Hello, I am further trying to understand basic circuitry. I have attached a pic below of my circuit. I understand the basics of ohms law but I start to get confused when it comes to parallel circuits.
My first question---Is forward voltage similar to voltage drop? The reason I ask is that I noticed some Led calculators online would use forward voltage OR voltage drop as a variable to calculate resistance in a circuit.
My second question--Is my drawing above correct in the fact the each led would take 3v leaving 6v remaining for the resistor? Because i remember that in parallel circuits each branch is equal to supply voltage.
My final and most important question Is how do I go about solving this on my own to figure out what resistance i would need? Would it be correct to say that an LED would have 3v across it so OHMS law 3 Divided by 18ma? I really am not sure, if someone could please break it down to me. Thanks a million.
If you say "forward voltage", then there also must be a "reverse voltage".
A part with unequal volt drop in different directions, like a LED/diode/etc.
I think "volt drop" is more used for parts that drop equally in both directions, like resistors/wires.
Drawing is ok, but the LEDs are usually not drawn as a jellyfish, but as a diode with two 'light' arrows coming out of it.
So you also see polarity.
The two circuits only share the same power source. You can see them as two independent circuits.
Volt drop across the two LEDs in series is about 6volt, and fairly independent of current through the LEDs.
Current in the circuit therefore depends mostly on the value of the resistor (with 6volt across).
Calculate resistor value with V/A, or (voltage across resistor)/(wanted current), or 6/0.018= ~333ohm.
A standard 330ohm resistor will do.
Red power LEDs have a Vf of ~2.4volt. Other (power) LEDs (blue, white) are usually ~3.3volt.
Leo..
My first question---Is forward voltage similar to voltage drop? The reason I ask is that I noticed some Led calculators online would use forward voltage OR voltage drop as a variable to calculate resistance in a circuit.
Yes... If the LED is operating "normally" at it's normal voltage. If there is 3V across the LED and 3V across the resistor, they both have a 3V "drop" across them.
The confusing thing about LEDs is that they are non-linear. All diodes are non-linear. That means the resistance changes when you change the voltage. Ohm's Law is still true but it's not easy to use Ohm's Law directly with the diode. That's why we make our calculations with the resistor and then the current through and voltage across the LED "fall into place".
Wawa:
If you say "forward voltage", then there also must be a "reverse voltage".
A part with unequal volt drop in different directions, like a LED/diode/etc.
I think "volt drop" is more used for parts that drop equally in both directions, like resistors/wires.
Drawing is ok, but the LEDs are usually not drawn as a jellyfish, but as a diode with two 'light' arrows coming out of it.
So you also see polarity.
The two circuits only share the same power source. You can see them as two independent circuits.
Volt drop across the two LEDs in series is about 6volt, and fairly independent of current through the LEDs.
Current in the circuit therefore depends mostly on the value of the resistor (with 6volt across).
Calculate resistor value with V/A, or (voltage across resistor)/(wanted current), or 6/0.018= ~333ohm.
A standard 330ohm resistor will do.
Red power LEDs have a Vf of ~2.4volt. Other (power) LEDs (blue, white) are usually ~3.3volt.
Leo..
Very good explanation. Makes much more sense. Is the reason you combine both LED voltages as one, because of KVL? Then whats remaining Must either taken up by more LED or a resister?
Keep in mind assumptions are being made as Vfwd may be 2.7v etc.
You seem very knowledgeable. I just read this snippet from the website you linked me
"A series resistor is not necessary if the voltage can be regulated to match the LEDs (Vf). One way to do this is to match a battery to the LEDs. If your LEDs forward voltage is 1.2v, you can string ten of them (10x1.2v=12V) and power them from a 12v battery and not need a series equivalent resistor.
My question is how would you calculate the math for resistance and current when there is no resistor. Thank again for the help
What you want to do is control the current flowing through the LED.
Modern LEDs are much brighter than those of yester year.
An 18ma current in an older LED might be need to get reasonable illumination.
This same 18ma in today’s LEDs might be 3 times what is needed.
For example, many high efficiency SMD leds are plenty bright at 5ma.
With that being said, powering a LED without a resistor can be done with a 'current source' set to the desired level needed.
Driving a LED directly with a 'voltage source' is not recommended, (your so called battery matching).
Think of a LED as needing current, not a voltage.
The forward LED voltage is a result of the solid state material not the resistance of the LED.
However, if you want to think in terms of resistance, let’s say you have a Vfwd of 2 volts for a current of 5ma.
2v / 5ma = 400Ω (but don’t think of the LED as a resistor).
One example of a constant current source of about 35ma.
.7v / 20 Ω = 35ma
