# Is this true about voltage?

I think something might have clicked a bit in my understanding today and I just want to make sure I have it right?

So without going into the concept of adding current, if I have an LED needing two volts hooked up to a 9 volt battery with a 7 volt resistor so that it's a total of 9 volts to balance out the circuit.
Does this mean that in terms of the batteries capacity even though the LED is only using two volts the battery will still be drained at a rate of 9 volts?

And if this is true then is this why you would use a relay instead of a resistor to step down the voltage so the battery lasts longer?

Sorry no such thing. A resistor is measured in ohms.

An LED uses normally 20mA, the resistor must also have the same current flowing through it. The extra energy involved from 9V as opposed to 5V is turned into heat in the resistor. So it is more wasteful.

Sorry I can't make any sense of this last bit. A relay is not something you would consider here

1 Like

No such thing but close. Since the LED drops 2v the resistor must drop the other 7 volts.

No, now your talking about current. The resistor can be chosen to allow only a couple of mA's or it can allow more based on the chosen ohm rating. If you chose a resistor that is too large(high Ohm's) then it will not allow enough current to light up the LED, If thr resistor is too small it will allow to much current and the LED will blow. But always the circuit will add up to 9v.

relay's do not step down voltage. they are just an electronically controlled switch.

They have the flow of electrons backward

1 Like

Volts is a measure of how much potential energy there is between the posts. It has nothing to do with how fast the battery is being drained. That is all to do with current.

Imagine a big water tank and a hose. We can raise the tank to make water spray out of the hose with more pressure. That's like voltage.

We can use a restriction on the hose to slow the flow down. That's current.

The thing that controls how fast the tank drains is the speed at which we let the water out. If the tank is higher then we need more restriction to keep the speed there. But the height of the tank is irrelevant, only the speed that we allow the water to drain.

It's current that drains the battery.

The resistor is wasting power (1) but without it, you'll get more current, pulling more total power/energy out of the battery and draining it faster, and possibly burning-out the LED.

Current "flows" (sort-of like water). The same current flows through the resistor and LED. And with series components, the voltage divides (so you can have 7V across the resistor and 2V across the LED).

High resistance is like a skinny pipe that restricts flow. resistance w (or no resistance) resistance is like a fat pipe and the electrons flow more easily.

Resistance is "the resistance to current flow".

Voltage is like water pressure... With more voltage you get more current (through the same resistance).

This is all described mathematically by Ohms Law: Current = Voltage / Resistance.

But the water analogy only goes so far... If you cut a pipe you get zero resistance and water flows-out all over the place. If you cut a wire you have infinite resistance (no connection) and no current flows.

No water resistance isn't really a problem (as long as you don't have a flood) but zero electrical resistance is a "short circuit" and you get excess current and your battery dies quickly or your power supply blows a fuse or burns-out.

You can have 7V across the resistor, but there are no "7V resistors". LEDs do have a "forward operating voltage" and they are non-linear (their resistance goes up as voltage goes down) so if the current is properly controlled/limited the voltage "magically falls into place".

A relay is an electrically-controlled and electrically-isolated switch. For example, a relay can be used to turn on/off AC power with the Arduino which can't directly-power much more than a little LED. In fact, it can't directly-power a relay coil and it needs a driver circuit, but the relay isolates the Arduino and you from lethal voltages.**

LEDs always require a resistor or some kind of current limiting/control because they are "odd" non-linear devices. Most "things" work with "constant voltage" but LEDs want "constant current".

(1) Power (wattage) is calculated as Voltage x Current. With 20mA through the LED and 2V across it, the LED is using 40mW. With 7V across the LED, it's dissipating 140mW (as heat).

Thanks, i guess knowing when i need to consider current vs volts, and espically when both are used is what is just going over my head.
Cause if a led says it needs 2v, and all you got is a 9v battery, and you can balance it out with a resistor, i am missing the part where that 7v in question is now current instead of volts.

I guess for more simplicity say its just LED, resistor and battery, no microcontroller mixed in to try and get my brain on par with this concept.

The 7v dropped across the resistor can be measured with a multimeter by placing the probes on both sides of the resistor. It is the size(ohm's) of the resistor that controls the current. Study the pic in post #4.

Ok, so to some level of sorta right concept, using wrong terms, in a nut shell i was kinda partly getting it in the back of my mind in terms of the addition of the resistor does increase the rate of drain on the battery?

Let's simplify it a little more and talk about two resistors in series which makes a Voltage Divider. If the resistors are equal the voltage divides equally. The resistances sum so the current depends on the voltage and the sum of the resistances. You can run-through the Ohm's Law calculations for the resistors together, or separately for each resistor.

The LED and resistor also make a voltage divider but the non-linearity of the LED makes it more difficult to understand. Ohm's Law still applies... It's a law of nature with man-made units-of-measure... But the resistance of the LED depends on the voltage across it.

@mythn7 one of the first things you will ever learn in electronics is usually Ohms law, it starts quite easy but can become complex. I would suggest reading and watching a few videos to become familiar with some easy Ohms law examples. Just google Ohms law there are thousands of examples. Here is a simple video to get you started https://www.youtube.com/watch?v=HsLLq6Rm5tU

EDIT when you start to feel comfortable with the concept then start to research voltage drop and voltage dividers

If the city water pressure, where it comes into your home, is 40psi, what is the water pressure at the end of a 500 foot garden hose.

> 40
< 40
= 40

Why ?

So in terms of the pic, where im confused using the water analogy is, a small pipe creates back pressure, as long as the hose doent blow, NONE of the water is wasted or transformed into something else. So in terms of capacity nothing is wasted.

But a resistor i understand takes the energy and turns it to heat for venting away, which to me seems more like a hose with a pinhole in it.

Where you are going wrong is thinking you can consider voltage and current separately. You must consider them both at the same time along with the resistance of a path.. There is no current without a voltage and a circuit path to follow.

Ohmâ€™s law formula. R = E/I

E = voltage in volts
I = current in amps
R= resistance in ohms

You want to select a resistor that will limit the current to a valu that lights the LED but does not allow enough current to damage the LED.

A typical LED maximum current is 20 mA (a mA is 1000th of an amp so 0.02 amps)

So you need to drop 7 volts across the resistor. Letâ€™s select 10 mA (0.01 amps) as our desired current. Substituting into the formula gives:

R= 7V/0.01A which results in a resistor value of 700 ohms. I believe the closest common 5% tolerance resistor is 680 ohms, which is close enough for LED work.

Look for an introductory electronics video on YouTube for a better explanation of Ohmâ€™s law. Use Ohmâ€™s law as a search phrase maybe.

EDIT: someone pointed out to me that I had the formula wrong. Due to a typo I had two â€śRâ€ťs in the formula and not enough â€śIâ€ťs. I have corrected it. Sorry for the confusion.

That is correct but if it is sized correctly the "pin hole" is very small while the energy saved by keeping the LED in check is much larger.

Actually comparing electricity to fluid flow doesnâ€™t get too far before the analogy becomes torturous, so it is better to not go down that path.

1 Like

All of the water/electricity that comes out of one end of the battery goes back into the other end of the battery. None leaks. It isn't a diversion to the environment that reduces the power delivered to the LED.

The energy dissipated in the flow around the circuit happens where the flow meets resistance.

The small pipe creates friction and a pressure drop. If you completely stopped the flow with a valve, there would be no back-pressure due to the small pipe, but only to the high resistance of the valve.

If you didn't have the pressure drop from the resistance of the small pipe, many more amps would flow through the smaller resistance in the rest of the circuit (the LED) and generate much more heat in the LED (for a limited time)

The power law (`P=I*E`) is what converts amps and volts to heat:

Numerically the heating at the LED is `AMPS^2*R_led` and the heat at the resistance is `AMPS^2*R_resistor' The power coming out of the battery is the sum of both. Increasing the resistance reduces the power drawn out of the battery and dissipated in the circuit.

Maybe it would help to think about it this way.

The voltage is not the capacity of the battery. A way to think about that is to think about your 9V. When it's halfway through its lifetime it still has close to 9V. It doesn't go down to 4.5V.

It's not the volts that you're using up.

There is an excess of electrons on one side of the battery and a deficit on the other. You are allowing electrons to move from one side to the other. The amount of useful work any one electron can do on its way round is related to the voltage.

(for the remainder I'm going to way oversimplify the quantitative side of the units and focus just on what they represent)

But it's the number of electrons in the battery that are available to be moved that is the capacity. Once you've moved all you can move, then the battery is dead. And each one of those electrons has the same amount of potential energy as when the 9V was fully charged.

For each electron, as he goes around the circuit, he has to do all of that work. 2 volts worth of work gets used up in the LED and 7V worth gets spent on the resistor so that he gets to the other side with 0.

And this is true for each of the electrons that are going to go across. When you've used up half of them, then you have half the electrons but they still have that same 9V worth of potential each.

It's when you run out of electrons that the battery is dead.

And the number of electrons making the trip per time is the current.

So if you're moving 2A of current you will drain the battery twice as fast as if you are moving 1A of current.

In reality the voltage of the battery does go down over time. But this has more to do with the chemical reaction that is driving the voltage. And it only goes down by some. It isn't the quantity that defines the lifetime of the battery.

For example, at half lifetime a 9V battery will have way more than 4.5V. It will still be very close to 9.