Please help transistor amplifier

[michael1978]

You talking about post 114? The resistance is 22K//4.7K ( if I ignore the input resistance of the transistor). In fact the input resistance is quite low to me.

Usually you want the input resistance to be a little on the high side so the stage driving this input don't have to drive a low impedance. Think if you have a stage like in post 114, if the collector is driving the following stage with low impedance like 1K. The collector resistance is 1K//1K=500 ohm. You lower the gain of the driving stage.

In both diagrams, the input resistance of the transistor is β(r'e + Re1).=β(20Ω)≈2K assuming β=100. This is quite low. Another way to look at it is the transistor in both case require 5mA/β= 50uA. The impedance of the input biasing network has to be low enough so it is not loaded down by the input of the transistor. Jony got into this a little in post 79 about the base current.

CAN I ASK YOU SOMETHING about impedance, my english is not so good, how you select input impedance, not to say in the end the input impedance was 50k of 10k etc, you select in the begin of how

 

[yungman]

For normal frequency ( not RF), you usually want it to be as high as practical. It is limited by the input requirement of the amplifier...Like in your case, the transistor Ib of 50uA and transistor input impedance. Jony in post 79 explained this and I explained about the input impedance in the post 150.

In your case, with drawing in both post 114 and 146, the emitter resistor Re1 is eithe 14 or 12Ω, that is very low. That is really the gating factor of the circuit. Also you are trying to get gain of 50 out of one transistor, this force you to have such a low Re1 in order to be able to use a collector resistor of 1K.

In real life design, if I want to have a gain of 50, I would do either one below:

1) Using Re1=150, Rc=10K, Ic=0.5mA, then use an emitter follower transistor to buffer the output.
2) Use two stage of this and divide the gain between the two stages. This is the prefer way.
3) Use JFET instead of BJT.

 

[michael1978]

For normal frequency ( not RF), you usually want it to be as high as practical. It is limited by the input requirement of the amplifier...Like in your case, the transistor Ib of 50uA and transistor input impedance. Jony in post 79 explained this and I explained about the input impedance in the post 150.

In your case, with drawing in both post 114 and 146, the emitter resistor Re1 is eithe 14 or 12Ω, that is very low. That is really the gating factor of the circuit. Also you are trying to get gain of 50 out of one transistor, this force you to have such a low Re1 in order to be able to use a collector resistor of 1K.

In real life design, if I want to have a gain of 50, I would do either one below:

1) Using Re1=150, Rc=10K, Ic=0.5mA, then use an emitter follower transistor to buffer the output.
2) Use two stage of this and divide the gain between the two stages. This is the prefer way.
3) Use JFET instead of BJT.

thank you for answer, and R1, R2? but first i am bussy with transistor, is difficult jfet to learn, and for the book, you can buy it chip used boooks in amazone, because i want to open one visa of mastercard

 

[yungman]

thank you for answer, and R1, R2? but first i am bussy with transistor, is difficult jfet to learn, and for the book, you can buy it chip used boooks in amazone, because i want to open one visa of mastercard

When you use a credit card to open an account with Amazon, make sure don't let your computer remember your password, and Amazon ask you whether you want to keep logging in even you close the page...DON'T. Always logout when you close the page and don't let the computer remember your pass word.

 

[michael1978]

When you use a credit card to open an account with Amazon, make sure don't let your computer remember your password, and Amazon ask you whether you want to keep logging in even you close the page...DON'T. Always logout when you close the page and don't let the computer remember your pass word.

yeh i know but there are a lot of hackers today, i have to open only with 200E if it is possible, because i think they will not open with low credit

 

[michael1978]

JONEY thank you for reply i am learning a lot from you, you are the beste;-) COMPLIMENT

HI Jony how are you, may i ask you something, the schematic of designin of amplifier , you make it with light spice of some other software , because you told me i use light spice,
greetings

 

[yungman]

yeh i know but there are a lot of hackers today, i have to open only with 200E if it is possible, because i think they will not open with low credit

Well, there is a balance of being safe and missing out. Write to Amazon and ask what is the best way. It is their utmost interest to keep it safe as their business depends on this too.

 

[michael1978]

Well, there is a balance of being safe and missing out. Write to Amazon and ask what is the best way. It is their utmost interest to keep it safe as their business depends on this too.

thnx for reply, yes i know, but i will go first to my bank to ask

 

[yungman]

The bank cannot help much, call the credit card company like VISA also.

Ask Amazon, they should be able to help you. I have very good experience with Amazon, I have been ordering from them for years, never have a single problem. Their service is second to none...absolutely none. One time I order a tiller...over 100lbs gas tiller for turning the ground, There was a small crack on some non critical area, I asked for exchange, they shipped me a second one immediately and arranged to pickup the first one. The kicker was I decided to keep the original, they said "no problem", they just arranged to pickup the second one!

I buy computers, guitar pickups, protein powders, gifts...I just paid $1600 to buy a new Nikon camera and Tamron lens from them just a month ago. I even paid a little more just to buy on Amazon because of their service. Oh Yeh, thousands of dollars of textbooks used from Amazon also. When they say is good condition, they ARE in good condition. Half the time, they look new.

One thing about studying, you need books, lots of books. I buy 7 to 8 textbooks on each subject to compare as no book is good in all topics, each has their strong and weak points. You need a few to get the whole picture. Malvino is one exceptional one. Can you imagine if I buy my two tall book shelves of new books?! That would have been very expensive. I just bought the Malvino 6th edition for $8 shipped!

 

[michael1978]

The bank cannot help much, call the credit card company like VISA also.

Ask Amazon, they should be able to help you. I have very good experience with Amazon, I have been ordering from them for years, never have a single problem. Their service is second to none...absolutely none. One time I order a tiller...over 100lbs gas tiller for turning the ground, There was a small crack on some non critical area, I asked for exchange, they shipped me a second one immediately and arranged to pickup the first one. The kicker was I decided to keep the original, they said "no problem", they just arranged to pickup the second one!

I buy computers, guitar pickups, protein powders, gifts...I just paid $1600 to buy a new Nikon camera and Tamron lens from them just a month ago. I even paid a little more just to buy on Amazon because of their service. Oh Yeh, thousands of dollars of textbooks used from Amazon also. When they say is good condition, they ARE in good condition. Half the time, they look new.

One thing about studying, you need books, lots of books. I buy 7 to 8 textbooks on each subject to compare as no book is good in all topics, each has their strong and weak points. You need a few to get the whole picture. Malvino is one exceptional one. Can you imagine if I buy my two tall book shelves of new books?! That would have been very expensive. I just bought the Malvino 6th edition for $8 shipped!

i will try, so chip you buy malvino, and do you have 7 edition, if you have 7 edition, which is better, 6 of 7 edition,

 

[yungman]

I have not received it yet. I must had the first edition in the late 70s! I long lost the book. The book is in my head and been using what I learned to do a lot of designs all these years. I just want to keep a copy for my collection, not that I need it. I since learn a whole lot more about transistors and op-amps. It is a very good introduction book for electronics.

 

[michael1978]

I have not received it yet. I must had the first edition in the late 70s! I long lost the book. The book is in my head and been using what I learned to do a lot of designs all these years. I just want to keep a copy for my collection, not that I need it. I since learn a whole lot more about transistors and op-amps. It is a very good introduction book for electronics.

you can find it in internet

 

[yungman]

you can find it in internet

I am old school, I want it to be in my hands!

I am keeping it as a collection, it's the very first book I used to really learn electronics, it has a special meaning for me. I studied electronics all by myself all these years, books are like my university. I have more books on the subjects I studied than the Stanford University library...A whole lot more. The only way for me is to buy online where I have the resource of the whole world. That's the reason I keep telling you to find a way to buy on Amazon, you can't just go to a book store in your area to look for books, they don't have the vast collection.

 

[michael1978]

I am old school, I want it to be in my hands!

I am keeping it as a collection, it's the very first book I used to really learn electronics, it has a special meaning for me. I studied electronics all by myself all these years, books are like my university. I have more books on the subjects I studied than the Stanford University library...A whole lot more. The only way for me is to buy online where I have the resource of the whole world. That's the reason I keep telling you to find a way to buy on Amazon, you can't just go to a book store in your area to look for books, they don't have the vast collection.

yes i know, because i buy some books here but pfff they don't explain good and you don't learn nothing special to give you some idea to make something, that book understand i think only the author;-), you have so much books, and if you want to buy at amazon, you have to look for used books

 

[yungman]

yes i know, because i buy some books here but pfff they don't explain good and you don't learn nothing special to give you some idea to make something, that book understand i think only the author;-), you have so much books, and if you want to buy at amazon, you have to look for used books

I bought 100% used, still cost me thousands! The new ones are mostly over $100, even used ones are $40 to $60 each. That's the reason why I was suggesting you to work it out on Amazon.

BTW, I received the Malvino, this one is old! But hey, it's $8.00 to the front door! I can't complain.

 

[michael1978]

I bought 100% used, still cost me thousands! The new ones are mostly over $100, even used ones are $40 to $60 each. That's the reason why I was suggesting you to work it out on Amazon.

BTW, I received the Malvino, this one is old! But hey, it's $8.00 to the front door! I can't complain.

yes i understand, i with try first with this book, and to look to open my card, because i don't have so much time, you download books from internet? do you know any site?

 

[michael1978]

We need Re2 resistor simply because it is almost impassible to build amplifier with desired low voltage gain and Ve > Vbe. Which is needed for good bias point stability.
But if you determined to use only Re1 resistor. We need to change DC bias method.
So we are not going to use voltage divider, instead of voltage divider we will use a collector-feedback bias circuit.
So for our example for Vcc = 10V and Rc = 1K;RL = 10K; Ic = 5mA and Av = 50.

re = 26mV/Ic = 5.2Ω

Re1 = (Rc||RL)/Av - re = 909Ω/50 - 5.2 = 12Ω

Vc = Vcc - Ic*Rc = 5V

Vb = Ic*Re1 + Vbe = 0.71V

And

Ib = Ic/hfe = 5mA/150 = 33.4μA

Rb1+Rb2 = (Vc - Vb)/Ib = 4.29V/33.4μA = 128KΩ

Rb1 = Rb2 = 128K/2 = 68KΩ

And simulation show that voltage gain is equal to AV = 51.8092[V/V]

hi Jony can you help me with this equation
on page 3 post 48
this one Re2 = ( 220 * 11.5 ) / (220 - 12.5) = 12.1 = 12Ω.
is not like you told me Re2 = ( 220 * 11.5 ) / (220 + 12.5) = 12.1 = 12Ω in place from - in plus +

 

[yungman]

hi Jony can you help me with this equation
on page 3 post 48
this one Re2 = ( 220 * 11.5 ) / (220 - 12.5) = 12.1 = 12Ω.
is not like you told me Re2 = ( 220 * 11.5 ) / (220 + 12.5) = 12.1 = 12Ω in place from - in plus +

I read the post 48, I am pretty sure that was not the description of this circuit. There is no Re2, the collector current setting is different.

 

[yungman]

yes i understand, i with try first with this book, and to look to open my card, because i don't have so much time, you download books from internet? do you know any site?

I type "free download solution manual Malvino..." and see what comes up!

It's a lot of leg work, you might go around in circles and they end up asking for money, then you have to quit and look for another site. There is no one site for this. But you'll likely find one if you work at it.

 

[michael1978]

I read the post 48, I am pretty sure that was not the description of this circuit. There is no Re2, the collector current setting is different.

look page 4 post 49, and you will see
this one Re2 = ( 220 * 11.5 ) / (220 - 12.5) = 12.1 = 12Ω.
is not like you told me Re2 = ( 220 * 11.5 ) / (220 + 12.5) = 12.1 = 12Ω in place from - in plus +
but i think in place from 12.5 is 11.5 jony i think he say to me i make mistake

 

[NascentOxygen]

look page 4 post 49, and you will see
this one Re2 = ( 220 * 11.5 ) / (220 - 12.5) = 12.1 = 12Ω.
is not like you told me Re2 = ( 220 * 11.5 ) / (220 + 12.5) = 12.1 = 12Ω in place from - in plus +
but i think in place from 12.5 is 11.5 jony i think he say to me i make mistake

...

re = 26mv/ic = 26mv/4ma = 6.5Ω

18Ω = (re + (re1||re2)) = ( 6.5Ω + (220||re2) )

re1||re2 = 18Ω - re = 11.5Ω

re2 = ( 220 * 11.5 ) / (220 [strike]- 12.5[/strike][/color] - 11.5[/color]) = 12.13 ⋍[/color] 12Ω.

 

[michael1978]

...

thnx for answer nascentoxygen

re = 26mv/ic = 26mv/4ma = 6.5Ω

18Ω = (re + (re1||re2)) = ( 6.5Ω + (220||re2) )

re1||re2 = 18Ω - re = 11.5Ω

re2 = ( 220 * 11.5 ) / (220 - 11.5) = 12.13 ⋍ 12Ω.
--------------------
re2 = ( 220 * 11.5 ) / (220 -11.5) = 12.13 ⋍ 12Ω.
why don't become plus + in place from minus - like this re1||re2 (re1*re2)/(re1+re2)
re2 = ( 220 * 11.5 ) / (220 -11.5) = 12.13 ⋍ 12
, of this is other formula
can you answer me

 

[michael1978]

hi Jony can you help me with this equation
look page 4 post 49
re2 = ( 220 * 11.5 ) / (220 - 11.5) = 12.13 ⋍ 12Ω.
why don't become plus + in place from minus - like this re1||re2 (re1*re2)/(re1+re2)
re2 = ( 220 * 11.5 ) / (220 - 11.5) = 12.13 ⋍ 12
, of this is other formula
can you answer me

 

[michael1978]

look page 4 post 49, and you will see
look page 4 post 49
re2 = ( 220 * 11.5 ) / (220 - 11.5) = 12.13 ⋍ 12Ω.
why don't become plus + in place from minus - like this re1||re2 (re1*re2)/(re1+re2)
re2 = ( 220 * 11.5 ) / (220 - 11.5) = 12.13 ⋍ 12
, of this is other formula
can you answer me

 

[NascentOxygen]

re2 = ( 220 * 11.5 ) / (220 - 12.5 - 11.5) = 12.13 ⋍ 12Ω.

In my post, the -12.5 is crossed out. It's replaced by -11.5.

 

[NascentOxygen]

why don't become plus + in place from minus - like this re1||re2 (re1*re2)/(re1+re2)

This expression has re2 on both sides. We need to extract it to one side, and if you do the algebra you will see how the - sign comes about.

re2 = ( 220 * 11.5 ) / (220 - 11.5) = 12.13 ⋍ 12

 

[michael1978]

In my post, the -12.5 is crossed out. It's replaced by -11.5.

yes i know but i ask somthing else
look this
re2 = ( 220 * 11.5 ) / (220 - 11.5) = 12.13 ⋍ 12Ω.
why don't become plus + in place from minus - like this re1||re2 (re1*re2)/(re1+re2)
re2 = ( 220 * 11.5 ) / (220 - 11.5) = 12.13 ⋍ 12
, of this is other formula
can you answer me

 

[michael1978]

This expression has re2 on both sides. We need to extract it to one side, and if you do the algebra you will see how the - sign comes about.

re2 = ( 220 * 11.5 ) / (220 - 11.5) = 12.13 ⋍ 12

yes you right i am learning a little bit math , from the book basic math for electronics

is not the same like this equation

re1||re2 (re1*re2)/(re1+re2)

 

[michael1978]

When we start design any circuit we need to know circuit specification.
If you have a 2mV input voltage and 100mV at output you need a amplifier with gain
Av = 100mV/2mV = 50[V/V] and Zin = 50K and Rload >2K.
Is not so easy to meet all this requirements whit this simple amplifier.
So I change them to
Voltage Gain= 50
Load Resistance= 10k ohm
Vce= 5V

First we need select BJT I choose BC546C with typical hfe = 520 and Hfe_min = 420

I start selection from Rc resistor.

Rc < 0.1Rload = 1KΩ

Additional I assume Ve = 1V

So

Ic = (Vcc - Vce - Ve)/Rc = (10V - 5V - 1V)/1KΩ = 4mA

next

Re1 = Ve/Ic = 1V/4mA = 250 but I chose 220Ω

Vb = Ic*RE + Vbe = 4mA * 220Ω + 0.65V = 1.53V (voltage at base)

Ib = Ic/Hfe_min = 4mA/420 ≈ 10μA (base current)

R2 = Vb / ( 5 * Ib) = 30K

R1 = ( Vcc - Vb) / ( 6 * Ib) = 150KΩ

So know if we want voltage gain 50V/V

Av = 50 = (Rc|| RL) / ( re + (Re1||Re2) )

( re + (Re1||Re2) ) = (Rc|| RL) / 70 = 909Ω/50 = 18Ω

re = 26mV/Ic = 26mV/4mA = 6.5Ω

18Ω = (re + (Re1||Re2)) = ( 6.5Ω + (220||Re2) )

Re1||Re2 = 18Ω - re = 11.5Ω

Re2 = ( 220 * 11.5 ) / (220 - 12.5) = 12.1 = 12Ω.

And now we have a circuit that we can change gain quite easily.

From
Rc/Re1 = 1K/220 = 4.5[V/V] if we remove Re2 and C2

to

Rc/re = 1K/6.5 = 153[V/V] if we short Re1.And normally to meet all your requirements we need to use more practical amplifier circuit or op amp.

JONEY why i get 0.200 Vac?,

 

[Jony130]

JONEY why i get 0.200 Vac?,

I don't know why. Maybe you made a mistake in simulation program.

 

[michael1978]

I don't know why. Maybe you made a mistake in simulation program.

yes Jony, i do it one more time and is now 0,68 Vac now smaller, of i have to change the emitter capacitor , because the value of capacitor i make 1000u, of transistor , which transistor you think to use do you know some type?
because input is 2mv*50gain=100MV
thnx for reply

 

[NascentOxygen]

The BC109C is a popular high gain general purpose transistor.

 

[michael1978]

The BC109C is a popular high gain general purpose transistor.

i will change now

i don't have that type of transistor

 

[michael1978]

look my circuit, i don't know where i make mistake, normal gain have to be 50*2Mv=100mV Peak

 

[NascentOxygen]

It's not a very good design. For bias point stability (against V_B and ß variations) you should allow a couple of volts across the emitter resistor; you have just 0.4V.

For maximum symmetrical output swing, you should allow approx as much voltage across R_C as V_CE. But you have ~1.8V and 7.8V, resp.

At what frequency are you measuring the ac gain?

 

[Jony130]

No the design is good. Simply the BJT that michael1978 use in his simulation has a very low current gain.
Ie = 0.4V/220Ω = 1.8mA

And Ib = IR4 - IR5

IR4 = 1V/30K = 33.4μA

IR5 = (10V - 1V)/150kΩ = 9V/150KΩ = 60μA

So Ib = 60μA - 33.4μA = 26.6μA

And Hfe = β = (Ie/Ib - 1) = 66.6

And this circuit was design for Hfe > 420

 

[michael1978]

It's not a very good design. For bias point stability (against V_B and ß variations) you should allow a couple of volts across the emitter resistor; you have just 0.4V.

For maximum symmetrical output swing, you should allow approx as much voltage across R_C as V_CE. But you have ~1.8V and 7.8V, resp.

At what frequency are you measuring the ac gain?

at 20kh
thnx for reply

 

[michael1978]

No the design is good. Simply the BJT that michael1978 use in his simulation has a very low current gain.
Ie = 0.4V/220Ω = 1.8mA

And Ib = IR4 - IR5

IR4 = 1V/30K = 33.4μA

IR5 = (10V - 1V)/150kΩ = 9V/150KΩ = 60μA

So Ib = 60μA - 33.4μA = 26.6μA

And Hfe = β = (Ie/Ib - 1) = 66.6

And this circuit was design for Hfe > 420

so result is correct 66.6, for ib like this is formula for current Ir5-Ir4? is not the same for ir4 and ir5 they are in serie ir4 and ir5
but Jony you told Ve to be 1v why is now Ve 0.4V?
thnx for reply

 

[michael1978]

so result is correct 66.6, or voltage gain is 66.6, if so 66.6*2Mv=133mv again the result is not corrcect
for ib like this is formula for current Ir5-Ir4? is not the same for ir4 and ir5 they are in serie ir4 and ir5
but Jony you told Ve to be 1v why is now Ve 0.4V?
thnx for reply

thnx for reply
]so result is correct 66.6, or voltage gain is 66.6, if so 66.6*2Mv=133mv again the result is not corrcect
for ib like this is formula for current Ir5-Ir4? is not the same for ir4 and ir5 they are in serie ir4 and ir5
but Jony you told Ve to be 1v why is now Ve 0.4V?
thnx for reply[/QUOTE]
thnx for reply

 

[michael1978]

thnx for reply
]so result is correct 66.6, or voltage gain is 66.6, if so 66.6*2Mv=133mv again the result is not corrcect
for ib like this is formula for current Ir5-Ir4? is not the same for ir4 and ir5 they are in serie ir4 and ir5
but Jony you told Ve to be 1v why is now Ve 0.4V?
thnx for reply

thnx for reply[/QUOTE]

hi Jony, can you answer me please

 

[Jony130]

thnx for reply
]so result is correct 66.6, or voltage gain is 66.6, if so 66.6*2Mv=133mv again the result is not corrcect
for ib like this is formula for current Ir5-Ir4? is not the same for ir4 and ir5 they are in serie ir4 and ir5
but Jony you told Ve to be 1v why is now Ve 0.4V?
thnx for reply

66 is a BJT current gain, not a amplifier voltage gain.
So in order to make this circuit work properly. You need change the BJT that you use in your simulation program. You need to find the BJT with Hfe > 300 in your simulation.

 

[michael1978]

66 is a BJT current gain, not a amplifier voltage gain.
So in order to make this circuit work properly. You need change the BJT that you use in your simulation program. You need to find the BJT with Hfe > 300 in your simulation.

ah 66 is current gain, so you mean to change type of transistor, but is possible for you to tell me any type of transistor which work, because there are a lot of type of transistor, and somebody he tell me one type of transistor but that transistor don't exist in my list of transistors
and yes Jony if you can tell me, i try to find self soms but i can't find it, that is the problem
thnx for reply

 

[Jony130]

BC548C
What simulation program do yo use ?

 

[michael1978]

BC548C
What simulation program do yo use ?

o joney i don't have, i have bc548A but not BC548C, i use B2 Spice A/V

 

[michael1978]

hey Jony i try with bc548A, and i get voltage gain of 84mv is this correct?

 

[Jony130]

Use BC547C

 

[michael1978]

Use BC547C

yes Joney i have that type i change, and now voltage gain is 94mv is correct now?

 

[Jony130]

Well first you need to learn how to use the simulation program.
I use B2 Spice V5 and as a BJT use BC547C. And AC sweep analysis show that voltage gain is equal to 50.1V/V
For 20mV at input I get 1V at output.

 

[michael1978]

Well first you need to learn how to use the simulation program.
I use B2 Spice V5 and as a BJT use BC547C. And AC sweep analysis show that voltage gain is equal to 50.1V/V
For 20mV at input I get 1V at output.

but this was disegn for 10v battery not 12v

 

[Jony130]

Well for Vcc = 10V the voltage gain is equal to 46V/V (0.9V at output for 20mV at input) because we don't included RL in our calculations when we solve for Re2.