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Author Topic: Arduino digital input, 80Vdc  (Read 5894 times)
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UPDATE !

Sorry to bring this up again, but I need a little further help.

As I said earlier, I finally build this system and works perfectly, however I am a bit concerned on the temperature the diodes are getting up to.

There are 2 diodes (1N4750A with a 27V drop). First one to drop from 72V to 45V and then to 18V.
With a 15mA current.

1st diode)
   Power Dissipated:   
P=I∙E=0.015∙(72.73-45.73)=0.405W=405 mW

2nd diode)
   Power Dissipated:   
P=I∙E=0.015∙(45.73-18.73)=0.405W=405 mW

This diodes are 1 Watt, however they get very hot to the touch.

Any suggestion on how to upgrade the system to prevent them getting so hot?

Thank you very much for your help smiley
« Last Edit: October 19, 2013, 07:34:52 am by Sergegsx » Logged

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Lower the current and/or lower the voltage drop over the zeners, (change from 2 to 3 zeners)

Pelle
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Lower the current and/or lower the voltage drop over the zeners, (change from 2 to 3 zeners)

Pelle

Hello Pelle,

I can not lower the current as it is already powering the minimum things requiered (ie: pcb status led + 4N25).
I can add a third zener diode but I wanted to know if there was any other option as i am already with too many components (2zener+2diodes) per channel. If i add a third diode means another 6 diodes in the pcb + having to make the pcb bigger.

any other option?

thanks for the help
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A fan? Moving air has a profound effect on heat dissipation.

The problem with power ratings is that they assume certain conditions that your setup may not include. A certain amount of free space around the component, mounting height above the board, not having more hot components nearby, those sorts of things.
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What I can see from your layout you have the indication LED in parallell with the optocoupler.
Connect the indication LED in series with the optocoupler and the current are half of what is now.

One less component (2,2KΩ)

Pelle
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What I can see from your layout you have the indication LED in parallell with the optocoupler.
Connect the indication LED in series with the optocoupler and the current are half of what is now.

One less component (2,2KΩ)

Pelle

Thats a very good idea Pelle !! Just cant believe how easy it was.

I did the simulation in livewire and we went from 402mW to 192mW !! Plus as you say, I can get rid of 6 resistors !!  smiley-money

So whats happening here is that by using only 1 resistor, the current "lost" in the extra resistor to drop the voltage is not required any more. The leds need very little current.
Is this correct? I just want to fully understand whats happening now.

Once again thanks ! When I fully understand whats happening I will modify my PCB to test it. Thank you  smiley


* before.JPG (46.46 KB, 744x538 - viewed 26 times.)

* after.JPG (44.48 KB, 730x548 - viewed 28 times.)
« Last Edit: October 19, 2013, 11:04:19 am by Sergegsx » Logged

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When the input is 80V you are running the opto isolators at more than 10mA , which is far more than they need. I suggest you increase the 2k2 resistors to 4K7 or 10K. If you do rewire the PCB to put the LEDs in series with the opto isolators, then this will make the LEDs less bright, but unless you view them in direct sunlight they should still be bright enough, especially if they are green ones.
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With the latest change, arent they running at seven mA? According to the simulation at least.
Could i get a one or two line explanation on the change of putting the led in series ? Is what i said earlier correct?
There are only 0.7 volts to the second led, should i worry about this?
Thanks everyone
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Your revised schematics need amending. If you are going to connect the indicator LED in series with the opto isolator, then the 1N4148 diode should be in parallel with the series combination of the indicator and the opto isolator. Otherwise, if the 80V input is reversed, you will blow the LED.

Another issue with both the original and the revised circuit is that if you reverse the 80V input, then almost the full 80V appears across the 2.2k resistor, and it will dissipate about 3W. This is a consequence of adding the zener diodes and reducing the series resistor.

You haven't told us what the forward voltage of your indicator LEDs is (or even what colour they are), so I can't calculate the exact current flowing through the circuit.
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Your revised schematics need amending. If you are going to connect the indicator LED in series with the opto isolator, then the 1N4148 diode should be in parallel with the series combination of the indicator and the opto isolator. Otherwise, if the 80V input is reversed, you will blow the LED.
Absolutely right, I forgot about that, Thanks for looking into it and telling me.
Attached the corrected schematic. (I left the SW2 switch to do some tests in the simulator)

Another issue with both the original and the revised circuit is that if you reverse the 80V input, then almost the full 80V appears across the 2.2k resistor, and it will dissipate about 3W. This is a consequence of adding the zener diodes and reducing the series resistor.
Anything that can be done with this? I understand what you are saying but I cant see this can be fixed, right?

You haven't told us what the forward voltage of your indicator LEDs is (or even what colour they are), so I can't calculate the exact current flowing through the circuit.
Sorry about that, its a regular 5mm green led. I dont know the forward voltage as I got them from the store without further information.  According to https://www.sparkfun.com/products/9592
Could we say "1.8-2.2VDC forward drop
Max current: 20mA
Suggested using current: 16-18mA
Luminous Intensity: 150-200mcd"

I really appreciate your help dc42, now and also in the past you have helped me.

btw. I dont need very bright leds on the PCB, just the minimum to see if the channel is ON or OFF. So the current through the LED+optocoupler can be lowered, 5mA is ok?


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« Last Edit: October 19, 2013, 02:36:05 pm by Sergegsx » Logged

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Skip the 1N4148 and put an 1N400X in serie with the zeners, no revers voltage and not possible to burn 2k2.
What current do the optokopplare LED need?
Look in the daasheet after CTR current transfer ratio.

LED current  x CTR = phototransistor current.

LED current 10mA x CTR 50% = possible phototransistor current 5 mA

If you use 10kohm pullup = 0,5 mA from optocoupler you only need 1 mA LED current.

Look in the datasheet  what CTR your optocoupler have and calculate the current needed.
Calculate the resistor with 60 volts input.

Pelle, sorry about the spelning, my Pad talkning swedish
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The full design is here, as you can see there are already 2 x 1n400x at the begining as protection. however, previously I was told to put also this  other diode across the optocoupler as it was a faster diode.  Is this correct? should I leave it or remove it?

Let me go through your calculations, CRT is new to me. After I read about it I will post again.
Thanks again pelle


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« Last Edit: October 19, 2013, 03:48:52 pm by Sergegsx » Logged

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Current for the 4N25

4N25 CTR  50 %

I dont understand very well your formulas, but:
if indicator LED current is choosen to be 10mA (maybe lower as i dont need it to be bright)

LED current  x CTR = phototransistor current.
LED current (10mA) x CTR 50% = possible phototransistor current 5 mA

If you use 10kohm pullup = 0,5 mA from optocoupler you only need 1 mA LED current.
I dont understand this last phrase.
as they are in series, if 10mA go through the status led, arent there going to be the same 10mA through the optocoupler led?


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We take it backwards.

What pullup resistans do you use? I prefer 10kohms
To make aurdino input low you need 0.5 mA thru the phototransistor.

You have to calculate with CTRmin = 20%

LED current must be 5 times higher = 2.5 mA

If you change pullup you have to recalculate.

Pelle
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The full design is here, as you can see there are already 2 x 1n400x at the begining as protection. however, previously I was told to put also this  other diode across the optocoupler as it was a faster diode.  Is this correct? should I leave it or remove it?

Without the zener diodes in the circuit, the 1N4148 in reverse-parallel connection provides the best protection. With the zener diodes added, you should either use a much higher value series resistor (so that the power dissipation isn't a problem even with the polarity reversed), or add back a single 1N400x diode as protection against reverse polarity. I would keep the 1N4148 as additional protection, because the capacitance of a 1N400x will let transients through.

Let me go through your calculations, CRT is new to me. After I read about it I will post again.

When doing calculations on the current transfer ratio, be sure to use the minimum guaranteed current transfer ratio, and check whether the Vce at which the current transfer ratio is measured is applicable to your circuit.
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