Discussion in 'General Electronics Chat' started by cjdelphi, Jan 27, 2011.

1. cjdelphi Thread Starter New Member

Mar 26, 2009
272
2
Let's call the load an LED (could as easily be a resistor)...

1 LED (or anything, let's say 100ma for argument sake)
----(14.5-2.4v)/0.100amp =121ohm resistor or 12.1v*0.1amp = 1.21 watts

A resistor of 121ohms and rated at say 2 watts connected to a 14.5v battery to light up a 100ma led...

So is the resistor going to heat up simply because of the current traveling through the resistor? or is the heat generated because it prevents more electrons flowing through when a load is applied? because without a load, the circuit would show 14.5v with no load and with a load (say an LED) the voltage would read in the 2.4-4.2v depending on the bandgap and the LED specs..

So why does the voltage drop when i measure it from the multimeter? should it not read 14.5v but only deliver 100ma to the load?....

2. Audioguru Expert

Dec 20, 2007
10,496
1,167
An ordinary 5mm LED is operated at 20mA or 25mA. It burns out if it is fed 100mA.
An LED limits its own voltage and if it is fed a higher voltage then its current will be much too high and it will burn out. The resistor in series limits the current, not the voltage.

The power heating the resistor is the voltage across it squared divided by the resistor value or the current in the resistor squared times the resistor value.

3. Wendy Moderator

Mar 24, 2008
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One of the common themes we get from beginners is LEDs and voltage (as opposed to current). There are LEDs that can take 700ma (or even 1.4ma), but they cost and are very bright. They are still white LEDs though, and drop the standard 3.6VDC. The resistor or other current source is the thing, not the LED.

Another common theme we are seeing with the high current LEDs is switching power supplys, basically a switching constant current sources. The efficiencies are much higher with this concept, as with voltage regulated SMPS. 9V @ 400ma may work out to 3.6V (LED Vf) @ 0.7A, for example. Because of the conversion vs. voltage drop they tend to run much cooler.

4. cjdelphi Thread Starter New Member

Mar 26, 2009
272
2
I was using an LED as an example of a load.... I never mentioned the type of LED either, 5mm or other (the ones I use handle up to 1amp @ 4.2volts)... I was merely using it as an example...

The question has nothing to do with LED's... it's why does the voltage drop when the max load is approached? What is the reason for the voltage drop? supply 100volts, supply 5ma, but if the max it can supply is 10ma, once you hit that 10ma, the voltage drops until it ends up being a couple of volts because the load is trying to consume more than the resistor can supply.

5. Wendy Moderator

Mar 24, 2008
21,351
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Thing I was saying, the LED is not the load. You MUST have a resistor in series with an LED, always, and it is the resistor that is the load. Without the resistor the LED is smoke.

The answer to your question is entirely Ohm's Law. It will accurately predict the current for a given voltage and resistance, just as your resistor is the load. If you put another resistor in series with the first resistor Ohm's Law covers that too.

6. cjdelphi Thread Starter New Member

Mar 26, 2009
272
2
what? I used it as an example, I have an SSC which takes 3.5 amps and is direct run off a lithium 4.2v (called Direct Drive in the flashlight world, no poofs of smoke anywhere to be seen because the LED handles 4.2v) battery ... but this is beside the point, I asked a question about why voltage drops when it can't supply the current to the load.

OK if you want to hijack the thread to make this all about LED's feel free, but my question is specifically about voltage drop and why it takes place (the LED was just a real world example but it could be a Bulb, electromagnetic, Resistor) anything...

Last edited: Jan 28, 2011
7. Wendy Moderator

Mar 24, 2008
21,351
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You keep trying to use semiconductors as the load, I assume you are talking SSR (SSC?), it too will smoke. I answered your question, probably during an edit. Do not assume someone is finished just because the post is up. It can take me 10 minutes to come up with a coherent answer.

Some limited cases a semiconductor can be a resistance (such as a BJT), not many. A resistor is almost always your load, and Ohm's Law is your guide.

You don't want to talk LEDs don't mention LEDs, our answers feed off of your setup conditions in the questions. You ask, we respond, that is how it works.

8. cjdelphi Thread Starter New Member

Mar 26, 2009
272
2
I thought i clearly stated the load could be anything...... i'd call an LED which can handle 1amp @ 4.2v's a pretty decent load would you not say? an SSC P7 (ssc is the manufacturer) like Cree.

[You keep trying to use semiconductors as the load, I assume you are talking SSR (SSC?), it too will smoke. I answered your question, probably during an edit. Do not assume someone is finished just because the post is up. It can take me 10 minutes to come up with a coherent answer.]

You really need to stop thinking you can smoke /all/ LED's by connecting them directly to a power source... yes pre 2003 and all the cheaper LED's (5mm) ones will smoke without a resistor (unless.. the voltage is lower than the forward voltage specs) ... all the newer higher power LED's can almost all certainly run without a resistor /providing/ it's heatsinked properly or you will get smoke. [providing the voltage is within spec, eg 4.2v lithium]

Last edited: Jan 28, 2011
9. Wendy Moderator

Mar 24, 2008
21,351
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Then you need to stop bringing up LEDs. I already told you they are not loads. The unit you show is a complete circuit, with drivers included. And I have answered your question, it is the resistors (or the total circuit). Do you understand Ohm's Law? It and the power equation are the answers you are asking for.

You don't want to talk LEDs but in the same post show pictures of a module using the same. Make up your mind please.

Ohm's Law:

E=I/R; E=Voltage, I=Current, R=Resistance

Power Equation:

P=EI; P=Power (in watts), E=Voltage, I=Current

And semiconductors like diodes will smoke without resistors. Why is this so hard to understand? Focus on the answers.

10. cjdelphi Thread Starter New Member

Mar 26, 2009
272
2
And semiconductors like diodes will smoke without resistors. Why is this so hard to understand? Focus on the answers.

Give me an example......

11. Commander#1 New Member

Jan 22, 2011
14
0

Hi, guys (and maybe gals):
From what I've read here, some explanations seem to be in order. It seems the biggest concern here is the
"voltage drop" - especially at the battery. All batteries are temporary storage devices - somewhat akin to
capacitors. All batteries have internal resistance. When current is drawn from the battery, it must pass
through this resistance. From OHM's law, a current passing through a resistance produces a voltage drop.
This voltage drop is subtracted from the battery's present voltage level as the current is being used. The greater
the current draw, the greater the voltage drop. An excellent example of this is the battery in your car. If you
connect a VOM to the battery, you will see somewhere around 14 volts. When you try to start the engine, you
will see the volt meter drop dramatically - to somewhere around 9 volts because, the starter motor is yanking
somewhere between 100 to 200 amps and the battery is rated for about 400 amps. All is fine and dandy as long as
the engine starts within the next 10 to 15 sec's. Grind that motor for 45 sec's and you will notice a distinct reduction
in starting RPM's. Why? Because you are reaching the limits of the battery. If you leave the meter attached and
simply turn on the over head lamp, the meter will barely move - because the lamp is pulling a very small current;
therefore, the battery will last just about all day.
Now a resistor is one of the few things where it's name tells you what it does - it resists the movement of current.
As such, a resistor will produce some amount of heat. To calculate how much heat is being wasted (and it is
wasted), measure the resistance value; measure the voltage drop across the resistor; use OHM's law to calculate
the current through the resistor (I=V/R - or - (old school- I=E/R)); calculate power(P) (P=V*I); power is measured
in Watts and we feel those watts as heat. A 100 watt light bulb puts out more heat than a 4 watt night light. Don't
believe me? Grab one sometime while it's on! Oh, yah!! As far as the resistor is concerned, get one rated double
(as a minimum) the power required and that will keep you out of trouble. You mentioned an SSC (hopefully you
mean a Solid State Controller. This idea would fit given the information you have provided). I have seen these
used in high speed electric model boats and combat robots. They are exceptionally well suited for anything
battery driven. The object of their use is plane old power management. Since the SSC uses a bit of power from
the battery just to operate, the combination of battery/SSC/load must be optimized for longevity with light weight
for the time frame involved.
So, yes, running current through a resistor will heat the resistor (depending upon resistance value and amount of
current) and battery voltage will drop depending - yes - on the amount of current demand. And, yes, you have
correctly determined the load (or bias) resistor and it's power rating for the given LED (". . for argument sake . ."
also known as 'theoretical'). Reality tends to be a little different from theoretical, but, the measuring process
is still the same.
I hope this is of some help to you. I also may have inadvertently stepped on some toes here, of which I sincerely
apologise. I Thank you for listening to this old codger. Until next time . . .

Phil Potter (Oh my gosh - the world is - - - round!?)

12. Wendy Moderator

Mar 24, 2008
21,351
2,918
Try connecting a regular diode or LED (2 different devices) across a battery without a resistor. Back bias is OK (though the LED may pop), forward bias lots of heat and smoke. It is the resistors in circuits that usually limit the current, whether it is a diode, transistor, SCR, TRIAC, whatever. The resistor is what sets up the currect flow in the vast majority of cases.

13. cjdelphi Thread Starter New Member

Mar 26, 2009
272
2
Direct Drive means (connect the LED directly across the battery without a resistor) the video shows it in action notice the lack of smoke?

I have a DD torch i'll rip it open and take pictures of it tomorrow.... i'll film it showing you exactly what you just asked for

14. beenthere Retired Moderator

Apr 20, 2004
15,815
293
There may be an internal resistor included in the package. A PN junction, once supplied with enough voltage to bias into conduction, has no means of limiting current. If the voltage source can supply more current than the LED (as this discussion seems to be focused exclusively on a specific type of LED) can"use". Excess current will heat and destroy the PN junction.

The direct drive LED must have current limiting built into the device. What does a meter indicate on a diode check? If it works on a 4.2 volt cell, what happens if attached to a stiff 6 volt supply?

Here is a link into our Ebook. It has a section on LED's. They all work the same way, even though the forward bias voltage may be different. Once conducting, current must be controlled.

15. Audioguru Expert

Dec 20, 2007
10,496
1,167
The CandlePower forum has flashlight guys talking about "how much would it pull from a 16340". They talk with many abbreviations.
I think an "it" is a high power Cree white LED with 4 chips in parallel as was shown here and maybe a "16340" is a lithium-ion battery cell that is 163mm long by 40mm in diameter. I think "pull" is the current.

A lithium-ion cell is charged to exactly 4.20V. Cree high power LED datasheets show a forward voltage of about 3.4V to 3.8V at 1A so maybe the battery protection circuit limits the current to 1.0A.