The circuit is constructed correctly, but the values cannot be confirmed until you provide the forward voltage drop and desired forward current for the LED. This is usually obtained from the LED manufacturers datasheet.
stuck with equivalent resistance
- Thread starter clangray
- Start date
Vf = 1.25 (range for IR)
I = 15ma (.015A)
Please expand on the series string. I thought we were working with parallel.
To calculate the Resistor value (assume the LEDs are the same type):
Vsup=5V <---supply voltage
Vf=1.25V <---LED Forward voltage drop
If=0.015A <---LED Forward current (15mA)
nLeds=2 <---Number of LEDs in string
Vd=nLeds x Vf=2.50v <---total LED voltage drop
Rlimit=(Vsup-Vd) / If = 166.66 ohms <---current limit Resistor value (or next larger standard size)
To calculate the power dissipation of the resistor to determine its wattage rating.
I=0.015 Amps (or 15mA) <---current thru the string
Rlimit=166.66 ohms <---current limit resistor
Pd=0.015^2 x 166.66 = 0.0375 Watts (or 38mW)
So... use standard value 0.125 Watt (1/8 watt) or higher resistor power rating. Tolerance=5% or less.
How you're adding an ON/OFF indicator should be included in your diagram just for us to be able to absolutely confirm you've done it correct or if there's an error.
The indicated voltages at the first LED is 2.46V and at the second LED is 1.23V. This would suggest a Vf of 1.23 volts per LED. Also indicated in the original post is a current of 15.3mA. So to me it looks like a 1.23Vf @ 15.3mA.
It depends on what you are wanting to know.
It appears that way.....but I don't want to guess.
The TS has provided specific values in post #22.
Here's a SERIES circuit consisting of a resistor and two LED's. In series means one after the otter, back to back.
The second image labeled PARALLEL (actually a series resistor and LED with a duplicate set parallel) shows two pathways of assumed equal values. The upper set and lower set are parallel to each other.
The third image shows YOUR series/parallel setup. Each branch consists of (in series) a resistor and two LED's. Each series is in parallel to each other.
The second image labeled PARALLEL (actually a series resistor and LED with a duplicate set parallel) shows two pathways of assumed equal values. The upper set and lower set are parallel to each other.
The third image shows YOUR series/parallel setup. Each branch consists of (in series) a resistor and two LED's. Each series is in parallel to each other.
I saw that. It could be a rounded value to the diagram or could be a typo. In the actual circuit the Vf appears to be 1.23Vf and the current is shown as 15.3mA. Calc's of the Vf, R & V agree with the 1.23Vf and 15.3mA reports.
Even if we were to argue the 20mV difference and 0.0003mA difference it's all too close to be considered a significant difference. Unless we were talking about VHF (Very High Frequency) or higher. (That subject, VHF, would be "Off Topic" so let's not go there please). The question is "Is my circuit correct?" Answer is "Yes". But we don't know how the TS has connected the ON/OFF indicator. It could be its own parallel circuit or the TS may have inadvertently included the indicator in series with the other LED's, which would change things considerably. BTW: An indicator would seem superfluous. Since if the LED's are lit one would assume the circuit is ON.
Your welcome to read his/her mind if you want....
I've been scolded both ways - for reading their mind and for taking them literally. So if the TS reported one factoid and then another, all we can do is wait for clarity.Your welcome to read his/her mind if you want....
Then is it a parallel strings arranghd in para
The LEDs are invisible b/c they are 940nm IR - hence the need for power status. If my setup is parallel/series as you indicated, which math governs that part of the circuit?
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A couple of ways to add a power indicator. One would be to put an LED in series with the supply and adjust the size of the current-limiting resistors to account for the voltage drop. The downside is that this LED has to carry all of the supply current. If it can't handle the full ~30 mA, then you can put a shunt resistor in parallel with it sized to pass the excess current around the diode when the diode is at its nominal forward bias voltage.
Note that you don't have a lot of excess voltage to work with. Your two LEDs in series drop about 2.5 V, so you need to go with an indicator LED that drops as little voltage as possible. There are red LEDs that have Vf of about 1.5 V. Even that only leaves you will about 0.5 V of headroom, meaning that your current limiting resistors would need to be about 33 Ω. The worrying aspect is that small changes in either the supply voltage or the LED forward voltages will have amplified impacts on the LED current.
Let's say that you use a red LED having a Vf of 1.5 V and decide that you want about 10 mA flowing through it. That means that your shunt needs to develop 1.5 V across it when the other 20 mA is flowing through it, making it about 75 Ω.
It sounds like you could really benefit from spending time learning basic electronic circuit analysis. It would make your like a lot simpler and enable you to do a lot more things.
Note that you don't have a lot of excess voltage to work with. Your two LEDs in series drop about 2.5 V, so you need to go with an indicator LED that drops as little voltage as possible. There are red LEDs that have Vf of about 1.5 V. Even that only leaves you will about 0.5 V of headroom, meaning that your current limiting resistors would need to be about 33 Ω. The worrying aspect is that small changes in either the supply voltage or the LED forward voltages will have amplified impacts on the LED current.
Let's say that you use a red LED having a Vf of 1.5 V and decide that you want about 10 mA flowing through it. That means that your shunt needs to develop 1.5 V across it when the other 20 mA is flowing through it, making it about 75 Ω.
It sounds like you could really benefit from spending time learning basic electronic circuit analysis. It would make your like a lot simpler and enable you to do a lot more things.
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Hello,
Is it only the two resistors you need to be in parallel so that the entire circuit is in parallel?
Since you know the voltage to the right of each resistor VR and the voltage on the left VL, you can calculate the current though each resistor alone. Since there are two strings and current splits in a parallel circuit, that would mean that the total current from the power source is two times the single string current.
Now since you parallel VR with one string with VR of the other string and the voltages are exactly the same and you need to supply two times the current, you need a resistor that is 1/2 of the value of either resistor.
Now you know the current and the resistor value, so you can use P=R*I^2 to get the total power in the now single resistor. You should also use a resistor that is two times this power calculation, which is P2=2*R*I^2 and that is to keep the resistor cooler.
This only works in theory though because it is not usually possible to get two LED's in series to have the same exact characteristic forward voltage. Since there will be some variation there, it is best to use two resistors as shown in your original schematic. If this is entirely a purely theoretical question though, then yes one resistor of 1/2 the value does the same thing when the two strings are put in parallel.
I've seen white LEDs burn out when they are connected in parallel. What happens sometimes is one LED draws more current and it gets hot, and then it starts to blink on and off, then eventually it blows out. After that, the second LED draws more current then it should, so it blinks then burns out. It's a pain in the neck when this happens with a flashlight.
One of the Streamlight flashlights did this not sure if they corrected that problem or not that was several years ago.
When you say now single resistor is this a real resistor or an abstraction?
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Update: I want to swap the indicator LED with an illuminated push-button power switch. I am looking in the 5V range. Would this be okay to put the push-button in a series configuration with the circuit, or is some other configuration warranted?
Hi,
On paper you can calculate it and in real life once you calculate it you can buy a resistor of that value and run the circuit. Dont forget though the paralleling of LEDs problem is still of concern.
I really am paying attention but what is the #1 concern abut the parallels? There has been a lot of information and I am a bit mired down. Ill reread the posts in the meantime -
My component use in the 2 strings is the same type. Each string will be attached individually to the same Voltage source. I believe I have my power ratings calculated correctly. I have reduced he voltage source to 5V from 15V - although it wasn't clear that this needed to be done absolutely but I feel better about it.
What else do I need to address all of the parallel concerns. It may be the writing on the wall to go with total series if possible.
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