Does this Circuit make sense?

[Raul]

Each series LED chain has two LEDs, which drop 3.2V each for 6.4V total. That leaves 5.6V across each current limiting resistor (assuming your 12V power supply is actually 12V not 13.8V). To get 20 mA current flow, R = V / I which is 5.6 / 0.02 which is 280 so I would use 270 ohm resistors (or two 560 ohm resistors in parallel) for each current limiting resistor. Power dissipation in each resistor will be 5.6 * 0.02 which is 0.112 watts so they won't get very hot.

You suggest a 270 ohm resistor

If I take my 12VDC power supply and subtract the 6.4 voltage drop I get 5.6
If I divide 5.6 by 20mA I get 0.28
Why would I not use a 280 ohm resistor? Why a 270?

Here is a 1/4watt 280 ohm flameproof resistor for $1.50 each.
That's pretty spendy when resistors are usually under 25 cents, but it's flameproof


About Voltage Drop:
Is Voltage Drop the same as Forward Voltage?
When I look at the specs for the LEDS I selected I don't see a specification for Voltage Drop
http://www.superbrightleds.com/more...60-degree-viewing-angle-4500-millilumens/341/
There is however a figure for Wattage Consumption 0.064 Watts
I have no clue how to use that bit of data.

I plugged the data I have into a resistor calculator and got this result:

Here is a chart where forward voltage is associated with LED color
http://www.oksolar.com/led/led_color_chart.htm

This Wiki Page
http://en.wikipedia.org/wiki/LED_circuit
says in pertinant part:

where power supply voltage (Vs) is the voltage of the power supply, e.g. a 9 volt battery, LED voltage drop (Vf) is the forward voltage drop across the LED, and LED current (I) is the desired current of the LED.

Which informs me that the terms are used interchangeably. IZZAT right?


On another note:
Do resistors often catch fire?
This a doll house it's made of wood, should I be looking at flameproof resistors or maybe I should just isolate the electronics away from the LEDS in a metal project box.

 

[(*steve*)]

270 is suggested because it is a commonly available value.

$1.50 is *way too much* to pay for a resistor when the specific properties are not required. (I'd be expecting 5c or 10c for a resistor, less if you buy some reasonable number of them).

Resistors don't catch fire unless you severely overload them. In this case, if you got a 1/2 W resistor it would not significantly overheat even if you shorted out most of your LEDs.

No need for isolation. At these currents and voltages all you need to do is to wire it up correctly and to make sure that the wires are insulated (get some heatshrink tubing for any exposed joins)

 

[Raul]

270 is suggested because it is a commonly available value.

That makes sense.

$1.50 is *way too much* to pay for a resistor when the specific properties are not required. [...]Resistors don't catch fire unless you severely overload them. In this case, if you got a 1/2 W resistor it would not significantly overheat even if you shorted out most of your LEDs.

thx

 

[KrisBlueNZ]

You suggest a 270 ohm resistor. If I take my 12VDC power supply and subtract the 6.4 voltage drop I get 5.6 If I divide 5.6 by 20mA I get 0.28
Why would I not use a 280 ohm resistor? Why a 270?

As Steve as already said, 270 ohms is a preferred value. I suggested two 560R in parallel because 560 ohms is a preferred value and the resulting resistance is 280 ohms. It doesn't need to be exactly 280 ohms though. The forward voltages are just approximate anyway.

About Voltage Drop: Is Voltage Drop the same as Forward Voltage? When I look at the specs for the LEDS I selected I don't see a specification for Voltage Drop. There is however a figure for Wattage Consumption: 0.064 Watts. I have no clue how to use that bit of data.

Yes, LEDs are operated with forward current, so the voltage across the LED (its 'voltage drop') is equal to the forward voltage. This is specified as 3.2V at the recommended operating current of 20 mA. The power consumption figure is just the product of the current and the voltage, from the formula P = I V (power = current x voltage). 0.02 amps multiplied by 3.2 volts is 0.064 watts. This figure is not really important.

I plugged the data I have into a resistor calculator and got this result:

What result?

Do resistors often catch fire? This a doll house it's made of wood, should I be looking at flameproof resistors or maybe I should just isolate the electronics away from the LEDS in a metal project box.

No, unless they are forced to dissipate a LOT of power and/or are near to something flammable. The worst reasonable case would be if the whole LED chain gets shorted out, and the full 12V supply voltage appears across the resistor. A 270 ohm resistor will draw 44 mA (from Ohm's law, I = V / R), which gives a power dissipation of 533 mW (from P = I V). For a small component like a resistor, this will cause a temperature rise in the region of 100 degrees Celsius, i.e. a temperature of around 120 degrees. You could use a 1 watt resistor, which is physically larger and won't heat up so much, and/or you could connect the resistor to thick tracks, e.g. several stripboard tracks soldered together, and use short leads on the resistor, so heat will be dissipated over a larger surface area. You can get fusible resistors but I don't think they would fuse at that temperature.

 

[Raul]

The forward voltages are just approximate anyway.

Oh - - wow. I really and truly thought the numbers were reported with such singular precision because they were precise.
As Rummy said: "There is the known and the unknown" I have a fair bit of the latter.

Yes, LEDs are operated with forward current, so the voltage across the LED (its 'voltage drop') is equal to the forward voltage.

Equal to as opposed to being the same thing?
So while the value may be close enough to say it's the same what is going on is different.
Correct me if I'm wrong, I'm extrapolating here:
Forward voltage is industry vernacular for "the voltage needed to power a circuit" ??
Voltage Drop is vernacular the "the power consumed by a circuit" ??
OK Great.
Now of course that just begs the question of why they are the same. Is it because of the inefficiencies in such a simple circuit and the LED and resistor are not so inefficient as to lose power in large enough amounts to be worth calculating for just a couple of iterations?

If one were powering a large number of LEDs, ERGO a lot of resistors in the circuit should one also think about voltage drop due to the resistor? It generates heat yes? That's power loss yes?

This is specified as 3.2V at the recommended operating current of 20 mA. The power consumption figure is just the product of the current and the voltage, from the formula P = I V (power = current x voltage). 0.02 amps multiplied by 3.2 volts is 0.064 watts. This figure is not really important.

Is there a configuration where it might be important?

What result?

That was my error. I had attached an image to a post – the post I believe you responded to. Then, I deleted the image and then deleted the whole post and re-typed it and re-posted.
The image was from the resistor calculator into which I plugged wrong data, but didn’t catch my error in time. Your reading of the post was probably just after I deleted the image, had but not deleted that line to which you were inquiring. Confusing huh?

No, unless they are forced to dissipate a LOT of power and/or are near to something flammable. The worst reasonable case would be if the whole LED chain gets shorted out, and the full 12V supply voltage appears across the resistor. A 270 ohm resistor will draw 44 mA (from Ohm's law, I = V / R), which gives a power dissipation of 533 mW (from P = I V). For a small component like a resistor, this will cause a temperature rise in the region of 100 degrees Celsius, i.e. a temperature of around 120 degrees. You could use a 1 watt resistor, which is physically larger and won't heat up so much, and/or you could connect the resistor to thick tracks, e.g. several stripboard tracks soldered together, and use short leads on the resistor, so heat will be dissipated over a larger surface area. You can get fusible resistors but I don't think they would fuse at that temperature.

Oh wow~!! That cursed “known and unknown” raises its head again.
I had thought the wattage of the resistor was a critical component of controlling power. So the 1/4 watt of the resistor I had selected is just a threshold below which one ought not to fall because of heat dissipation issues?
Is there a point at which the wattage of the resistor becomes to great - - I mean besides the unavailability of large wattage small resistors?

My original plan was to solder the resistor to the wire very near to the LED. Then heat shrink the LED and leads from the LED and solder connections. This would give me a rather flexible assembly. Then push the wiring and leads and resistor through holes cut in the panels of the doll house into the cavity and glue the light fixture (it’s also just wood) using a temperature sensitive glue; such as hide glue, which comes apart easily at 140 F in case repairs were later needed.

I could also pot the resistor to the fixture using a fast cure epoxy.

 

[BobK]

The "forward voltage" across an LED is called that because it refers to the voltage at a specific current in the forward direction, i.e. with the anode at a higher potential than the cathode. A diode would also have a "reverse voltage" which would be different at a given current, and would apply when the cathode was at a higher potential than the anode. A resistor has no forward or reverse, it acts the same in either direction, thus the voltage across it is simply called voltage drop. So "forward voltage" and "voltage drop" are analogous but there is this subtle difference that "forward voltage" only applies if the current is going in one direction.

Bob

 

[Raul]

The "forward voltage" across an LED is called that because it refers to the voltage at a specific current in the forward direction, i.e. with the anode at a higher potential than the cathode.

Higher potential being more positive?
I read the wiki page on potential. Well, I looked at the page. Reading connotes apprehension and I did very little of that.

A diode would also have a "reverse voltage" which would be different at a given current, and would apply when the cathode was at a higher potential than the anode.

Reverse Voltage in this case being when the input voltage polarity is flipped - yes?
I’m guessing that’d be different from the REVERSE as it is used in this image and description found here:
http://en.wikipedia.org/wiki/Diode#Current.E2.80.93voltage_characteristic
In that paragraph the author is discussing the energy needed to turn on a diode – yes? I couldn’t quite get what he meant by “reverse” and “breakdown” in the image there. If I had to guess: I'd speculate that he was describing conditions during the build up to the diode turning on.

thus the voltage across it is simply called voltage drop. So "forward voltage" and "voltage drop" are analogous but there is this subtle difference that "forward voltage" only applies if the current is going in one direction.

Thx

Is the term Forward Voltage used for things other than Diodes and Resistors? When I google it I get mostly discussion about diodes.

Somewhere I read that Voltage Drop in a Diode is the delta in voltage between anode and cathode of the diode as the diode passes current.
If so, I should be able to test an unknown LED by powering it up till it lights up and placing a Multimeter across it set to VDC. Except, how would one know which resistor to use? Guesswork? Some convention?

 

[(*steve*)]

This may help:

https://www.electronicspoint.com/got-question-driving-leds-another-work-progress-t228474.html

 

[KrisBlueNZ]

Oh - - wow. I really and truly thought the numbers were reported with such singular precision because they were precise.
As Rummy said: "There is the known and the unknown" I have a fair bit of the latter.

Well, the forward voltage specification at a particular current is fairly accurate but will vary somewhat with temperature, from batch to batch, and from component to component within a batch. It's not a tightly controlled parameter; LEDs are current-driven, and the voltage across the LED at a particular current is not critical.

[forward voltage vs. voltage drop are equal]
Equal to as opposed to being the same thing?

For an LED that's emitting light, they ARE the same thing. I worded that wrong and misled you, sorry.
The forward voltage for a semiconductor junction, when it's being supplied with a particular amount of current, is the voltage that will be across the junction. This is the same as the voltage "dropped" by the LED.

[...]
If one were powering a large number of LEDs, ERGO a lot of resistors in the circuit should one also think about voltage drop due to the resistor? It generates heat yes? That's power loss yes?

If you're powering a lot of LEDs in series, you only need one resistor. The resistor "absorbs" the remaining voltage, that is, the power supply voltage minus the sum of all the forward voltages, and this determines the current, through Ohm's law: I = V / R. A resistance of 'R' ohms with 'V' volts across it will pass a current of 'I' amps.

Power is lost in the LEDs AND in the resistor. Power is calculated as P = I V (power = current x voltage). The current in all parts of a series circuit is the same, so power dissipated in each component is proportional to the voltage across it.

Current is determined by the resistor (that's why it's called a current limiting resistor). Power is dissipated in each LED according to its characteristics, and the resistor dissipates the remaining power in the circuit because it drops the remaining voltage. I think of a resistor as a tension spring. It takes up the slack distance (voltage) and sets the current (tension).

[...]I had thought the wattage of the resistor was a critical component of controlling power.

No. The wattage is a rating, for the maximum power that the resistor is designed to dissipate. The actual power it dissipates (is forced to dissipate) is P = I V and is determined by the circuit. If a resistor is forced to dissipate too much power it will get too hot and may fail.

So the 1/4 watt of the resistor I had selected is just a threshold below which one ought not to fall because of heat dissipation issues?

Pretty much.

Is there a point at which the wattage of the resistor becomes to great - - I mean besides the unavailability of large wattage small resistors?

Well, you can get some pretty big resistors! Large ones are usually wirewound - they are made from resistive wire. Size is needed because of the amount of heat generated, which has to be dissipated somehow.

My original plan was to solder the resistor to the wire very near to the LED. Then heat shrink the LED and leads from the LED and solder connections. This would give me a rather flexible assembly. Then push the wiring and leads and resistor through holes cut in the panels of the doll house into the cavity and glue the light fixture (it’s also just wood) using a temperature sensitive glue; such as hide glue, which comes apart easily at 140 F in case repairs were later needed.
I could also pot the resistor to the fixture using a fast cure epoxy.

I think you're obsessing a bit here. If you insulate your wires and connections properly, there's no reason why the resistor would dissipate much more power than you calculate. If you were using high power LEDs running at hundreds of milliamps or amps, then you need to be concerned about dissipation in the current limiting resistor or circuit.

If you want to reduce the resistor temperature, you need to increase the surface area that is thermally connected to it. This is what a heatsink does. I wouldn't encapsulate the resistor. If necessary, use a larger one.

 

[Raul]

I think you're obsessing a bit here.

Yah I may be, but when I'm largely uninformed about a thing, I prefer to err on the side of caution. I found some 1/2 watt 270 ohm resistors that should do nicely.

If you're powering a lot of LEDs in series, you only need one resistor. The resistor "absorbs" the remaining voltage

and limits current.

The resistor is acting like a narrow orifice through which only so much power can pass in any given time frame. Which is why I can use a PS with ampere rating higher than the sum of in the circuit diodes can use. - yes?

I grasp this in a very superficial way. I say this because intuitively, I expect the first diode in the series to be over burdened by power and the one at the end to be starved.
So for me, it's like that algebra problem that I never understood at all, with terms that all escaped me entirely, and the mechanics of which remained a mystery, but by sheer wrote memory I could parrot the solution with all the comprehension of a parrot. Mere sounds strung together with no meaning.

This is also where, for me, the explanation of Amperes being like the water in a hydraulic system fails entirely. In a linear system ( in series ) the water is drained away by each demand point leaving less pressure and less flow, and less water for the next in line. The pressure and flow at the beginning of the system is higher than it is at the end. To achieve any semblance of even pressure one must design and install a plenum of some sort to create a large distribution area (in parallel) that evens the flow and pressure out among all the individual demands.

Current is not like this - or so I take it - but I don't see how.

Voltage is like this because just like the demands on a hydraulic line, each demand point ( diode) consumes some power leaving less for the next one down stream?

I struggle with the mechanics of why series wiring is so very different from parallel wiring.

I can parrot things like "In a series circuit the current is the same for all elements." and "In a parallel circuit the voltage is the same for all elements." from the Wiki page on Series and Parallel Circuits
http://en.wikipedia.org/wiki/Series_and_parallel_circuits
But articulating it from wrote is not comprehension.
I'd like to have a mental visual of what is happening at the mechanical level: why do the electrons behave so differently?

Goldfish - - - red plastic castle - -

 

[BobK]

This is also where, for me, the explanation of Amperes being like the water in a hydraulic system fails entirely. In a linear system ( in series ) the water is drained away by each demand point leaving less pressure and less flow, and less water for the next in line. The pressure and flow at the beginning of the system is higher than it is at the end. To achieve any semblance of even pressure one must design and install a plenum of some sort to create a large distribution area (in parallel) that evens the flow and pressure out among all the individual demands.

Current is not like this - or so I take it - but I don't see how.

That is becuase you are using the hydraulic metaphor incorrectly. There are no leaks in the system (a better plumber than I must have put it together!). The flow of water in a series of connections is always the same, the pressure can change as it goes through different devices, but not the flow.

In a parallel connection, the water comes to a Y, and can take two different paths. The flow of water is the two paths can differ, but the sum of the two flows is the same as the flow coming in.

Bob

 

[KrisBlueNZ]

The resistor is acting like a narrow orifice through which only so much power can pass in any given time frame. Which is why I can use a PS with ampere rating higher than the sum of in the circuit diodes can use. - yes?

No. The LEDs are in series, so the SAME current flows through each of them. Not just the same AMOUNT of current; actually the same electrons! So you don't add the currents together; there is only one current flow. You do, however, add the VOLTAGES (forward voltages) together, since the LEDs are in series. The difference between the sum of the LED forward voltages and the power supply voltage is the amount of voltage that the resistor will drop; the resistor's behaviour is to give a fixed ratio between voltage and current, so the voltage across the resistor determines the current in the whole string.

I grasp this in a very superficial way. I say this because intuitively, I expect the first diode in the series to be over burdened by power and the one at the end to be starved.

No, not at all. The same CURRENT is flowing in each of the LEDs.

Here is an analogy I've come up with that may help you understand this better. This analogy is not compatible with standard analogies; it is a different paradigm altogether.

In this analogy:
Voltage = distance
Current = tension (mechanical)
A resistor is represented as a tension spring.

Each LED in the chain has a roughly constant forward voltage, so it can be represented (roughly) by a piece with a fixed length. These pieces are joined end-to-end, because the LEDs are connected in series, as a string (like Christmas lights). At one end, a tension spring (the resistor) is connected. Then the whole series string is stretched to a distance that represents the power supply voltage.

As the string is stretched, the resistor stretches (increases its length - increases its voltage across it) to "take up the slack". The more you stretch the combined string, the more the resistor "stretches". This causes the tension (current) to increase. This tension (current) is equal at all points in the string, and this tension (current) is determined by how much the resistor (spring) is stretched, i.e. the voltage across it.

The forward voltage of each LED is not exactly constant; it increases somewhat with increasing current. So as you stretch the string more, the distance (voltage) across each LED will increase somewhat, because of the increasing tension (current) in the string. But the resistor is much more springy, and "takes up the slack". As long as a significant amount of the total voltage is lost in the resistor (i.e. the spring is stretched to a significant extent), small variations in the LED forward voltage (distance across each LED) do not make much difference to the current (tension); it is mostly determined by the characteristics of the resistor (spring).

Again I want to emphasise that this analogy is not compatible with conventional analogies of pressure and flow in water pipes.

This is also where, for me, the explanation of Amperes being like the water in a hydraulic system fails entirely. In a linear system ( in series ) the water is drained away by each demand point leaving less pressure and less flow, and less water for the next in line. The pressure and flow at the beginning of the system is higher than it is at the end. To achieve any semblance of even pressure one must design and install a plenum of some sort to create a large distribution area (in parallel) that evens the flow and pressure out among all the individual demands.

The traditional analogy is accurate, but I find it harder to understand than my analogy. The fact that you don't understand it properly yet prompts me to suggest my analogy instead, since it works quite well for a lot of commonly seen circuit configurations (topologies).

Current is not like this - or so I take it - but I don't see how.

I hope my analogy helps. I'm calling it the "DTS" analogy - Distance, Tension and Stretchiness, which correspond to voltage, current and resistance, respectively.

But articulating it from wrote is not comprehension.

I know. I love Wikipedia, but I do find that their articles on electronics tend to be somewhat abstract and to rely heavily on mathematics without providing a way to understand the system at a "gut level". Hence my DTS analogy.

Goldfish - - - red plastic castle - -

Yeah...
What?! :--)

 

[Raul]

That is becuase you are using the hydraulic metaphor incorrectly. There are no leaks in the system (a better plumber than I must have put it together!). The flow of water in a series of connections is always the same, the pressure can change as it goes through different devices, but not the flow.

Ahh. That’s different, I was thinking that the water got tapped off like a bucket gets filled at station #1 etc. Proves that one can never quite communicate perfectly. “Now hormone squirts from cell to cell, THAT’s communication.” (Frank Herbert “The Dosadi Experiment”)

in a series, the SAME current flows through each of them. Not just the same AMOUNT of current; actually the same electrons!

Oh my. Now that puts a different complexion on things.

As the string is stretched, the resistor stretches (increases its length - increases its voltage across it) to "take up the slack". The more you stretch the combined string, the more the resistor "stretches". This causes the tension (current) to increase. This tension (current) is equal at all points in the string, and this tension (current) is determined by how much the resistor (spring) is stretched, i.e. the voltage across it.

I like that. You should stick with it.

Wikipedia, […] articles on electronics tend to be somewhat abstract and to rely heavily on mathematics

I think that the authors are just showing off. They don’t get paid so their only compensation is the snootiness factor: “Hey look at how blindingly brilliant I am.” That sort of thing.
Imagine the poor sodding bastard who puts on his resume that he wrote wiki articles?


I got my power supply. It’s 12.23 Volts.
So I went looking for Resistors. I live in the woods so it’s all online for me.
I found bags of ten and twenty resistors being sold for under a buck. I purchased fuseable ones at MCM electronics http://www.mcmelectronics.com and they ship USPS so it’s cheap.
I got a bag of ten 1 watt 15 ohm flameproof fusing resistors for $0.32, not each, that’s $0.32 for the whole bag, and a bag of twenty 270 ohm ½ watt fuseable resistors for $0.97 ( for another LED project) plus some other crap.