PIC input using PHOTOTRANSISTOR - Electronic Circuits Projects Diagrams Free

aljamri · 3rd October 2007, 05:09 PM

Hi,
I've got a some questions about PIC input using PHOTOTRANSISTOR OPTOCOUPLER of type CNY-17 Data sheet file address:

http://www.tranzistoare.ro/datasheets/270/101963_DS.pdf'

In the attached circuit, how much is the required input voltage ( Pin 1 & 2 ) to the transistor to conduct the required voltage to make RTCC reads logic 1

aljamri · 3rd October 2007, 05:17 PM

Sorry this is the circuit

ericgibbs · 3rd October 2007, 06:05 PM

hi,

This is for +5V supply to Rs.

blueroomelectronics · 3rd October 2007, 06:06 PM

It's in the datasheet, typical 1.35V
Don't forget to add a current limiting resistor or you can damage the LED.

aljamri · 3rd October 2007, 08:56 PM

Thanks for replys,

if my Supply voltage is 2V ac as i measured it ( I am not sure i have to check again if it is ac or dc ), so:

- is that enough ?,
- is (ac dc) makes any differents

Note:
I'll be able to measure it again and check for ac ro dc.

thanks

aljamri · 3rd October 2007, 09:08 PM

Quote:

Originally Posted by blueroomelectronics

It's in the datasheet, typical 1.35V
Don't forget to add a current limiting resistor or you can damage the LED.

But it is written as Emitter Input Forwarding Voltage, and my voltage will be applied across the photodiod.

ericgibbs · 4th October 2007, 08:07 AM

Quote:

Originally Posted by aljamri

Thanks for replys,

if my Supply voltage is 2V ac as i measured it ( I am not sure i have to check again if it is ac or dc ), so:

- is that enough ?,
- is (ac dc) makes any differents

Note:
I'll be able to measure it again and check for ac ro dc.

thanks

Hi,
>>> But it is written as Emitter Input Forwarding Voltage, and my voltage will be applied across the photodiod.

The opto emitter is a LED device, which when driven by the correct voltage/current it emits 'light'.
Its the photo detector diode [or transistor] in the opto which detects and amplifies the 'light', then it goes to the PIC input.

A opto emitter will work from a range of 'dc' voltages, BUT it must be connected to the voltage via a resistor, else it will be damaged.

Use the simple formula to calculate the value of the series resistor.
If the emitter requires say 30mA and it has a forward voltage drop of 1.2V at this current and its connected to a +2Vdc supply, then...........

Series Resistor = [2-1.2]/.03 = 27R

IF its 2Vac then check the datasheet to see if the emitter diode can accept a 2Vac reverse voltage.

Is this clear.?

eblc1388 · 4th October 2007, 08:38 AM

For AC operation, it is good practice to connect a reverse polarity diode across the emitter LED, to take care of any possibility of voltage spikes coming from the AC mains.

aljamri · 5th October 2007, 01:36 AM

Quote:

Originally Posted by ericgibbs

Use the simple formula to calculate the value of the series resistor.
If the emitter requires say 30mA and it has a forward voltage drop of 1.2V at this current and its connected to a +2Vdc supply, then...........
Series Resistor = [2-1.2]/.03 = 27R

i measured across the diode ( pin 1 & 2 ) and got 0.8 volt.

Quote:

Originally Posted by ericgibbs

IF its 2Vac then check the datasheet to see if the emitter diode can accept a 2Vac reverse voltage.

Is this clear.?

yes its clear and by the way, the input voltage is ac

ericgibbs · 5th October 2007, 08:05 AM

Quote:

Originally Posted by aljamri

i measured across the diode ( pin 1 & 2 ) and got 0.8 volt.
yes its clear and by the way, the input voltage is ac

hi,
If its 2Vac and you have a 27R resistor is series with diode emitter and the emitter spec sheet says
the nominal forward voltage drop is 1.2V, you should have 1.2V across the emitter.

You must take into consideration that the emitter is only passing current for the positive half cycle of the 'ac' input
and blocking on negative half cycle.

How did you measure the 0.8V, was the meter on 'dc' or 'ac',
its best to use an oscilloscope this will give a 'true' indication of whats across the emitter diode.

I hope you didn't connect the emitter without a series resistor, if you did, the emitter may be damaged.

Do you get any pulses out of the detector output?

aljamri · 5th October 2007, 12:28 PM

Quote:

Originally Posted by ericgibbs

How did you measure the 0.8V, was the meter on 'dc' or 'ac', .

meter was on ac (as i remember)

Quote:

Originally Posted by ericgibbs

its best to use an oscilloscope this will give a 'true' indication of whats across the emitter diode.

i'll try to use our old oscilloscope, i hope it is still working

Quote:

Originally Posted by ericgibbs

I hope you didn't connect the emitter without a series resistor, if you did, the emitter may be damaged.

no

Quote:

Originally Posted by ericgibbs

Do you get any pulses out of the detector output?

my project in my office/lab and i'll not be able to go there until next Sunday, i'll write all your comments and try to take notes. i'll let you know about any progress.

Thanks

ericgibbs · 5th October 2007, 12:35 PM

hi,

>> meter was on ac (as i remember)

Across the emitter diode the waveform will not be sinusoidal therefore you will not get a true 'ac' voltage reading..

It will swing below ground by 2Vrms and above ground by 1.2Vdc.

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3rd October 2007, 06:06 PM	(permalink)
blueroomelectronics	It's in the datasheet, typical 1.35V Don't forget to add a current limiting resistor or you can damage the LED. __________________ Bill Smart Kits build Smart People http://www.blueroomelectronics.com

3rd October 2007, 08:56 PM	(permalink)
aljamri	Thanks for replys, if my Supply voltage is 2V ac as i measured it ( I am not sure i have to check again if it is ac or dc ), so: - is that enough ?, - is (ac dc) makes any differents Note: I'll be able to measure it again and check for ac ro dc. thanks __________________ Its what your friend in Your mind, what you in your friends mind

4th October 2007, 08:38 AM	(permalink)
eblc1388	For AC operation, it is good practice to connect a reverse polarity diode across the emitter LED, to take care of any possibility of voltage spikes coming from the AC mains. __________________ L.Chung

5th October 2007, 12:35 PM	(permalink)
ericgibbs	hi, >> meter was on ac (as i remember) Across the emitter diode the waveform will not be sinusoidal therefore you will not get a true 'ac' voltage reading.. It will swing below ground by 2Vrms and above ground by 1.2Vdc. __________________ Eric "Good enough is Perfect" PIC tutorials: Gramo's: www.digital-diy.net/ Bill's: www.blueroomelectronics.com/