How Does a Transistor Function as a Switch in Different Circuit Configurations?

[michael1978]

please can you help, explain me in easy way this circuit
http://obrazki.elektroda.pl/2237673700_1387888073.jpg

normal must be
swith in position A
q1 on
q2 off
q3 on

switch in position B
q1 off
q2 on
q3 off

 

[Jony130]

This circuit work as you expect.

Also I attach the simulation file for Circuit Wizard.

 

[michael1978]

Hey jony

Thank you very much,
but what you did?
Where i make mistake?

Your circuit is working perfect like in my book

can you teach me this where i make mistake thnk

BUT YOU CIRCUIT IS NOT LIKE MINE IN BOOK? in the book they did it with resistor two transistor the third transistor they did with lamp so they work like yours, but to me they work on on off not like in book, but in the book they did tell which value of reisistor i must put in circuit? so i don't know where i make mistake if you can teach me where i make mistake and how work thnk for time

 

[Jony130]

Simply you forget about Q2 base current and also that light bulb has low resistance.
In your diagram Q1 is CUT-OFF completely Q3 is also CUT-OFF completely because Q2 is in saturation.
So he short Q3 base to GND. Or if we use a different point of view.
Q1 is off because Q1 base don't see any close path from Vcc.
But Q2 is "ON" because there is a path for the Q2base current from Vcc through light bulb to Q2 base-emitter and back to GND. And this flow of a Q2 base current will "ON" the Q2 and Ic current will start to flow from
Vcc--->BL2 ---> collector-emitter---->GND
And Q3 is also cut off because Q2 steals all the current from Q3 base current (short Q3 base to gnd).

So as you can see despite the fact that Q1 is OFF. There will be a current flow through the BL1 bulb.
Of course this current is a Q2 base current but still the bulb will light because Q2 base current is determined by BL1 resistance and OHMS law.
Ib2 = (Vcc - Vbe2)/Rbulb.

 

[michael1978]

yes jony that is true what you say, but how can i make to work this circuit like in my book, will bulb, like you make your circuit? i read the book complete elecronics, because in the book he don't say the value of resistance, is possible this circuit to work like your which you make it? and how to do it? i relly don't understand in the book he don't make mistake, and also when i simulate my circuit, i put the mouse to emiter of Q1 is Ov 0a, in q2 there is 0v but there is current, and in Q3 is the same like in Q1, can you make this circuit to work like yours thank you very much johny for reply and explain

 

[Jony130]

Pleas read again my previous replay and try to answer this question.
What causes that the BL1 light bulb glow?

 

[michael1978]

Pleas read again my previous replay and try to answer this question.
What causes that the BL1 light bulb glow?

SO when is switch in position A there is base current ok like you say the transistor is on and BL1 is on glow, and transistor Q1 is like switch close so Q2 is off no base current, and is like switch open Q2, and Q3 is on because there is current on base, it takes from collector and ground current from Q2, like you say they work in cutoff and saturation, am i right?

i rely don't understand do you rembembe you show how to work like a amplifier, no i don't know how to work like switch,

i don't understand this book, he show how work like switch three transistor like i show you,
but what do you think jony, how where is mistake

thnx for reply and advice have a nice sleep

 

[Jony130]

But as you can see in the diagram You and I have a switch in position B,
And in my post I described how this circuit work when switch is in B position. And I ask you why BL1 light bulb glow when we have a switch in B position. And I don't see any mistake in circuit. And everything work as is should.

 

[michael1978]

but as you can see in the diagram you and i have a switch in position b,
and in my post i described how this circuit work when switch is in b position. And i ask you why bl1 light bulb glow when we have a switch in b position. And i don't see any mistake in circuit. And everything work as is should.

one more thing ? Why the bulb not glow in q3 and q1 yes? so when switch in position b the q1 collector take current from base of q2 of what how it work?, but why q3 he don't take current from q2? because i make four transistor and three transistor bulb are on and the last is off, what is this? how you calculate this, i don't understan is that possible that q1 take current from base of Q2

can you please explain this circuits how it work, how you calculate PLEASE JOHNY THANKS
Second if you want to make transitor to work like a switch there is only one way which you give me circuit to add resistor in serie in base?

 

[Jony130]

Ok I will try to explain you how this circuit work.

First we have a switch in A position.
And the situation look like this

We start our analysis at Q1 base. Q1 base see a path for his current from Vcc.
And this is why Q1 will turn ON. I repeat -> Q1 "ON" because there is a path for the Q1 base current from Vcc through R1 --->Q1 base-emitter and back to GND.
I mark Q1 base current in red.

Vcc --- > R1 ---> Q1 base-emitter junction --- > GND.

And this base current Turn ON the transistor. So Collector current can flow in this circuit.

Vcc--->BL1--->Q1 collector-emiter --->GND

Q2 is CUT-OFF because Q1 shorts Q2 base to ground via saturated (Full "ON") Q1.

Q3 base also see a path from Vcc via BL2. So Q3 base current start to flow in this circuit:

Vcc --->BL2--->Q3 base-emitter junction--->GND

This current will TURN ON Q3 so collector current will start to flow:


Vcc--->BL3 ---> Q3 collector-emitter--->GNDNow we flip the switch into B position. And the situation look like this:

Q1 is CUT-OFF because Q1 base is short to ground via 1K resistor (Q1 base don't see any close loop path to Vcc ).

But now Q2 will see a path for his current from VCC. And this base current will start to flow in this circuit :
Vcc --->BL1--->Q2 base-emitter junction--->GND

This base current will TURN-ON Q2 so that Q2 collector current will start to flow:

Vcc--->BL2--->Q2 collector - emitter--->GND

Q3 also will be in CUT-OFF because Q2 transistor collector-emitter saturation voltage (Vce(sat)) is lower then 0.6V needed to TURN ON Q3 (Q3 base is shorted to ground via saturated Q2 ).

 

[michael1978]

Ok I will try to explain you how this circuit work.

First we have a switch in A position.
And the situation look like this

We start our analysis at Q1 base. Q1 base see a path for his current from Vcc.
And this is why Q1 will turn ON. I repeat -> Q1 "ON" because there is a path for the Q1 base current from Vcc through R1 --->Q1 base-emitter and back to GND.
I mark Q1 base current in red.

Vcc --- > R1 ---> Q1 base-emitter junction --- > GND.

And this base current Turn ON the transistor. So Collector current can flow in this circuit.

Vcc--->BL1--->Q1 collector-emiter --->GND

Q2 is CUT-OFF because Q1 shorts Q2 base to ground via saturated (Full "ON") Q1.

Q3 base also see a path from Vcc via BL2. So Q3 base current start to flow in this circuit:

Vcc --->BL2--->Q3 base-emitter junction--->GND

This current will TURN ON Q3 so collector current will start to flow:


Vcc--->BL3 ---> Q3 collector-emitter--->GND


Now we flip the switch into B position. And the situation look like this:

Q1 is CUT-OFF because Q1 base is short to ground via 1K resistor (Q1 base don't see any close loop path to Vcc ).

But now Q2 will see a path for his current from VCC. And this base current will start to flow in this circuit :
Vcc --->BL1--->Q2 base-emitter junction--->GND

This base current will TURN-ON Q2 so that Q2 collector current will start to flow:

Vcc--->BL2--->Q2 collector - emitter--->GND

Q3 also will be in CUT-OFF because Q2 transistor collector-emitter saturation voltage (Vce(sat)) is lower then 0.6V needed to TURN ON Q3 (Q3 base is shorted to ground via saturated Q2 ).

johny thnx for time, i understand now,


the path of current and voltage...

but like i see this work like this on off on, off on off is that right of maybe i am wrong?

my circuit is working on on on , on on off, i mean the buble glow,

what do you mean by ;;;Q2 is CUT-OFF because Q1 shorts Q2 base to ground via saturated (Full "ON") Q1.

so q2 is off and buble don't glow

why the buble don't glouw on off on, off on off?
and how the software calculate the voltage and current
do you know to calculate , kan you tell me please? the voltage and current of all circuit?

second they work like you explain me ok,
so switch a...
current on buble 1 is ok glow in second is zero current but the buble glow?, and third is ok current buble glow, why the second buble glow? what is current and voltage

when i switch to b
first buble glow current is zero, second buble glow there is a current, and third buble is off there is zero current, why the buble one glow? what is voltage and current

i send you my circuit? they have to work like you say,
you told me very good how they work but the buble don't work

so johny what do you think ? what is the problem in circuit?


johny do you know which book is good to learn electronics and to not boring you

 

[Jony130]

but like i see this work like this on off on, off on off is that right of maybe i am wrong?

Yes, transistor are Q1-ON; Q2-OFF; Q3-ON or if we have a switch in B position we have
Q1-OFF; Q2-ON; Q3-OFF

what do you mean by ;;;Q2 is CUT-OFF because Q1 shorts Q2 base to ground via saturated (Full "ON") Q1.

Do you know what "short" is in electronic ?
Read this
https://www.physicsforums.com/attachment.php?attachmentid=65135&d=1388163392
We can also say that: Q2 is CUT_OFF because Q1 transistor collector-emitter saturation voltage (Vce(sat)) is lower then 0.6V needed to TURN ON Q2. And this is why Q2 is CUT-OFF.

so q2 is off and buble don't glow

No, Q2 is OFF but current will still flow through BL2 so the bulb will glow.
Haven't you noticed that when Q2 is OFF, the Q3 base current will now flow through the bulb?

why the buble don't glouw on off on, off on off?

Because you forget that base current also flow through the light bulb.
And if this base current is large (low resistance) the light bulb will glow.

and how the software calculate the voltage and current
do you know to calculate , kan you tell me please? the voltage and current of all circuit?

In circuits when transistor work as a switch (transistor is in saturation region of in cut-off) we can easily calculate all current in the circuit. All we need to know is the resistance and Vcc voltage.
The base current is equal to:
Ib = (Vcc - Vbe)/Rb where:

Vcc --> power supply voltage

Vbe --> Base-emitter diode forward voltage drop (0.6V ... 0.8V)

RB --> base resistor resistance in our circuit Rb is 1K for Q1 base and for the rest of a transistors RB = light bulb resistance.

And collector current is equal to

Ic = (Vcc - Vce(sat))/Rc

Vce(sat) - Collector-emitter saturation voltage (0.2V typical).

Rc --> resistance connect between Vcc and collector in our case light bulb resistance.

so switch a...
current on buble 1 is ok glow in second is zero current but the buble glow?, and third is ok current buble glow, why the second buble glow? what is current and voltage

Are you sure that current is zero for second bulb (BL2)? You still forgetting about base current.
Once more take a look at this

As you can see we have a switch in A position. So Q1 is ON ; Q2 is OFF and Q3 is ON.
And second bulb glows, because Q3 transistor base current is now flowing through the light bulb. The Q3 base current is mark in green on the diagram.

when i switch to b
first buble glow current is zero, second buble glow there is a current, and third buble is off there is zero current, why the buble one glow? what is voltage and current

Because now despite the fact that Q1 is OFF the BL1 current is not equal to 0A.
As you can see here in red

Q2 transistor base current is now flow through first light bulb.
Additional the light bulb resistance is ""low" so the base current is high:
Ib = (10V - 0.7V)/100Ω ≈ 93mA and this is why light bulb glow (shines).

they have to work like you say,
you told me very good how they work but the buble don't work
so johny what do you think ? what is the problem in circuit?

But everything work as is should be. And to fix this issue with the bulbs simple reduce the base current.

johny do you know which book is good to learn electronics and to not boring you

Can you tell me the name of the book you are reading now?

try this books
https://www.amazon.com/dp/0071360573/?tag=pfamazon01-20
https://www.amazon.com/dp/0596153740/?tag=pfamazon01-20
https://www.amazon.com/dp/0132549867/?tag=pfamazon01-20

 

[michael1978]

Yes, transistor are Q1-ON; Q2-OFF; Q3-ON or if we have a switch in B position we have
Q1-OFF; Q2-ON; Q3-OFF

Do you know what "short" is in electronic ?
Read this
https://www.physicsforums.com/attachment.php?attachmentid=65135&d=1388163392
We can also say that: Q2 is CUT_OFF because Q1 transistor collector-emitter saturation voltage (Vce(sat)) is lower then 0.6V needed to TURN ON Q2. And this is why Q2 is CUT-OFF. No, Q2 is OFF but current will still flow through BL2 so the bulb will glow.
Haven't you noticed that when Q2 is OFF, the Q3 base current will now flow through the bulb?


Because you forget that base current also flow through the light bulb.
And if this base current is large (low resistance) the light bulb will glow.


In circuits when transistor work as a switch (transistor is in saturation region of in cut-off) we can easily calculate all current in the circuit. All we need to know is the resistance and Vcc voltage.
The base current is equal to:
Ib = (Vcc - Vbe)/Rb where:

Vcc --> power supply voltage

Vbe --> Base-emitter diode forward voltage drop (0.6V ... 0.8V)

RB --> base resistor resistance in our circuit Rb is 1K for Q1 base and for the rest of a transistors RB = light bulb resistance.

And collector current is equal to

Ic = (Vcc - Vce(sat))/Rc

Vce(sat) - Collector-emitter saturation voltage (0.2V typical).

Rc --> resistance connect between Vcc and collector in our case light bulb resistance.


Are you sure that current is zero for second bulb (BL2)? You still forgetting about base current.
Once more take a look at this

As you can see we have a switch in A position. So Q1 is ON ; Q2 is OFF and Q3 is ON.
And second bulb glows, because Q3 transistor base current is now flowing through the light bulb. The Q3 base current is mark in green on the diagram. Because now despite the fact that Q1 is OFF the BL1 current is not equal to 0A.
As you can see here in red

Q2 transistor base current is now flow through first light bulb.
Additional the light bulb resistance is ""low" so the base current is high:
Ib = (10V - 0.7V)/100Ω ≈ 93mA and this is why light bulb glow (shines). But everything work as is should be. And to fix this issue with the bulbs simple reduce the base current. Can you tell me the name of the book you are reading now?

try this books
https://www.amazon.com/dp/0071360573/?tag=pfamazon01-20
https://www.amazon.com/dp/0596153740/?tag=pfamazon01-20
https://www.amazon.com/dp/0132549867/?tag=pfamazon01-20

HI JOHNY, thnx for answer

i am reading complete electronics self teaching guide with projects by earl boysen, harry kybett,
but now i want to start to read pracitcal electronics for inventors, is this book good?

johny now i understand i did not know that transistor can work also like this ?
when switch in postion b normal transistor is off but the bubl glow and hi can find current through the base of second buble transistor to emitter gnd, , is like this? , and also second transistor is on because he find current thrug the buble one transistor one is okso now i have problem with calculating

where they come this value
i send you a file, and look

johny goodnight

 

[Jony130]

but now i want to start to read pracitcal electronics for inventors, is this book good?

I don't know I never read this book.

johny now i understand i did not know that transistor can work also like this ?

But this has nothing to do with transistors.

when switch in postion b normal transistor is off but the bubl glow and hi can find current through the base of second buble transistor to emitter gnd, , is like this?

Which transistor on your diagram is "normal"? Which bulb is glowing?
Why you don't use component name designators? It is so hard to write Q1 transistor is OFF or BL1 bulb is glowing?

, and also second transistor is on because he find current thrug the buble one transistor one is ok

I really don't understand what you are trying to say.

so now i have problem with calculating

where they come this value
i send you a file, and look

Hmm, as you may know transistor are nonlinear device and this is why equation describe transistor are also nonlinear. Normally, it is impossible to find analytical solution for this type of a circuits. We can use numerical method or use iteration method. But we often use hand calculation. And in this case we are force to assume some transistors parameters needed for calculation. And this is why we almost always assume Vbe value between 0.6V...0.7V. We have the same situation with other BJT parameters for example Hfe or Vce_saturation voltage.
Of course in data sheet we can find approximate values ​​of the parameters which we are interested.

So for your circuit.
Also do you know what type of a bulb you use in simulation? Or what type of a BJT you are using?

 

[michael1978]

i don't know i never read this book.


But this has nothing to do with transistors.



Which transistor on your diagram is "normal"? Which bulb is glowing?
Why you don't use component name designators? It is so hard to write q1 transistor is off or bl1 bulb is glowing?


I really don't understand what you are trying to say.

Hmm, as you may know transistor are nonlinear device and this is why equation describe transistor are also nonlinear. Normally, it is impossible to find analytical solution for this type of a circuits. We can use numerical method or use iteration method. But we often use hand calculation. And in this case we are force to assume some transistors parameters needed for calculation. And this is why we almost always assume vbe value between 0.6v...0.7v. We have the same situation with other bjt parameters for example hfe or vce_saturation voltage.
Of course in data sheet we can find approximate values ​​of the parameters which we are interested.

So for your circuit.
Also do you know what type of a bulb you use in simulation? Or what type of a bjt you are using?

jony how are you? Tonight is good , happy new year

But i think ideal transistor ? Buble 6v 60ma i think 100ohm

HOW TO DO CALCULATING PLEASE? HOW COME THIS VALUE SIMULATE SELF AND YOU WILL SEE I WILL SEND YOU FILE PLEASE


One more time happy new year? Eennjjooyy

 

[Jony130]

Ok. If we assume R_bulb = 100Ω . Don't forget that light bulb resistance is also voltage dependent (nonlinear). Light bulb resistance change when you apply different voltage to it.

So for switch in B position Q2 base current will be around

Ib2 = (Vcc - Vbe2)/R_bulb ≈ (10V - 0.7V)/100Ω ≈ 93mA

And Ic2 = (Vcc - Vce(sat)) ≈ (10V - 0.1V)/100Ω ≈ 99mA

That's all we can do, we are forced to assume Vbe and Vce(sat) values, because we don't know nothing about this ideal BJT use in simulator.

 

[michael1978]

Ok. If we assume R_bulb = 100Ω . Don't forget that light bulb resistance is also voltage dependent (nonlinear). Light bulb resistance change when you apply different voltage to it.

So for switch in B position Q2 base current will be around

Ib2 = (Vcc - Vbe2)/R_bulb ≈ (10V - 0.7V)/100Ω ≈ 93mA

And Ic2 = (Vcc - Vce(sat)) ≈ (10V - 0.1V)/100Ω ≈ 99mA

That's all we can do, we are forced to assume Vbe and Vce(sat) values, because we don't know nothing about this ideal BJT use in simulator.

THNX JOHNY FOR ANSWER...Ic2 = (Vcc - Vce(sat)) ≈ (10V - 0.1V)/100Ω ≈ 99mA, VCE(sat) 0.1? HOW COME THAT IS 0.1 VOLT? it most be 0v of not?
AND I SIMLATE THIS FILE SO VBE=114mv q1, vbe 1.01vQ2, q3=146Mv

see the picture down

 

[Jony130]

VCE(sat) 0.1? HOW COME THAT IS 0.1 VOLT? it most be 0v of not?

Who told you that?
Vce(sat) is always greater than zero.
See this 2N2222 datasheet
http://www.fairchildsemi.com/ds/PN/PN2222A.pdf

AND I SIMLATE THIS FILE SO VBE=114mv q1, vbe 1.01vQ2, q3=146Mv

?? We use uppercase (capital) letters for the quantity symbols for voltage, current and power.
Also capital letter M is use in electronic as a Mega = 1 000 000 = 1M

 

[michael1978]

Who told you that?
Vce(sat) is always greater than zero.
See this 2N2222 datasheet
http://www.fairchildsemi.com/ds/PN/PN2222A.pdf


?? We use uppercase (capital) letters for the quantity symbols for voltage, current and power.
Also capital letter M is use in electronic as a Mega = 1 000 000 = 1M

HI
I RELLY DIDT KNOW, YOU MEAN THIS 2 parameters


I know johny, about letters, but i writte like that, but i will write good now,
you know Johny i am an amateur i like to learn electronics, the last circuit which i send you, , why Vbe have that value? how do you calculate

Johny thnx for answer goodnight
but

 

[Jony130]

HI
why Vbe have that value? how do you calculate

I did not done any Vbe calculation, it is impassible to calculate Vbe and Vce(sat) voltage.
Why ? Because we don't know nothing about transistor you use in simulation. In real life we have very similar situation, and this is why we almost always use approximate value, around 0.7V.
Assuming 0.7V (or 0.65V or whatever between 0.5V...0.8V) across a forward biased silicon junction is just a rough approximation, or rule of thumb. It may be good enough for making rough estimates of circuit conditions.
Try red this :

http://electronics.stackexchange.co...ant-0-7-for-a-transistor-in-the-active-region

 

[michael1978]

I did not done any Vbe calculation, it is impassible to calculate Vbe and Vce(sat) voltage.
Why ? Because we don't know nothing about transistor you use in simulation. In real life we have very similar situation, and this is why we almost always use approximate value, around 0.7V.
Assuming 0.7V (or 0.65V or whatever between 0.5V...0.8V) across a forward biased silicon junction is just a rough approximation, or rule of thumb. It may be good enough for making rough estimates of circuit conditions.
Try red this :

http://electronics.stackexchange.co...ant-0-7-for-a-transistor-in-the-active-region

hey Johny thnx for reply, i read but is difficult to understand, but i understand you, did you hear about this book, Electronic Principles by Albert malvino i like to buy this book but i don't know is good of not. i hear from one friend is good

 

[Jony130]

I measure Vbe(base-emitter voltage) vs Ib (base current) for BC337-40
The power supply voltage was Vcc = 10V

RB = 680kΩ...Vbe = 0.614V...Ib = 13.8µA

RB = 470kΩ...Vbe = 0.616V...Ib = 20µA

RB = 220kΩ...Vbe = 0.624V...Ib = 42.61µA

RB = 100kΩ...Vbe = 0.639V...Ib = 93.61µA

RB = 50kΩ...Vbe = 0.659V...Ib = 187µA

RB = 10kΩ...Vbe = 0.719V...Ib = 928µA

RB = 5kΩ...Vbe = 0.748V...Ib = 1.85mA

RB = 2kΩ...Vbe = 0.787V...Ib = 4.6mA

RB = 1kΩ...Vbe = 0.819V...Ib = 9.18mA

RB = 500Ω...Vbe = 0.856V...Ib = 18.29mA

RB = 200Ω...Vbe = 0.989V...Ib = 45mA

And as you can see the base current change 45mA/13.8μA = 3260 times. But Vbe change only by 375mV. This very small change in Vbe compared with the huge changes in the base current can be ignored in some cases. And this is why we use Vbe= 0.6...0.7V in hand calculations.
Also if you want to fix the issue with the bulbs, simply add resistor in series with the transistors bases. This resistors will reduce the current that is flowing through the bulbs. So the bulbs will give no light.

did you hear about this book, Electronic Principles by Albert malvino i like to buy this book but i don't know is good of not. i hear from one friend is good

Yes I also heard that this is a good book, but I never read it. Why you don't buy book in your your native language?

 

[michael1978]

I measure Vbe(base-emitter voltage) vs Ib (base current) for BC337-40
The power supply voltage was Vcc = 10V

RB = 680kΩ...Vbe = 0.614V...Ib = 13.8µA

RB = 470kΩ...Vbe = 0.616V...Ib = 20µA

RB = 220kΩ...Vbe = 0.624V...Ib = 42.61µA

RB = 100kΩ...Vbe = 0.639V...Ib = 93.61µA

RB = 50kΩ...Vbe = 0.659V...Ib = 187µA

RB = 10kΩ...Vbe = 0.719V...Ib = 928µA

RB = 5kΩ...Vbe = 0.748V...Ib = 1.85mA

RB = 2kΩ...Vbe = 0.787V...Ib = 4.6mA

RB = 1kΩ...Vbe = 0.819V...Ib = 9.18mA

RB = 500Ω...Vbe = 0.856V...Ib = 18.29mA

RB = 200Ω...Vbe = 0.989V...Ib = 45mA

And as you can see the base current change 45mA/13.8μA = 3260 times. But Vbe change only by 375mV. This very small change in Vbe compared with the huge changes in the base current can be ignored in some cases. And this is why we use Vbe= 0.6...0.7V in hand calculations.
Also if you want to fix the issue with the bulbs, simply add resistor in series with the transistors bases. This resistors will reduce the current that is flowing through the bulbs. So the bulbs will give no light.


Yes I also heard that this is a good book, but I never read it. Why you don't buy book in your your native language?

Johny thnx for reply i understan now, but the problem in my language we don't have so much book in electronics, also in Belgium i try to find book in biblothek and in store but they explain not so good clear i mean...for example i start one course via online in holland basic electronics, but this nothing they teach you a little bit, not to do some experiments, they don't go deep in electronics, i see the last time not so long Basic Electronics, i read i little bit but pffff nothing
have a nice sleep thnx