NPN & PNP

[rsdoran]

thanks rsdoran

I need to know, did this really clarify it for you?

 

[magdyfayad]

thanks rsdoran

in the fact no , it is not suficient for me , but i did not want to ask more questions ??

i attached a file and contain one question , if posible answer it and explain in details the differencd between comon emitter ( transistor is good as a switch ) and common collector ( transistor not good as a switch ) and why transistor not good as a switch at common collector ??

thanks

 

[rsdoran]

I will look at it and respond tomorrow BUT your question said common emitter and common collector, that (use of common) implies they are connected to the common ( 0V in most cases).

 

[magdyfayad]

you can look to the file attachement and describle please in details what are the differences betwenn Q1 , Q2 and between Q3 ,Q4 as i know Q3 , Q4 is not good as a switch , why ??

so i want to know what the voltage applied between the emitter and PLC input for Q3 ,Q4 in both cases Q3 , Q4 are on and are off

thanks

 

[keithkyll]

A transistor is normally used as an amplifier. To use it as a switch, you need to turn it on at maximum - turn it on fully.
The voltage between Base and Emitter determines the Collector voltage.
With NPN, the Base must be higher than the Emitter.

For Q2, a Base voltage of 7 to 10 volts will be enough to turn it on fully. In this circuit, you have Voltage Gain.
For Q3, the Base voltage needs to be closer to 24 volts to turn it on fully. 24.7 volts I think. In this circuit, you have Current Gain.

To understand, you need to know amplifier theory, and what it takes to put a transistor into full saturation. The switch transistor in your sensor is most likely a Darlington. This is 2 transistors in one package. A Darlington has very high gain. Study 'Darlington Pairs' and 'Transistor Gain' for more info. With an NPN Darlington, there is so much gain that only a few volts are needed to turn it on fully. A Darlington is used with 5 volt logic.

Ron is better at explaining this. Wait for his answer. For now, use the terms above as Google study points. Study how a Darlington makes a good switch.

 

[rsdoran]

OK, let's get some things understood, the terms common collector or common emitter IMPLIES they are connected to the common.

Q1 (PNP) has the emitter connected to V+ and the emitter to a LOAD, proper wiring, therefore when the transistor is triggered there will be current through the LOAD

Q2 (NPN) has the emitter connected to the common (0V) so when it is triggered there will be current flow BUT note that on the NPN collector you will read ALMOST zero with a meter while with the PNP you will read almost source voltage.

Q3 and Q4 are wired backwards, this can get tricky depending on device and situation.

As I said before, in many cases they will work opposite of how they normally do, now they will pass current when NOT triggered.

Q3 and Q4 should be passing current to the load when not triggered

THE arrow is the EMITTER, the other side is the COLLECTOR.

I can not believe Keith said wait for my answer, he is far better at this then I am.

 

[parky]

If you use an npn output to the plc you will need a pull up resistor (look at previous reply showing the two types, effectively you would connect the plc input at the junction of the collector & the resistor but the logic will be inverted, theextra resistor will be needed as npn sensors are open collector in general & the plc will not have a pull up resistor

 

[panic mode]

some of the statments are not correct. keithkill provided most accurate reply.
here is the picture:

each of the transistors (discussing bipolar types!) will ONLY pass current when energized. to make transistor work as a switch, base current has to be high enough to cause saturation, one has to forward bias base-emitter PN junction (like a diode). this means that for NPN transistor, potential of the base must be higher than emitter, for PNP it must be lower than potential of emitter. Vce or voltage drop across saturated transistor is low (50-100mV). For Si transistor, Vbe is about 0.7V (for Ge is less, somewhere about 0.3V or so). Assuming Si products (most common these days) we can stick with 0.7V for Vbe.
In a circuit where power rails are at 0VDC and 24VDC, to turn on Q3 as a switch, base must be at potential Vb=24VDC-(Vce-Vbe)= 24.65V and for Q4 this would be Vb=-0.65V (since Vce and Vbe are negative).

Either way driving Q3 or Q4 to saturation requires voltage that is OUTSIDE supply voltage range (24.65V>24V and -0.65V<0V) which is not very practical (voltage drop on transistor would be big and worse - load dependant).

Note, when working with transistors it is quite common to see that some specimens conduct after emitter and collector have been swapped around. But it is not transistor that conducts, it is the optional diode that was added for protection. Signal transistors don't have this becasue any leakage would potentially affect signal quality but many power transistors (switching type) have this (not just bipolar, this is common for power MOSFETs as well)

 

[panic mode]

for for those who would like to play with circuits but don't want to spend money on actual parts, breadboard, scope etc (or don't want to get burned by soldering iron), try SwitcherCAD, it's free and powerfull:

http://www.linear.com/designtools/software/index.jsp

 

[magdyfayad]

Thanks all​

But PLC input can operate from 10 : 30 vdc as we did not need 24 vdc exactly , so if a voltage drop on the transistor is high but the PLC input did not less than 10 vdc , i think there is no any problem , yes or not ??

​

So we did not need to additional supply voltage 24.65 vdc more than the original supply voltage 24 vdc , as the orginal supply voltage may be work well as PLC input work from 10 : 30 vdc , is it correct or not ??​

Thanks again​

 

[keithkyll]

Yes and correct. I will assume 5 volt logic. For Q3, the Collector voltage range will be 0 - 4.3 volts. For Q4, subtract this voltage from the supply. That means with a 24 volt supply, the Collector voltage range will be 23.3 - 19.3 volts. (Correct me if I'm wrong guys).
This is why we say it won't work. It's not a switch in that configuration (Q3 and Q4) - it's a current amplifier. Q3 might do something. Q4 will be on all the time.
If you are planning to add a second transistor on the output of your sensor to invert the logic, then it might work because you will have 24 volts for the base voltage. You still need a resistor on the base. The voltage difference between Base and Emitter is ALWAYS .7 volts (for silicon transistors). This important rule alone should make it clear why Q3 and Q4 won't do what you want.

Also, your drawing are Theoretical for a switch. When someone gives an example for a switch, they show one transistor. In actual practice, 2 transistors (in one package) are used. I'm not sure what would happen with Q3 and Q4 if they were Darlingtons.

OT,
(Ron, sorry if I put you on the spot Buddy. You're always good at providing great detail.) I suspect this is being translated into Arabic, so I am trying to be careful with the words I use in case he is using an automatic translator. Those really can make technical English hard to understand.
Maybe if we say it enough ways, it will sink in.

 

[rsdoran]

The problem is that it is not as straighforward as looking at the OP's diagram with Q3 and Q4 wired, no a transitor would not work in that case BUT we are not talking about a transistor we are talking about transistorized devices.

In some cases these devices can work in place of one another i.e. NPN could be used in place of a PNP but the action is reversed, sometimes without a pull up/down resitor. The line techs at the xmas paper plant use to do it to me all the time because they would run out of the right sensor.

The problem with sensors is there are a multitude of varieties and can be normally open, normally closed, and/or configurable.

For the most part understanding that the emitter on PNP should be connected to V+, think doubl P as in positive, and the emitter on NPN should be connected to V-, think double N for negative you will not have problems.

Anytime you have to use the wrong sensor i.e. an NPN where a PNP was designed to be you have a problem to begin with.

Here are basics for using a pull up/down resistor. http://www.seattlerobotics.org/encoder/mar97/basics.html

Also may help to take a look at the different sensors available, I offered this before: http://mysick.com/partnerPortal/eCat.aspx?go=FinderSearch&Cat=Gus&At=Fa&Cult=English&FamilyID=242&List=1&Category=Produktfinder&Selections=4969%2c5489

There are other sites that offer more information:
http://www.balluff.com/NR/rdonlyres/E5FE86F8-9E1F-4022-92B6-B0284B1E61AA/0/Plc_inpu.pdf

http://www.balluff.com/Balluff/Website/Templates/Searches/TechnicalReferenceSearch.aspx?NRMODE=Published&NRNODEGUID=%7bD38D9840-513A-4B7F-A516-2063E3844402%7d&NRORIGINALURL=%2fBalluff%2fus%2fProductsChannel%2fTech%2bRef%2f%3fmenuLevel%3d2&NRCACHEHINT=Guest&menuLevel=2

http://www.turck-usa.com/Support/Technical_Information.htm

 

[magdyfayad]

Thanks ALL

This site : http://www.turck-usa.com/Support/Technical_Information.htm

the above site is useful

 

[RoTaTech]

When using 3 wire sensors the Black is always the output.

Not always. If one has worked with sensors long enough, one may remember the old (North American standard?) Red = +Vcc, Black = Common, White = signal

 

[rsdoran]

Not always. If one has worked with sensors long enough, one may remember the old (North American standard?) Red = +Vcc, Black = Common, White = signal

My answer was in reference to a question the OP originally asked but either edited or I imagined it, do not see it now so not sure BUT originally it was asked what happens with a 3 wire when you swap the wires around i.e. brown and blue or black and brown, something along those lines.