embarrasingly easy newbie question

[captaincaveman]

this is the one thing i always struggle to get with ohms law

say you have a 12v supply and want to put a L.E.D across it how do you select the resistor you need to reduce the voltage to say 2V for L.E.D?knowing the max forward current is 30mA


with ohms law R=V/I, therefore will 12v/20mA(20mA as mid way current) give you the resistor you need? This is where i get confused, cause im trying to get 2v across the resistor. and what would you choose the power rating of resistor based on?

sorry if this is too simple, but its a sticking point

 

[Hero999]

The voltage drop across the LED doesn't vary much when the current flowing through it changes, it doesn't bahave like a resistor. Different colour LEDs have different voltage drops and are typically as follows: infrared 1.2V, red 1.8V - 1.9V, orange 1.9V - 2V yellow 2.1V, green 2.2V- 3.5V, blue/white 3.5, and voilet-ultravoilet 3.5V-4.2V, it's a good idea to look at the datasheet.

Calculating the series resistor value is fairly straight forward, you simply subtract the voltage drop from the supply then use ohm's law, as follows:

R = (Vsupply - VLED)/ILED

 

[dknguyen]

Diodes are not resistors (most devices aren't!) Once a diode begins to conduct (the forward voltage exceeds a certain threshold) or the reverse voltage exceeds what the diode can block (it will breakdown and will conduct in the reverse direction) the voltage drop across it remains fairly constant with current.

So, knowing this, if you stick a resistor in series with the diode, you know that the voltage drop across the diode must be Vdiode (it's called the forward voltage drop in the datasheet), therefore you know that the remainder of the power supply voltage will appear across the resistor: Vresistor=Vsupply-Vdiode. From this you obviously know the voltage across the resistor, and can easily do Vresistor=IR to get whatever current you need in the circuit.

 

[captaincaveman]

The voltage drop across the LED doesn't vary much when the current flowing through it changes, it doesn't bahave like a resistor. Different colour LEDs have different voltage drops and are typically as follows: infrared 1.2V, red 1.8V - 1.9V, orange 1.9V - 2V yellow 2.1V, green 2.2V- 3.5V, blue/white 3.5, and voilet-ultravoilet 3.5V-4.2V, it's a good idea to look at the datasheet.

Calculating the series resistor value is fairly straight forward, you simply subtract the voltage drop from the supply then use ohm's law, as follows:

R = (Vsupply - VLED)/ILED

so you take the VLED and ILED directly from manufacturer then, gotcha now, thanks that cleared that, never could get it, but didn't know that equation.

one more quicky, what if you have a circuit that works on 9v and you want to supply it from 12v, how do you find the resistor then?

 

[Nigel Goodwin]

so you take the VLED and ILED directly from manufacturer then, gotcha now, thanks that cleared that, never could get it, but didn't know that equation.

Not quite, ILED is the current YOU want!, the datasheet will tell you the MAXIMUM permitted.

[qyoue]

one more quicky, what if you have a circuit that works on 9v and you want to supply it from 12v, how do you find the resistor then?[/QUOTE]

Same formula! - simply insert 12V instead of 9V in the (VERY simple) equation.

 

[captaincaveman]

do you mean in the R=(Vsupply-VLED(but 9v circuit)/ILED(current of circuit)? will that therefore put 3v across the resistor to leave the 9v across the original circuit?

 

[Nigel Goodwin]

do you mean in the R=(Vsupply-VLED(but 9v circuit)/ILED(current of circuit)? will that therefore put 3v across the resistor to leave the 9v across the original circuit?

'VSupply' is the supply voltage, previously it was 9, with a 12v supply it becomes 12 - there's only one resistor!.

 

[captaincaveman]

yeah, sorry my wordings not brilliant. so it would be

R=12V-9V/I(of original circuit) is that right? and would that mean there would be 3v across the new resistor?

 

[Nigel Goodwin]

yeah, sorry my wordings not brilliant. so it would be

R=12V-9V/I(of original circuit) is that right? and would that mean there would be 3v across the new resistor?

No! - why have you added the 9V as VLED?. VLED is the voltage drop across the LED, and will be around 2V or so. No NEW resistor, just one to replace the original.

If you wanted to add a resistor in series with the original one, then the voltage drop would indeed be 3V across it - but why use two resistors instead of one?.

 

[captaincaveman]

No! - why have you added the 9V as VLED?. VLED is the voltage drop across the LED, and will be around 2V or so. No NEW resistor, just one to replace the original.

If you wanted to add a resistor in series with the original one, then the voltage drop would indeed be 3V across it - but why use two resistors instead of one?.

no sorry i confused ya, it got the equation with the led, this was on about another circuit, supplyed by 9v not the LED one, if you had a circuit say a timer curcuit that ran at 9v and you wanted to connect to 12v, then does that equation work to drop the voltage the 3v needed to run the 9v original circuit?
as in

Vsupply-V(of original 9v timer)/I(of 9v timer circuit)

just trying to get a hang of general way of dropping voltages for any given circuit with a series resistor, so i could run the 9v timer circuit on 12v, or 24v or whatever supply i got, or if i had a 3v circuit i wanted to run off 12v etc

would that equation suit all if i knew the current and voltage of any circuit?

hopefully i've put that better

 

[audioguru]

Why won't your 9V timer circuit work directly from a supply of 12V? Most will.

You can't reduce the supply voltage for an entire circuit with just a resistor. Its current is not constant so the voltage at the end of the resistor won't be constant. A voltage regulator would work to reduce the supply voltage from 12V down to 9V for the circuit.

 

[captaincaveman]

Why won't your 9V timer circuit work directly from a supply of 12V? Most will.

You can't reduce the supply voltage for an entire circuit with just a resistor. Its current is not constant so the voltage at the end of the resistor won't be constant. A voltage regulator would work to reduce the supply voltage from 12V down to 9V for the circuit.

it's a hypothetical circuit, its something i've always struggled on how to reduce voltages on circuits that i want to work off of car/bike battery that wont work directly, i was hoping on a general solution, could a voltage divider be used? or would it be a regulator on all these kinda circuits? what about zener diodes? when can they be used as regulators? or am i asking too many questions?

 

[audioguru]

The output voltage of a voltage divider changes when the input voltage changes and when the load current changes.

A zener diode and the resistor that feeds it are a voltage divider. When the load doesn't draw much current, the voltage will try to rise but the zener will conduct more current to ground to keep the voltage about the same. The current to ground is wasted.

A voltage regulator is an automatic series variable resistor. It uses a transistor inside to divide the voltage so that its output voltage doesn't change when the input voltage or the load current change. It has very little wasted current.

 

[captaincaveman]

The output voltage of a voltage divider changes when the input voltage changes and when the load current changes.

A zener diode and the resistor that feeds it are a voltage divider. When the load doesn't draw much current, the voltage will try to rise but the zener will conduct more current to ground to keep the voltage about the same. The current to ground is wasted.

A voltage regulator is an automatic series variable resistor. It uses a transistor inside to divide the voltage so that its output voltage doesn't change when the input voltage or the load current change. It has very little wasted current.

thanks for that , but could i still use the voltage divider, as the fluctuations might not be that important if they are within tolerences of the circuit on the output of the divider, and not too concerned by the losses on some circuit cause they would be tiny when the engine is running i assume. just if i can help it i wouldn't want to have to use a regulator is not totally critical to circuit and have always got loads of resistors kicking about

you've been really helpful guys, not just on this but it also unlocks other things too

 

[Hero999]

https://www.electronics2000.co.uk/data/itemsmr/potdiv.htm

**broken link removed**
Say the series resistor is R1, your circuit is R2, the supply is V1 and V2 is the output voltage across your circuit. Remember your circuit doesn't behave like a resistor, its resistance changes very widly when it switches (by a factor of 100 or so), so the voltage across your circuit will vary just as much. This is why a single resitor is useless for reducing the voltage in most applications, you can only rely on it if you're certain the device has a fixed resistance.

 

[captaincaveman]

https://www.electronics2000.co.uk/data/itemsmr/potdiv.htm

**broken link removed**
Say the series resistor is R1, your circuit is R2, the supply is V1 and V2 is the output voltage across your circuit. Remember your circuit doesn't behave like a resistor, its resistance changes very widly when it switches (by a factor of 100 or so), so the voltage across your circuit will vary just as much. This is why a single resitor is useless for reducing the voltage in most applications, you can only rely on it if you're certain the device has a fixed resistance.

sorry if im annoying you, you said a single resistor is useless, but are you saying the divider is ok?

 

[dknguyen]

He is saying that connecting a single resistor in series with a circuit and relying on the circuit's resistance to be constant so that the same voltage drop will always occur across the resistor (to reduce the voltage across the load) will not work because the resistance of the circuit changes by a LOT during operation.

With the resistive divider, the division ratio between input and output is:
(R2||e)/(R1+(R2||e)),

where e is the load resistance and should be very large, so large that R2||e ~ R2. || means parallel (you do know the equation for parallel resistors right?). Just think about it or look at the equation, for parallel resistors, the smaller resistor dominates and when the two resistors are very very very far apart, the smaller resistor completely dominates the equivelant resistance (a 1k resistor siting in air is like two resistors in parallel, 1k and a very very large resistor due to the high resistance of air- what do you get when you measure the resistance? just 1k from the smaller resistor).

So you can see here that the division ratio in the divider is less sensitive to changes in the load/circuit resistance than with just a single resistor. For a voltage divider to do it's job, the load resistance (between V2 and 0V) must be much less than R2 so as to have a neglible amount of current from the divider flowing into the load. This is why it is less sensitive than the single resistor- as long as the load resistance is much less than R2 it will work fine. With the single resistor any change in load resistance will cause an equally large change in the reduction ratio....however the divider it is not infinitely resistant to changes in load resistance. Notice that in the schematic, between V2 and 0V there is no wire? This is because it assumes that the circuit's input resistance is VERY high compared to the resisitve divider. The lower it gets, the more it skews the reduction ratio since current will leak out of the divider into the circuit.

(You can't provide power using the divider for this reason- it only provides a voltage, no current. As soon as a current flows into the circuit through V2 the reducton ratio is skewed and you get a LOT of current wasted that justs go straight from V1 to 0V).

But for your application a divider, it won't work since you are trying to provide current to the load, not just a bias voltage (some devices like MOSFET transistors in amplifiers are voltage controlled and need to be biased at a certain voltage and have very high gate resistances (the gate is a terminal used to bias the transistor) and therefore consume almost no current to bias, so dividers are used in these cases. As described earlier by me and several others, there are better ways to do it.

 

[captaincaveman]

thanks, got it now, didn't think about the circuit being a second resistor in parallel with R2 being a problem, but seen it now. also i see what you mean about the current too, instead of using the divider to put a certain voltage into an input of a circuit that already has a supply, im trying to supply the whole circuit from the same type of divider.

understand now where i was going wrong, cheers guys

 

[kchriste]

captaincaveman,
You also may interested in "Thevenin's Theorem" which allows you to simplify voltage divider circuits to make it more obvious how they will behave under various load conditions. See link:

 

[captaincaveman]

thanks, very useful