Ohm Explanation

[IlluminatingBikr]

I could really use some /ubbthreads/images/graemlins/help.gif understanding resistance and ohms.

Right now I understand the concept of resistance, but am not grasping the more technical parts of it.

 

[BF Hammer]

I'll try to keep this simple. Resistance is any part of an electric circuit that consumes electrical power. It can be as simple as a common resistor, a load (lamp, motor, ect.) or take the form of impedance in an AC circuit, where capcitors and coils act to resist electric current flow. Multiple sources of resistance in series with each other can be considered one single resistance equal to the sum of all the individual resistances (one path for current, lots of barriers to the current). Multiple sources of resistance in parallel with each other increase the current flow (multiple paths for current, voltage can push current past all resistance at once). The sum of all the current flowing through the resistance in each parallel path is equal to the total current measured at the source.

If you compare electrical forces to a typical plumbing system, current would be the water flowing in the pipes, the pipes would be the conductors (wires), water pressure is voltage, and and mechanical device driven by the water would be the resistance. I suppose the ground connection would be the sewage system. Let's say you're driving 3 turbines, if you use one pipe and connect them in series, the water must flow through all three devices, and the pressure is reduced at each device dependent on the rate of water flowing past each. Use three pipe paths (parallel) the pressure is equal for each turbine. This analogy can be extended more, but it is helpful for picturing how electical forces work and interact.

 

[Tomas]

Here's a reasonable place to start: Ohm's Law.

If you do a search (for example, with Google ) for Ohm's Law, you will find a large number of references and such. Some are better than others in presenting the information, and each person has a different way that a presentation will "click" with them.

Sometimes one can look at 20 different explanatations of the same thing, and not have it make sense. Then, on that 21st time, something is said in just the right way for it all to suddenly gel.

When I was giving seminars way back when, or even earlier when I was a trainer, I used to try different ways of presenting the information until I had that Ah-ha!!! look from each pair of eyeballs. It was a really good feeling when the "students" would obviously "get" it.

Hope this helps start you on your way! Remember, if an explanation doesn't make sense, there's a good chance it's that particular explanation, not you. Try another one! /ubbthreads/images/graemlins/grin.gif

 

[Lurker]

In a flashlight circuit with 4 cells and one white 5mm LED and a resistor in series, does the resistor consume battery power or conserve it (in addition to saving the LED)? What would happen to run time if you used an extremely high value resistor or an extremely low value resistor?

 

[BF Hammer]

The purpose of the resistor will be to prevent too much current from flowing through the LED. In addition to the information provided, you need to get the specs on the LED for forward bias voltage drop, maximum current value for the LED, and voltage of each battery cell. Standard alkaline cells are 1.5V each when fresh so 1.5 x 4 = 6V. The voltage drop of old standard 5mm red LED's is around .7V, the LED in question is just plain unknown. Subtract the voltage drop (my example) from the source voltage, you get 5.3V. Divide the max current into 5.3 and the answer is the minimum resistance required measured in ohms. This will give brightest recommended output, and minimum run time. Increasing the resistance dims output, extends battery run time. Decreasing resistance will overdrive the LED, perhaps giving brighter output and shorter overall life of the LED (or you turn it into a smoke-emitting diode). That's how you use Ohm's law to figure the current-limiting resistance required for an LED.

What about direct-driven LED circuits? Batteries have an internal resistance that increases as power is consumed from them. Some LED's have a high enough forward bias voltage drop in combination with certain types of batteries with a high enough internal resistance to allow the LED to illuminate without being damaged due to too much current.

 

[Doug Owen]

[ QUOTE ]
Lurker said:
In a flashlight circuit with 4 cells and one white 5mm LED and a resistor in series, does the resistor consume battery power or conserve it (in addition to saving the LED)? What would happen to run time if you used an extremely high value resistor or an extremely low value resistor?

[/ QUOTE ]

The answer to the first part is yes both times. Resistance consumes power anytime current is flowing. The formula is 'I squared R', that is the current in Amps squared (times itself) times the resistance in Ohms is the power consumed in Watts. However, in limiting the current typically the total run time is much improved so it conserves in that sense (even though contributing real energy loss in terms of heat generated in the resistor).

In your example, consider the battery 6 Volts and the forward voltage drop (Vf) of the LED to be 3 Volts, this means there is 3 Volts left 'across the resistor. If we use a 100 Ohm resistor, Ohm's Law tells us we have (3 Volts divided by 100 Ohms), .030 Amps. This is 30 miliamps, typically written 30 mA, a moderate overdrive for typical 5 mm parts. If instead we use a larger value, say 1000 Ohms we get (3 divided by 1000), .003 Amps, 3 mA, a tenth the former current. Still a useful level, BTW. Lower values, say 10 Ohms gets us .3 Amps, 300 mA. This is 'Luxeon country', our poor 5 mm part checked out first chance.

It's interesting to note that in each case the voltage across the resistor and LED is the same, 3 Volts each. And the current is the same, since they're in series, so the power consumed by each is the same. This automatically means 50% efficiency.

You can fairly easily recalculate things to see what's happening as the battery discharges. For instance at 5 Volts, only 2 Volts 'across the resistor' current levels (and therefore light output) is down by a third, but battery life is up a simliar (but not exactly equal) amount.

Run time depends on the current drawn (pull the power out at a higher rate and the battery dies sooner) and what your tolerance for 'not quite dead' is. If you keep the current low enough, very very long run times are possible. Direct drive and similar set ups often have this characteristic: very bright but short lifetime that quickly fades to progressively dimmer and dimmer output. This allows the maker to claim very impressive run times that most users never find.

Remember, there is always some resistance, in the battery, connections and internal to the LED device. We're just adding more resistance to the total.

Cheers,

Doug Owen

 

[Lurker]

Thanks for the great explanations, guys. Doug, in that example, since the resistor is consuming some power and heating up, would it make sense to replace it with a load that would produce light, such as another LED (or multiple LEDs) in series? Would that work to keep current down to safe levels and produce more total light over the battery life than the single LED with a resistor? What would be the drawback to that besides the monetary cost of additional LEDs?

Or maybe some combination of LEDs in parallel? Or reducing the cells from 4 to 3?

I am wondering because there are a lot of LED lights designed with this kind of circuit (such as the TurtleLight 1, Dorcy LED, PT Attitude, PT Impact 1 and 2, etc.) and I am trying to understand how using a resistor can possibly be the most efficient solution when trying to drive a 5mm LED on approximately 4 cells without a DC/DC converter.

Thanks.

 

[Doug Owen]

[ QUOTE ]
Lurker said:
Doug, in that example, since the resistor is consuming some power and heating up, would it make sense to replace it with a load that would produce light, such as another LED (or multiple LEDs) in series? Would that work to keep current down to safe levels and produce more total light over the battery life than the single LED with a resistor? What would be the drawback to that besides the monetary cost of additional LEDs?

Or maybe some combination of LEDs in parallel? Or reducing the cells from 4 to 3?


[/ QUOTE ]

Spot on, you've broken the code. Congratulations.

This is just the sort of thing the DD advocates are looking at. The rub comes from the real world (as it so often does), in this case what happens as the battery voltage changes. Remember, alkalines are down to 1.2 Volts per cell at the *half used* point, the maker considers end of life .8 Volts! Such a DD light would be out of light with 90% of the battery capacity unused. To avoid this, DD designs must seriously overdrive with fresh cells (so they at least put out some light with weak ones) with all the attendant trade offs. Adding extra resistance helps here, notice that in the calculation we just did, at the 'half used' point, 1.2 Volts we still had over half the original light out. Even when the cells are flat, we have four times .8 Volts, 3.2. Two tenths of a volt across the 100 ohms is 2 mA, still a useful level for some uses (it will of course, be fading fast).

Fewer cells makes this tougher. 3, while more efficient overall, will still stop with more useful energy left per cell.

Other batteries are worth consideration. For instance NiMH (and NiCd) cells have very stable voltages as they age (and generally lower and more stable internal resistance) at about 1.2 Volts per. So a pack of 3 gives us 3.6 Volts (and a bit more typically) through 90% or more of the usable life.

The next step is probably a low drop out regulator (a simple sort of DC/DC converter), sort of a 'smart resistor' if you will that adjusts the effective total resistance to set the current to exactly what we want and keep it there as the battery ages and the rest changes as it will (within the limits of available energy of course). Such a circuit can be as simple as a single transistor, a current regulator IC with 3 leads, 3 resistors and a small capacitor, totaling less than $2 and presenting few 'challenges' WRT building (no PCBs, bitty parts, etc.). Using 3 NiMH cells and such circuits I typically get 50 solid continuous hours of uniform light at 30 mA from 1700 mAH cells (nearly full capacity) at *over 90% efficiency*. The key is 'low headroom' (close match between the battery voltage and the requirement of the LED) and 'low drop out' (the ability of the regulator to work with very little voltage drop, under .1 Volt in this case). Funny bit is, when it shuts down regulation at the end of the battery, it's essentially a Direct Drive set up......

BTW, I think simple resistored lights are so popular because they're cheap and simple. IMO, not always good qualties......in flashlights or women.....

Doug Owen

 

[Lurker]

Thanks again, Doug. I followed most of that and feel enlightened.

And I can appreciate "cheap and simple" as much as the next guy. /ubbthreads/images/graemlins/naughty.gif

 

[IlluminatingBikr]

/ubbthreads/images/graemlins/eek.gif That is a lot of information for me to digest, especially during summer. /ubbthreads/images/graemlins/grin.gif Thank you very much for the help. /ubbthreads/images/graemlins/bowdown.gif

 

[IlluminatingBikr]

Ohm\'s Law Application

I think I have one more question now.... /ubbthreads/images/graemlins/help.gif /ubbthreads/images/graemlins/help.gif

Lets talk about the Blaster VI. It runs on 9 volts. You can use either a 2.2 ohm resistor, or a 0.5 ohm resistor.

Ohm's law states that I=V/R, where V=Voltage, I=Current, and R=Resistance.

In the first scenario, with the 2.2 ohm resistor, the formula is I=V/R = 9 volts/2.2 ohms = 4.09 and I am wondering specifically what the 4.09 is. Is that amps going to the LS?

In the second scenario, with the 0.5 ohm resistor, the formula is I=V/R = 9volts/0.5ohms =18? Is that possible that with the 0.5 resistor, the LS is receiving 18 amps? I don't think that is possible.

Could somebody please help me. Maybe I am misunderstanding what I is equal to in the formula. /ubbthreads/images/graemlins/confused.gif

 

[Jonathan]

Re: Ohm\'s Law Application

When you use Ohm's law, the I is current. R is the resistance, and V is the voltage _across_ the resistor, _not_ the total supply voltage, but the voltage across the resistor alone.

However you are not using Ohm's law correctly. If you were to connect a perfect 9V supply directly to a perfect 2.2 ohm resistor, _with nothing else_, then you would get 4.09A of current flowing through the circuit. Similarly, with a 0.5 ohm resistor, you would get 18A of current. In this particular circuit the resistor is connected to the supply, and the voltage across the resistor is simply the supply voltage.

But you will note that in the above paragraph I left out the Luxeon!!!!

To add in the Luxeon, you connect the supply to one terminal of the resistor, then the resistor to the Luxeon, then the Luxeon back to the supply. In a circuit, _all_ of the voltages have to add up to zero, so the voltage across the resistor is just the supply voltage minus the voltage across the Luxeon. To a first approximation, you treat the Luxeon as a fixed voltage load in your circuit.

So _very approximately_ the voltage that the resistance 'sees' is V(supply) - V(led), and the application of Ohm's law would be I = (V(supply) -V(led))/R

A better approximation would consider the fact that both the supply (batteries) and the load (Luxeon) are not fixed voltage sources, but in fact have some internal resistance. You can approximate a real battery as a perfect voltage source which decays over time, in series with a resistance which increases over time. You can approximate the Luxeon as a voltage load in series with a low value resistance.

In this case your circuit would be approximated as:
I = (V(supply) - V(led) / (R(resistor) + R(supply) + R(led) )

-Jon

 

[IlluminatingBikr]

I really appreciate all of your help, and I am sooo close to understanding this. I have noticed in the past, if somebody shows me something like a formula two or three times, I can grasp the formula. Later on I can usually figure out by myself the components of the formula, how it works, and why it works.

I think I would grasp Ohm's Law if somebody could write out the formula for ohm's law, and plug in the variables for the Blaster VI, with the 2.2 and 0.5 ohm resistors, and solve the formulas. I think this would really help me a lot. Thanks for all of you help. I am forever grateful.

 

[Doug Owen]

[ QUOTE ]
Illuminatingbikr said:

I think I would grasp Ohm's Law if somebody could write out the formula for ohm's law, and plug in the variables for the Blaster VI, with the 2.2 and 0.5 ohm resistors, and solve the formulas. I think this would really help me a lot. Thanks for all of you help. I am forever grateful.

[/ QUOTE ]

Fair enough.....'one more time, bartender....'.

First off, the part of Ohm's Law we're using is "Voltage in Volts divided by resistance in Ohms equals current in Amps".

To add to the fun, "Voltage in Volts times current in Amps equals power in Watts".

Let's assume the Vf is 7.0 Volts (probably pretty close, this would be a 'mid U bin' part) and our battery is fresh at 9.0. Let's also assume we have total of say an ohm in battery, other contacts, and our LED, so the extra .5 Ohms brings the total to 1.5 in the first case.

Nine minus 7 is 2 Volts across the total resistance (no matter where in the circuit it is). 2 Volts divided by 1.5 Ohms gives us a current of 1.33 Amps (about twice max). At this point our LED is dissipating 1.33 Amps times 7 Volts, 9.33 Watts. We need some serious heatsinking.

Now with 2.2 Ohms, we have 3.2 total. Dividing our 2 Volts by this 3.2 gives us .625 Amps (625 mA). Now we have .625 times 7 or 'only' 4.375 Watts to deal with (and roughly half the light out, with over twice the battery life).

You can change the battery voltage to reflect discharge, since Li cells like the 123 go down fairly linearly from 3.0 to 2.5 Volts before 'dropping off the edge', 2.75 V per cell (8.25 Volts total) is a good 'half used up' number. This means we now have 833 and 390 mA respectively. Even with the advantage of Lithium cells, it's clear that simple LED circuits like this have their limitations. This is about 'half bright', but the current is down, therefore the rate of discharge meaning it's going to take way longer to drain the remaining capacity than it did to get to the 'half dead' point. Consider at the 2.5 Volt point we pass only 333 and 156 mA, a quarter the initial. This is the time we'd like to lower the resistance and get some more light at the expense of the last of the battery (just what a true regulator does).

Doug Owen

 

[shiftd]

Oh, so that is where the Vf from the luxeon binning useful for. I thought the Vf only good to determine which led is brighter at certain voltage level, assuming all things being equal; never expect that it has such a deep calculation behind it.
thanks for the knowledge. this one shall be put on my tome of knowledge /ubbthreads/images/graemlins/grin.gif

 

[IlluminatingBikr]

[ QUOTE ]
Doug Owen said:

Fair enough.....'one more time, bartender....'.

[/ QUOTE ]

Thanks Doug, now I get it!!! Thank you very very much for your help. /ubbthreads/images/graemlins/bowdown.gif /ubbthreads/images/graemlins/bowdown.gif /ubbthreads/images/graemlins/bowdown.gif /ubbthreads/images/graemlins/bowdown.gif /ubbthreads/images/graemlins/bowdown.gif

 

[Doug Owen]

[ QUOTE ]
shiftd said:
Oh, so that is where the Vf from the luxeon binning useful for. I thought the Vf only good to determine which led is brighter at certain voltage level, assuming all things being equal; never expect that it has such a deep calculation behind it.


[/ QUOTE ]

Yup, that's a place it's useful.

The big V means we're looking at a voltage, the small f means it's the forward (as opposed to reverse) voltage. The missing bit is 'at what current?' since Vf is dependent on current (goes up with more current). In the case of LEDs this is usually 'full blast'. So this tells us that at the specified maximum current, 750 mA, there will be 7.0 Volts 'across the device'. Nichia and similar small guys are just rated at their appropriate current, usually 25 mA.

You can easily measure the Vf of the actual device you have, of course. Simply 'light it up' to the desired current level and measure the voltage across the leads of the LED.

Again, this value goes up with current increases (which is what makes constant voltage drives as stable as they are), but it goes *down* with temperature. As the device gets hotter, it's Vf drops something like .002 Volts per degree. This can mean more current and 'thermal run away'. Bummer that part.....

Doug Owen

 

[Doug Owen]

[ QUOTE ]
Illuminatingbikr said:


Thanks Doug, now I get it!!! Thank you very very much for your help.

[/ QUOTE ]

Hey, no charge.....

Careful, grasshopper, there's a hook in there.......

Doug Owen