Why is there an extra diode in this sketch?

[Femme_physics]

Hey guys, I'm trying to wrap my head around this electronic schematics of a PLC, because I've started ladder-programming it. Can anyone explain to me why are there TWO diodes before the transistor? I get that one diode shows an "active" or "inactive" state when it's turned on or not. But why the other diode before the transistor?

Attached is an image of the sketch as well as the entire PDF with the PLC specifications

 

[Jony130]

Well, answer is very simple. Your PLC use galvanic isolation/separation at the input in fact almost all PLC's use galvanic separation.
And this is why they use photocoupler at the input. And because the input is source/sink type they use bidirectional input optocoupler
http://www.ixysic.com/home/pdfs.nsf/www/LDA110.pdf/$file/LDA110.pdf

 

[jim hardy]

Perhaps there's a subtlety...
Jony's optocoupler is specified "bidirectional",
both of those diodes are light emitting,
one of them will emit light for positive input current, the other for negative.
So the PLC will detect input of either polarity.
The manual shows only positive input, a +24 volt supply is wired there, but it looks like the circuit designer provided for the input to sink current of either polarity. Good thinking, i'd say.

old jim

 

[Femme_physics]

Thank you for the clarification!

One thing is still unclear. Why is the diode near Y0 is pointing up and not down? I mean, since C0 = Ground, Y0 should theoretically never be turned on because the current will always prefer to flow where there is no resistance.

 

[Jony130]

This is a protection diode, so-called a back EMF protection diode.

 

[Baluncore]

LEDs are particularly sensitive to reverse breakdown voltages. For that reason a protection diode will be used where a long line with voltage transients may drive an LED.

But in this case the two parallel zigzag arrows, representing photons from the LEDs to the phototransistor, show that it is a bi-directional coupler. One diode conducts on positive inputs, the other on negative. In this case each diode protects the other from reverse breakdown.

 

[Windadct]

To continue Jony130s comment - the Diode in the output is sometimes called the Free Wheeling Diode - if the load (shown as a resistor) has any inductance, like a small relay coil, when the output transistor turns off - the inductor has current flowing that can not be instantaneously stopped (it has energy stored in it's magnetic field has to go somewhere). This diode allows the current to continue flowing (free wheel) until all of the energy is dissipated, this typically happens very quickly but without the diode the voltage generated at the at the transistor can cause it to fail.

Example - if the relay coil has 20mA of current, and 0.5mH Inductance, -- and using the switching time of 250nS for a 2N222 Transistor we use:

V = L * dI/dt = 0.25H * ( 20mA / 250nS) = 40V ..this does not sound like much and I just grabbed the numbers. does not include field wiring etc. and this is added to the DC voltage already in the circuit - so pretty easy to get 50 to 100 V across the transistor otherwise... it it typical good practice to place a diode across the coil itself as well - but this diode is to protect the Transistor - because not all electricians/technicians/engineers are perfect.

.

 

[Mike_In_Plano]

After reading through the data sheet, I'm disagreeing in part or whole with the suggestions.
While some opto-isolators have two LEDs such that one or the other will activate regardless of the applied polarity, there is no reference to this behavior in the datasheet.
With the open collector outputs, the diodes are not freewheeling as they don't connect to the positive power supply.
So, I believe all extra the diodes are a measure of protection against accidental miswiring. This fits well with the general intent that PLCs be tough, general purpose interfaces.

 

[jim hardy]

While some opto-isolators have two LEDs such that one or the other will activate regardless of the applied polarity, there is no reference to this behavior in the [PLC] datasheet.

hmmmm you know what ? you're right. That "bidirectional" was from Jony's isolator datasheet, not the PLC datasheet . my mistake.

Commendable attention to detail, Mike...

old jim

 

[Jony130]

After reading through the data sheet, I'm disagreeing in part or whole with the suggestions.
While some opto-isolators have two LEDs such that one or the other will activate regardless of the applied polarity, there is no reference to this behavior in the datasheet.

So how can you explain this ?

 

[jim hardy]

Well what do you know. Good eye, Jony. That adds considerable weight to the notion they'll take current in either direction as valid input.

Someplace else they defined the switch points as 14 and 16 volts, nominal 15 with two volts hysteresis.

 

[Windadct]

On the output - I stand corrected - in this case this diode is not going to see Free Wheeling - seems to be primarily for reverse voltage protection on the Transistor ( Mis-wiring, or mis-operaion in the field wring, possibly V-spike in field wiring etc.) As shown the field circuit would not free wheel back though this diode.

 

[Windadct]

On the output - I stand corrected - in this case this diode is not going to see Free Wheeling - seems to be primarily for reverse voltage protection on the Transistor ( Mis-wiring, or mis-operaion in the field wring, possibly V-spike in field wiring etc.) As shown the field circuit would not free wheel back though this diode.

 

[jim hardy]

Interesting deed - in the square just above where Jony found the source/sink note,
is a cryptic note about output:

max inductive load: 7.2W 24V

... Watts of inductance ?? Watt's up with that ?
That is just the 0.3 amps max ouput transistor current

next line says "output protection internal none"
a roundabout way of specifying that one must add his own protection for inductive loads

Sigh , one must be SOOOO very literal when reading computer specifications. They're always short on verbs , to my thinking..