back to the basics

[zachtheterrible]

:lol: this is driving me nuts. ive never been able to figure this out:

say that ive got a 10v source. and ive got an LED that has a forward voltage of 2v (so therefore i need to have only 2v going to it). how do i use ohm's law to calculate the value of resistor that i need in order for there to be 2v @ the LED? take a look @ my picture.

i can figure out voltages in a series of resistances but this . . . :evil:

thanx

 

[checkmate]

When working with diodes, always work with current constraints rather than voltage constraints. If you look at the I-V characteristics of the diode, when switched on, the current can vary greatly with negligible change in voltage. Since light intensity of the LED is related to the current flowing through it, working with voltages are quite pointless.

Read the specs on the recommended forward current. And choose a suitable resistor that will drop the rest of the voltage given that current.

 

[zachtheterrible]

thanx checkmate. actually, its not just for the LED, its for electronics in general, i was just using it as an example. i would really like to know how to do this.

 

[checkmate]

There is no universal law for electronics. In most circuits, the voltages and currents are either too complicated to calculate theoretically, or can't be done at all. That is why control theory spans an entire field in engineering.

 

[zachtheterrible]

you mean it is impossible to tell me wut resistance i need in order to get 2v @ that part of the circuit!?!?

 

[checkmate]

you mean it is impossible to tell me wut resistance i need in order to get 2v @ that part of the circuit!?!?

I've already told you the solution in my first post above.

 

[Nigel Goodwin]

you mean it is impossible to tell me wut resistance i need in order to get 2v @ that part of the circuit!?!?

To be a bit more helpful, you will drop 2V across the LED, as the supply is 10V that leaves 8V across the resistor. So using simple ohms law, you now know the voltage (8V), and you just use that and the current you require to work out the resistor value you need.

But you're not calculating to get 2V - you're calculating for the current you require - the 2V is an intrinsic property of the LED.

 

[Styx]

Also to expand on what Nigel just stated (simple Ohm's law - 2V across LED implied 8V across the resistor) and that LED's are current devices..

BUT 2V across the LED implies that is is drawing a large amount of current (w.r.t. device rating) to cause the diode drop to rise to 2V. MY rule of thumb that is pretty good is for safety and resonable brightness a diode drop of 1.2V for 10mA. From that I know what resistor I need

It might be advisable to put a diode in series with the LED to increase the voltage drop across that part of the ciruit IF 2V is soo critical

 

[stevez]

If you have an LED and have no other information you might have to make assumptions on the voltage and current in order to calculate the proper series resistor. Making assumptions can be risky and working with the mfr's specs is the correct way to do it. The risks are that the LED won't behave as expected in terms of visibility, life (it could fail immediately), etc.


In the absence of this information I usually start with 2 volts and 10 ma - adjusting from there based on visibility of the LED. If you had a 12 volt supply you'd calculate the series resistor so that it took a 10 volt drop at 10 ma of current. When I do this I test it with my variable supply starting with a lower voltage and adjusting upward until the LED is as bright as I think it should be (just a guess and it could be wrong!) then I measure the voltage drops and make adjustments. With a fixed supply you could just start with a slightly higher series resistance and work your way down from there. I would not consider this good design practice - it's just a way for a hobbyist to get by with a junk box full of parts.

 

[Nigel Goodwin]

Also to expand on what Nigel just stated (simple ohm's law - 2V across LED implied 8V across the resistor) and that LED's are current devices..

BUT 2V across the LED implies that is is drawing a large amount of current (w.r.t. device rating) to cause the diode drop to rise to 2V. MY rule of thumb that is pretty good is for safety and resonable brightness a diode drop of 1.2V for 10mA. From that I know what resistor I need

It might be advisable to put a diode in series with the LED to increase the voltage drop across that part of the ciruit IF 2V is soo critical

I think you all need to get a grip on reality here!.

We're talking about a series resistor for an LED!, it's going to make sod all difference if it's 1V drop or 2V drop!.

Just to do the calculations, 2V drop and 10mA is 800 ohms, if the LED then only drops 1V - the current will be 11mA - BIG DEAL!. As 800 ohms isn't a prefered value anyway, we'd use either 820 (E12 series) or 750 (E24 series) ohms (or 1K? or 680?).

Just grab a resistor with a brown or red stripe at the end and see how bright it is? :lol:

It's hardly a crucial design problem!.

 

[Dr.EM]

I have to agree with this^

In electronics at college, they teach a lot of theory like this. I would never do it out of choice though, I would do as you said, get a resistor that sounds about right and change it until the LED at a suitable brightness. That said, there probably are times when this kind of theory would be useful, in precision work etc.

 

[stevez]

It's helpful to see both the theoretical 'good design practice' methods and the simplified methods. As a hobbyist I can often get by with the simplified methods - for most of my work it's good enough. I do appreciate knowing the shortcomings of the methods. For more critical work one has to use good judgement in knowing when simplified methods are appropriate - and when they are not.

 

[Edmond]

OHM law in practice

U1=1,6...2V and I=2...20mA, depends on the LED.
U1=1,8V and I=10mA are a general values for a LED.

 

[bmcculla]

If the manufacturer gives you a nice IV curve like checkmate posted but with the axis labled its easy to find the voltage drop for the LED at whatever current you want. It's always best to give a little margin so you arn't stressing the LED.

If you want to get maximum performance from your LED without worrying about the voltage drop you can use an opamp as a current source to drive the LED with a very predictable current regardless of the voltage drop. This is a good way to vary the brightness for sensor apps.

 

[zachtheterrible]

that little picture that i drew @ the begining of the post is PURELY for example. the LED could be a buzzer, or it could be a part of a circuit that requires 2 volts, WHATEVER!! so now that we're not talking solely about an LED, does what has been said by everyone still hold true?

wut i think is being said is that if the right amount of current is achieved, voltage will kind of "fall into place". is this true, or did i miss the whole point :roll:

thanx for your patience :lol:

 

[checkmate]

Then please refer to my 2nd post, or get a book on electronics. Like I said, there's no simple ohm's law for everything.

 

[Nigel Goodwin]

that little picture that i drew @ the begining of the post is PURELY for example. the LED could be a buzzer, or it could be a part of a circuit that requires 2 volts, WHATEVER!! so now that we're not talking solely about an LED, does what has been said by everyone still hold true?

wut i think is being said is that if the right amount of current is achieved, voltage will kind of "fall into place". is this true, or did i miss the whole point :roll:

Ohms law works with voltage, current, and resistance - so you've got to know two of those to calculate the third. In a purely resistive network there's no problem, you can easily work it out (although it may involve some complicated maths in complicated networks!).

However, when non-resistive devices become involved it all goes pear shaped

Your LED example is fairly easy, because you 'know' the voltage drop across the LED you can very simply calculate the voltage across the resistor - which in your example would be 8V. This gives you one of the three, you need one more to calculate the third - in this case you're trying to find the resistor value to give a certain current, so you decide what current you want (10mA, 20mA - 5A!) and then simply calculate the resistance from ohms law.

Obviously the current you decide on could be fairly crucial, depending on the application - the first two of my 'suggestions' would be fine, the third would be slightly inappropriate :lol: The design process involves a lot of decisions about which values to choose - often you only need to choose one, and the rest are calculated from there.

As for replacing the LED with 'something else' it would depend entirely what it was - if it was a resistor, you would know it's value (read the stripes!), and as you want 2V across it you could easily calculate the current through it. This would also give you the current through the other resistor, and you could calculate the value required from that current and 8V (the voltage across it).

A buzzer may well be a more complicated device altogether, and a motor certainly would be - both will have varying 'resistances' depending on their loads and voltages applied.

So there's no 'magic' method, you simply apply ohms law with a full knowledge of the components you're using - otherwise you're likely to do something really stupid :lol:

This is why electronics courses aren't just a few weeks long, and why many years of experience make even more difference!.

 

[zachtheterrible]

i think that i am starting to catch the drift :lol:
i had the wrong idea about voltage.

thanx :lol: