in series, the voltage equation becomes: 3 V - Vf - 33*I -Rbulb*I=0. I is the current going through the circuit. 33 is the resistance of the included resistor on the LED module. The LED is basically such type of device that it will turn on as soon as there's current going through it and the source voltage is higher than the Vf, while on the other hand bulb won't be visible until the power dissipated reaches required level. That's one of the reasons why LEDs are replacing bulbs in many devices as indicator lights. –  AndrejaKo Jan 22 at 22:54

## 6 Answers

For the parallel connected LED and lamp, each has the entire battery voltage across.

When series connected, the voltage across each must sum to the battery voltage.

Without any more information than is given, the most likely answer is that the voltage across the lamp, which must equal the battery voltage minus the voltage across the LED, is insufficient to produce visible light.

While typing this answer, I see that you've added some pictures. It appears that the total battery voltage is about 3V. Given that many LEDs have a forward voltage in excess of 2V, this leaves less than 1V across the bulb.

Do you have a voltmeter with your kit? If so, measure the voltage across the lamp for the series connection.

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Brilliant thanks! –  Mark McLaren Jan 22 at 22:23
I think a more important point is the fact, mentioned by Spehro in a comment, that the incandescent lamp requires 300 mA, while a LED typically uses less than 20 mA, so the LED is limiting the current in the circuit to such a low value that the lamp won't make visible light. If you use two lamspps in series, rather than a LED and a lamp, both lamps should glow dimly, as they will have 150 mA or so. –  Peter Bennett Jan 22 at 22:51
@PeterBennett, LED current can vary widely whilst the LED voltage is approximately constant so I don't think your reasoning here is correct. The LED (ignoring any built in resistor), is limiting the voltage across the lamp and the lamp resistance (which varies with temperature) sets the series current. –  Alfred Centauri Jan 22 at 22:57
@Peter: a LED is not a resistor and it certainly doesn't "limit the current". The resistor is placed in series to limit the current to avoid burning the LED, but the current will solely depend on the voltage drop over the resistor, not the LED. As Alfred explained, a LED has a nearly constant voltage drop which results in a very low voltage being applied to the lamp, but doesn't provide linear ohmic resistance like a lamp. The lamp, on the other hand, is a purely (well, mostly) resistive load, where the current linearly depends on the applied voltage. –  Groo Jan 23 at 9:30
@Groo: I was perhaps a little sloppy in my explanation. Spehro mentions in his comment that the led part includes a 33 ohm resistor. Since the LED assembly works without damaging the LED when connected directly to the 3 V battery, it should be designed to draw less than 30 mA at 3 V, and would therefore limit the current to less than that when connected in series with the 300 mA lamp. –  Peter Bennett Jan 23 at 17:15

The LED drops so much voltage that there is very little left for the light bulb.

You have only two 1.5V batteries, which, in series, are barely enough for the LED forward voltage.

Incandescent bulbs quickly go dim when the power which they dissipate is reduced: power is voltage squared, divided by resistance.

For this very reason, dimming incandescent bulbs does not save much energy. Only a small fractional decrease in the dissipated wattage dims a light bulb almost all the way.

Filaments generate mostly heat, and only a small fraction as visible light. This is very sensitive to temperature, which is very sensitive to dissipated power.

Try looking at the lamp in a dark room; you may be able to see a faint red glow. Also, the light from the LED may be preventing you from seeing whatever dim glow the bulb is putting out, even in a dark room. Cover the LED, too.

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Great answer too, much appreciated. –  Mark McLaren Jan 22 at 22:32

There MUST be a resistor in series with the LED. An LED is a diode, and diodes rapidly increase the current they pass as the applied voltage rises above a certain point, well below 3V. So without a current-limiting resistor, the LED would pass so much current it would burn out.

The previous answers that say the LED drops the voltage are correct, but the drop is across the combination of the LED and the hidden resistor. The light bulb just adds a little more resistance, which reduces the current a little but only make the LED a tiny bit dimmer. But the light bulb is robbed of the minimum voltage it needs to light.

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Most correct answer. –  krs013 Jan 23 at 5:40
It may be worth mentioning that the internal "hidden" LED resistor is mostly relevant for the first OP's setup (the parallel circuit), which is where the LED is basically connected directly to 3V. The remaining 1V is then spread over the wiring and (luckily) that internal resistor which limits the current to avoid burning the LED. In the second circuit, it doesn't change the situation by much (presuming it's 33ohm vs lamp's 300ohm). –  Groo Jan 23 at 9:46

An additional and interesting factor is that the incandescent lamp's filament resistance when cold is roughly 1/10th of its resistance when hot.

Instead of 10 ohms, the cold unlit lamp is probably closer to 1 ohm.

At the 20mA of current needed to fully light the LED, the lamp filament dissipates only 400 microwatts of power. (Power = the square of current times resistance: $P = I^2 \cdot R$)

The series lamp is little more than a curly bit of wire completing the LED light circuit.

On the other hand, the positive-temperature-coefficient feature of incandescent lamps can be useful; see HP's early audio oscillator and read about Wien bridge oscillators.

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Welcome to EE.SE! Since this isn't really an answer to the question you should have posted it as a comment rather than an answer. –  Joe Hass Jan 30 at 21:46
It is an answer. It explains that at what's likely the operating current of the circuit, the lamp filament is never heated sufficiently to light. It further introduces the concepts of power and temperature coefficient, both unique and relevant contributions to the question. –  Phil Frost Jan 31 at 11:35
@PhilFrost Sorry, but I don't see anything like "heated insufficiently to light" in the answer. While you might infer that from the answer, I don't think the OP would. This "answer" doesn't even mention the LED at all. –  Joe Hass Jan 31 at 11:49
@JoeHass yes it does: "At the 20mA of current needed to fully light the LED". While I agree the point may not be clear to a six year old child or his father experimenting with SnapCircuits, it does attempt to answer the question. The proper course of action is to suggest ways for improvement, not to flag it for deletion. See “not an answer” vs. “not a good answer”, but this answer is not even "not a good answer", it's just a little brief for the likely level of understanding of the OP. –  Phil Frost Jan 31 at 12:01
Answering vs. commenting: I wasn't allowed to comment with 0 rep points. I looked for a balance between restating all the principles already clearly discussed and accepted by the OP and the insufficiency of just adding new information. To state that the LED is not mentioned at all is missing the fact that the LED is mentioned twice (maybe not in enough detail?); the answer was intentionally focused on the unlit incandescent lamp; said lamp's unexpected darkness being the source of the OP's question. –  SteveRay Feb 4 at 5:54

One way to Visualize, and therefore understand what is going on in the circuit, this being a very simplified explanation or analogy , would be by thinking that LED does not Pass enough current to light the lamp. Put simply, It has much more resistance than the lamp in essence. If you placed two identical type lamps in series, they should both glow equally bright, as bright as the current provided by the power supply allowed, As an really simple analogy, two 115 volt AC lamps of the same wattage placed in series would require + - 230 Volts to fully illuminate.

However the LED It has much more resistance than the lamp, or more precisely it is a semiconductor and has a forward voltage drop.

Simplified: A forward voltage drop across the LED of roughly 2 volts at say 20mA current draw, (I can't recall the precise numbers right now ) and that looks to be a 3 volt battery configuration, which would leave only about one volt for the lamp. Also the LED will not Pass along enough current to heat the filament (which is classified as an inductor and the LED a semiconductor, so my analogy is imprecise), but the lamp probably would not fully illuminate even if the supplied voltage was higher to compensate for the forward voltage drop of the LED.

I hope that makes sense. I apologize if not, but I am rushed and only saw this as I was about to have to get off line and do something else.

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Saying that LED does not "pass enough current" is misleading. a LED does not limit the current in any way, it simply provides a nearly constant voltage drop, which in turn affects the rest of the circuit. Remove all resistors from the circuit, and the LED will pass any imaginable current until it releases the magic smoke. –  Groo Jan 23 at 9:42

Great question! It's parallel versus series circuits, as noted. In parallel, both the lamp and the LED modules get a full 3 Volts. In series, they have to share the 3 Volts, so each gets some. If it were 2 of the same kind of bulb, each would get 1/2 of the voltage. 3 bulbs in series, each gets 1/3, and so forth. The LED module makes it more complex. But let's do the biggest issue first.

Each bulb is a resistor. Connect to the intended voltage, and current flows, entirely defined by the Voltage and Resistance. Mr. Ohm noted that $\dfrac{E}{I}=R$, by which he meant, for our purposes, Volts divided by Amperes equals Ohms. ("E" and "I" are physicist-shorthand for charge and current, "R" is resistance.) @Spehro quoted the lamp current as 300 milliamperes, .3 Amps. 3 Volts divided by .3 Amps equals 10 Ohms: $\dfrac{3V}{0.3A} = 10 \Omega$.

$\dfrac{V}{A}=R$, multiply both sides by $\dfrac{1}{V}: \dfrac{V}{AV}=\dfrac{R}{V}$. Drop $\dfrac{V}{V}$ and you get $\dfrac{1}{A}=\dfrac{R}{V}$ or $A=\dfrac{V}{R}$. Two bulbs in series, $10 \Omega + 10 \Omega = 20 \Omega$. $\dfrac{3V}{20\Omega} = .150A$. Half the current. $RA=V$, $10 \Omega \times .15 A = 1.5 V$, half the voltage per bulb.

The LED module makes it worse because the LED has a forward voltage drop of more tha 1 Volt and 33 Ohms in series with it. So the current is $A=\dfrac{3V - 1.xV}{10 + 33 \Omega}$, something in the range of $\dfrac{1.9V}{43 \Omega}$ to $\dfrac{1.1V}{43 \Omega}$. At most, less than 0.05 Amperes, perhaps as little as 0.025 Amperes. Between 1/6 and 1/10 the current that the bulb got in the parallel circuit!

For discussion purposes, red LEDs have a forward voltage drop of 1.(something) volts, green are around 2 Volts, blue even higher. Of course they have finite resistance values, but it's easiest to think of them as simply removing that fixed voltage. The current can then be calculated as the remaining voltage across the resistor. If there isn't a resistor, there needs to be some more elaborate way to limit the current.

For more fun, put an appropriate size electric motor in series with a bulb, note how bright the bulb is, then put some sort of load on the motor- a finger pressing gently, a fan blade or paddle to move air, etc. See any changes?

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