Go Down

Topic: Trying to understand current dispersal control across a parallel circuit  (Read 2883 times) previous topic - next topic

dwightthinker

Sorry, that's not completely true. 
This is true if Leds are close on the same Wafer.
 There are differences between distant areas on the same wafer.
 

Only one rule = never put LEDs in parallel.


In any case, never put LEDs in parallel if out of a hand full of LEDs.
Success is a false sense of future success.
The difference is often in a few 10s of mV. Adding a small resistor in
series with each is most likely enough.
Dwight

DrAzzy

Only one rule = never put LEDs in parallel.

Your rule is inconsistent with observed reality.

LEDs of the same type (mfg/model/lot) can be paralled just fine, and this is done widely in production by organizations that are way too sloppy to be matching LEDs from the same part of the wafer or whathaveyou. Go take apart any commercial LED array (flashlight, floodlight, streetlight if you can get your hands on one, aftermarket LED replacements for incandescents) - LEDs get put in parallel all the goddamned time, and when the LEDs are all of the same type, it works well enough that it's fine. It works particularly well with the 1W and 3W dies, where the dynamic resistance in the operating regime is relatively high.

It can, of course, only be done with very similar leds, and you should be aware that LEDs are not all the same, so that you can react appropriately if you discover that you have a batch of LEDs are significantly different from eachother and it impacts functioning.

I think this was much more of a problem in the past - LED process control has improved greatly.
ATTinyCore for x4/x5/x61/x7/x8/x41/1634/828/x313 megaTinyCore for the megaavr ATtinies - Board Manager:
http://drazzy.com/package_drazzy.com_index.json
ATtiny breakouts, mosfets, awesome prototyping board in my store http://tindie.com/stores/DrAzzy

Frostline

I still am having difficulty understanding why my answer for total resistance in circuit 1 in the OP is incorrect when supposedly my answer for total current was correct.

Total voltage / total current gives resistance ;)
Assuming the battery resistance is 0 ohms.

So lets have more circuits.



Information for the above circuits:
  • Supply voltage is 5V.
  • Forward voltage of red LED is 2V.
  • Forward voltage of blue LED is 3V.


Question 1: For circuit 1 would the Rtot always be equal to 220 ohm if the normal operating parameters of the resistor are followed?
Or to ask somewhat differently, will changing voltage only change the current of circuit 1 while the resistance remains constant?

For the current for circuit 1 it is simply an application of Ohm's Law.
I=V/R
I=5/220
I=22.7mA



Question 2: The forward voltage of the LED has to be considered when determining the current of circuit 2. Yes/No?

I=(Vin - Vf) / R
I=3/220
I=13.6mA

For circuit #3 determining the current should work the same as in circuit 2.

I=(Vin - Vf) / R
I=2/150
I=13.3mA

Question 3:What is the Rtot of circuit 3?


larryd

Quote
I=(Vin - Vf) / R
I=2/150
I=13.3mA

Question 3:What is the Rtot of circuit 3?
If the supply voltage is 5 volts and the calculated supply current draw is 13.3 ma:
R = V/A = 5/.0133 = 376 ohms = Rtot
150 ohms for the resistor and 376 - 150 = 226 for the LED
No technical PMs.
If you are asked a question, please respond with an answer.
If you are asked for more information, please supply it.
If you need clarification, ask for help.

Frostline

If a LED has that much internal resistance why would it need a current limiting resistor in series?

larryd

A LED is a non linear solid state device.
Connecting a voltage to the LED greater than its Vf will damage the structure and it will be destroyed.
Connecting a voltage less than Vf results in no current flow (just leakage current) through the LED.
Therefore you must limit the current through the LED when you have a supply voltage greater than Vf.
You limit the current with a series resistor.



No technical PMs.
If you are asked a question, please respond with an answer.
If you are asked for more information, please supply it.
If you need clarification, ask for help.

Frostline

I still don't quite see how a LED can have that much resistance. At 5V the LED alone would be limiting the circuit current to 22.1mA
A small internal resistance for the LED makes more sense to me.
Interesting conversation about it here.

larryd

@Frostline

The point is a LED is complicated to model.

I am giving you an equivalent resistance.

Wait till you get into inductive and capacitive reactance and resonance.

When dealing with non linear or reactive components, you can only calculate equivalent resistance.

Try this, put your device (LED in this case) into a closed black box that you cannot see into.
You are allowed the measure the applied voltage and the current draw.
How do you figure out the resistance of what's in your box?

Question:
1. How much resistance does a 1N4007 diode have when it is forward biased?
2. When it is reversed biased?
3. How much resistance does a fresh 9 volt carbon battery have?
4. What is the total circuit resistance in the schematic below. Itot = 106.67ma


Click this LINK to see what is in the black box.

.
No technical PMs.
If you are asked a question, please respond with an answer.
If you are asked for more information, please supply it.
If you need clarification, ask for help.

dwightthinker

The resistance of the LED changes with the voltage across it.
At 5V, across the LED, is might drop to 5 ohms. The current then
would surely burn it up.
This is what is meant by being non-linear.
A resistor keep almost the same resistance, regardless of the voltage
across it ( within reason ). It is called a linear device.
Diode in forward conduction are non-linear. As the voltage goes up the
resistance goes down.
To understand better, you might look up "load line".
If you had an exact plot of your LED, voltage against current,
you could plot a load line to hit the exact, say 20ma.
Do also remember, everything has a tolerance. When it is said
that the LED has X.X volts at 20ma, it will never be exact.
Again, you can use the load line plot to see how that effects
the amount of current in the LED.
Include the tolerance of the resistor that you are using for
more effects.
Drawing a load line drawing you can also see that the regulation
of current to the LED is better, the more voltage you are dropping
across the resistor.
If say you had a LED that ran at 3.3V and you had a supply at 3.6V,
the tolerance of a 10% resistor would make a much greater than
10% change in current.
Dwight

Frostline

I think what is making this so difficult for me to grasp is there is no mention of resistance really on LED datasheets.  
If it (the LED) had  resistance characteristics  why is there no graph of resistance vs voltages beyond Vf? Or at least a mention of of it somewhere?
Like this datasheet goes so far as to mention the capacitance of the LED.  But no ohm values to be found.
There is a power value 60mW, but no mention of what values were used to arrive at that other than ambient temperature(which does not seem rather helpful).

It seems like the consensus is to just force Ohm's Law to work.  Since the resistors can be measured for resistance and the current and voltage can be measured any discrepancy is just shrugged off as  equivalent resistance so the Ohm's Law formula is happy.

Since LarryD has been quite sporting in trying to answer my repetitive non-understanding of all this I will attempt to try and answer the four questions posed to me.
Quote
1. How much resistance does a 1N4007 diode have when it is forward biased?
2. When it is reversed biased?
3. How much resistance does a fresh 9 volt carbon battery have?
4. What is the total circuit resistance in the schematic below. Itot = 106.67ma
1. Using this datasheet the diode has a forward voltage of 1V @ 1A. 
R=V/I so 1 ohm
Though I figure that is incorrect. Since the same logic would not work in circuit 3 example from my last set of circuits. Using values from that example of Vf of 2V and a circuit current of 13.6mA I get 147 ohms for the LED in the circuit, not the 226 you said. 

2. DS says reverse bias voltage is 700V  RMS(not sure what that means) and only reverse current value I see is at peak of 5uA.  The 700V 70% of the peak reverse bias voltage listed at 1000V, no idea if that relationship also works for current but maybe it does so non-peak current of .0035A.  So 700/.0035 = 200k ohms.

3. No ideas how to even begin to calculate that.  Assume manufacturing discrepancies play a huge role.
Reading though this I am going to say 35Ω

4. I am going to say it cannot be determined.  What if there was an 8.2V battery under the box?


larryd

The questions were loaded.
1. How much resistance does a 1N4007 diode have when it is forward biased?
Like a LED, you really cannot say for sure what the resistance is.
If you were told what the current flow was then you might use ohms law to calculate an equivalent resistance.
But you weren't, the best answer is as you did, make some assumptions.
Full points for using the data sheet and basing your answer on quoted voltage to current characteristics.

2. When it is reversed biased?
This is a bit tricky.
A reversed biased diode has all the applied voltage dropped across it.
As you pointed out there is a reversed biased leakage value on the data sheet.
This is 5uA at the DC Blocking Voltage, for a 1N4007 this is 1000VDC.
You could say the reversed biased resistance was 1000V/5uA = 200,000,000 ohms = 200Meg
That is, an equivalent resistance of 200Meg at the rated PRV (peak reverse voltage).
In practice, at 5V, a reversed biased 1N4007 has a leakage current of zero amps.
(I know this from experience)
5V/0A = infinite (for all intents and purposes)

3. How much resistance does a fresh 9 volt carbon battery have?
The point here is, we sometimes miss the fact that an electronic device may have resistance.
Ex: a capacitor has an ESR (effective series resistance) .
Note: as a battery is discharged the internal resistance goes up leaving less voltage for the load.
Full points.

4. What is the total circuit resistance in the schematic below? Itot = 106.67ma
Back to the box.
With the applied voltage of 5V and a total current of 106.67mA your equivalent total resistance would be 5/160.67mA = 46.7 ohms.
At best, if all the 150 ohm resistors were in parallel this would give you 150/5 = 30 ohms leaving 16.7 inside the box.

Click the LINK below the drawing to see what was in the box.

Some conclusions:
- There is a difference between equivalent resistance and resistance.
- Equivalent resistance can vary as circuit conditions change where resistance will not.
- Data sheets are important!
- Measurements may be necessary to dissect a circuit which quite often is a black box to the technician.
- It appears, you have quick mind and you will do well in electronics!

Edit:
Let's say there was a LED in the box instead of the 1.8 volt battery and the LED forward voltage drop was 1.8V.
It would be correct to say the power dissipated by the LED would be W=1.8v X 106.67mA = 190mW.
If this was a 16.7 ohm resistor it would also dissipate the same 190mW.
The power calculation is valid maybe you can considered the equivalent resistance calculation as valid ;)



No technical PMs.
If you are asked a question, please respond with an answer.
If you are asked for more information, please supply it.
If you need clarification, ask for help.

dwightthinker

The reason they don't have voltage to resistance plots is because it
is more useful to have voltage to current plots.
As I said, with such a plot, one can draw a linear resistance
load line to see what resistance is best at some driving voltage.
Knowing the resistance at any particular current is just looking at the
plot and applying ohms law.
Knowing the resistance at any one current or voltage is of little use
at predicting what it will be at any other current or voltage.
Take the time to learn what a "load line" is and it will start to make sense.

A LED diode is not a silicon 1N4007 rectifier diode. Why would you expect its
voltage/current curve to match an LED diode. They are made from
completely different materials.
They do look some what similar but don't match.
Why would you think that 1V at 1A is not 1 ohms. It always is in
my book. A 1N4007 has little that is similar to a LED other than
they are both a type of diode. Silicon diodes do emit light though.
It is a 10.5 microns as I recall ( in the lower infrared band someplace ).
Dwight


Paul__B

I think what is making this so difficult for me to grasp is there is no mention of resistance really on LED datasheets.  
If it (the LED) had  resistance characteristics  why is there no graph of resistance vs voltages beyond Vf?
Because while we are discussing the finer conceptual points of ESR or "effective series resistance" and that only because you mentioned it, that is basically irrelevant to the practical operation of a LED.

There are two ways of operation of an LED.  Either you calculate, knowing the approximate voltage drop of the LED, the value of a series resistance corresponding to the specified current and your particular supply voltage, or else you design a circuit to regulate the current over a very wide range of applied voltages, so that the current will be as specified under all conditions.  For either of these, it is only required to know the specified (actually, maximum) current for the LED, and its usual voltage drop in operation.

Therefore those are the only (electrical) things specified in the datasheet.

Paulcet

Another reason you can not apply Ohm's law to LED (and other devices) directly, is because Ohm's law is linear and LED are non-linear

Paul__B

#29
Feb 01, 2016, 10:32 pm Last Edit: Feb 01, 2016, 10:40 pm by Paul__B Reason: Too good to pass up the pun!
It's not about Ohm's law at all, it is about understanding the nature of the component, so Ohm's law is irrelevant.

Ohm's law applies perfectly well to calculating the resistor when you follow the correct procedure.

Go Up