Resistance is Futile!

[WoodHokie4]

This has absolutely nothing to do with brewing, though I will probably have a few handy while working on this project.

Can anyone help me out with some resistance calculations for this wiring diagram? It seems pretty straight-forward, but since my diodes are all of different voltage drops, any of the standard calculators I've been able to see online are not helpful.

Like I said, nothing to do with brewing, but I figured, who better to ask than brewers?

Thanks in advance!

 

[Btaz]

I think your diagram is showing the voltage that you want on top ob the diode (I'll call this Va. If so, then for each path the equation is R = (Vcc-Va)/I where Vcc = 9V, Va and I vary with the path.

 

[Btaz]

Though looking at this a bit more I'm not sure this is going to do it an you will only see about a 0.7v drop across the diode. Meaning Va will not ever be the desired voltage.

 

[WoodHokie4]

Thanks Btaz. If this is the case, would my resistance be as follows from left to right:
350, 345, 270, 345, 340, 270 (ohm)

I wasn't sure because some of the parallel diagrams i've seen show really low resistor values (like 43ohm).

 

[WoodHokie4]

Though looking at this a bit more I'm not sure this is going to do it an you will only see about a 0.7v drop across the diode. Meaning Va will not ever be the desired voltage.

So, should I up my Vcc voltage? or not bother with resistors?

 

[Btaz]

What are you trying to do with each leg?

 

[WoodHokie4]

What are you trying to do with each leg?

each leg?

I'm trying to put together a lighting assembly and just want each diode to light up.

 

[gorrie16]

Without the resistors on each leg, you would not be able to control the 20mA going through each leg as the resistors are sized to created that current. Also, without the resistors, all your diodes would be tied to the same positive voltage and ground, in turn fighting to output different voltages at the same node. Something will be bound to give up the smoke. The resistance values posted above are correct.

 

[WoodHokie4]

Without the resistors on each leg, you would not be able to control the 20mA going through each leg as the resistors are sized to created that current. Also, without the resistors, all your diodes would be tied to the same positive voltage and ground, in turn fighting to output different voltages at the same node. Something will be bound to give up the smoke. The resistance values posted above are correct.

Resistors on both legs of the diode? All the diagrams I've seen have resistors on the (+) leg only and sharing a common (-) leg. It looked like the resistors were put in place to normalize the voltage across each diode. Is there something I'm missing?

 

[doug293cz]

Where did the voltage drops across the diodes come from? Are they the spec'ed drop at 20mA, or something else? As was noted previously, the values in the diagram seem high compared to "normal" diode forward drops.

 

[Btaz]

Ah, if these are LEDs then the past should have a forward voltage and forward current specified. Assuming the numbers in the diagram are these, then the equation I posted and your calculations should be correct.

 

[WoodHokie4]

Yes, these are LEDs. The numbers came from the specs the manufacturer gave

 

[mattd2]

Might be a stupid question but... why are you using different spec'd LEDs?

 

[WoodHokie4]

6 different colors. Each color has different specs

 

[evanmars]

350, 345, 270, 345, 340, 270 is correct.
'Each leg' refers to each LED/resistor combo.
You know the current flowing through each leg. 20mA. Since you also know the voltage drop over each LED, i.e. 2.0V for the first one, you know the voltage across each resistor, i.e. 9V-2V = 7V
Ohm's law state R = V/I, so to find the required resistance - 7v/.02a = 350

 

[gorrie16]

Resistors on both legs of the diode? All the diagrams I've seen have resistors on the (+) leg only and sharing a common (-) leg. It looked like the resistors were put in place to normalize the voltage across each diode. Is there something I'm missing?

Yea, no need to have resistors on both sides of the diodes, unless you wanted to create a second voltage drop. The resistors are in the circuit to "make up" the voltage drop needed between the + and the various diode drops. Voltage has to be equal at parallel nodes.

 

[WoodHokie4]

So, I ended up using 400ohm resistors on each diode and everything lit up great...even with an old 9v battery. I'm still not 100% sure that the voltage was correct on listing I found and my multimeter died, so no way to check. But hey, I'm glad it worked!

 

[passedpawn]

350, 345, 270, 345, 340, 270 is correct.
'Each leg' refers to each LED/resistor combo.
You know the current flowing through each leg. 20mA. Since you also know the voltage drop over each LED, i.e. 2.0V for the first one, you know the voltage across each resistor, i.e. 9V-2V = 7V
Ohm's law state R = V/I, so to find the required resistance - 7v/.02a = 350

^^ This is the answer.

The voltage drop should come from the datasheet. They usually spec the forward voltage for 20mA, so you're good there. Note that at any other current, you'll have to refer the the I/V curve in the datasheet, as the forward voltage changes with current.

If you drive all the LEDs at the same current, you might find that some look brighter than others. 1) LEDs have an efficiency rating, which indicates how much light they give per amp. Even LEDs in the same color can have much different efficiencies, 2) many LEDs often have molded lenses that focus the light. Consider this. 3) The eyeball is more sensitive to certain colors (red/orange) than others (green). So you might need to push the current on some of them.