Please help transistor amplifier

In summary: So, if you choose Ie=1mA, you only have 25Ω to play with, which means you want rl to be very large value, which means you want to use a Darlington transistor.In summary, if you want to design a CE amplifier with a gain of 50, you can use two stages, each with a gain of x10 and x5 respectively. Alternatively, you can use one stage with a gain of 50, but it may be more difficult to design and may not be as reliable. It is important to consider the input and output impedances of the circuit in order to achieve the desired voltage gain.
  • #141
Read the bias scheme in Malvino. Re2 is to stabilizing the collector current. This is called voltage divider bias. R1 and R2 form a voltage divider to give about 1.7V at the base. The emitter is about 0.7V below so the emitter is at about 1V. You have Re1+Re2 about 200 ohm ( don't be picky), so you are setting up the emitter current ( collector current) of about 5 mA. This 5 mA will give r'e ≈ 5 ohm. The point of using voltage divider bias is to control the current through the BJT by the voltage drop across the Re2. This make it a lot more predictable. Read Malvino, it's all there.

The C2 is to bypass Re2 at frequency you want to amplify. At high frequency, Re2 is being shorted out by C2, so the resistance on the emitter side is r'e+Re1=19 ohm or 20 ohm.

No matter how the circuit looks like, in common emitter BJT circuit, the gain is the total impedance at the collector divided by the total impedance at the emitter. AND it is inverted. It is just that simple.

I just went back and looked at page 5, 6 7 of this thread, Jony130 really...REALLY spent the time explained to you on this. You really need to spend less time asking question here and more time reading the book over and over, work out the problems. I don't have the book, I think it has answers even if you can't find the solution manual. I learn all these from Malvino in 1979 without the solution manual. You are spending almost a month on this and I am sure it's cover in only about 4 to 5 pages in Malvino. I am sure anything I said at this point, HAD been totally covered by Jony130. Read the book over and over and over until you understand. Or read this thread over and over. Write out everything Jony130 wrote step by step to verify that you are following, don't just read, write it out.
 
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  • #142
yungman said:
Read the bias scheme in Malvino. Re2 is to stabilizing the collector current. This is called voltage divider bias. R1 and R2 form a voltage divider to give about 1.7V at the base. The emitter is about 0.7V below so the emitter is at about 1V. You have Re1+Re2 about 200 ohm ( don't be picky), so you are setting up the emitter current ( collector current) of about 5 mA. This 5 mA will give r'e ≈ 5 ohm. The point of using voltage divider bias is to control the current through the BJT by the voltage drop across the Re2. This make it a lot more predictable. Read Malvino, it's all there.

The C2 is to bypass Re2 at frequency you want to amplify. At high frequency, Re2 is being shorted out by C2, so the resistance on the emitter side is r'e+Re1=19 ohm or 20 ohm.

No matter how the circuit looks like, in common emitter BJT circuit, the gain is the total impedance at the collector divided by the total impedance at the emitter. AND it is inverted. It is just that simple.

I just went back and looked at page 5, 6 7 of this thread, Jony130 really...REALLY spent the time explained to you on this. You really need to spend less time asking question here and more time reading the book over and over, work out the problems. I don't have the book, I think it has answers even if you can't find the solution manual. I learn all these from Malvino in 1979 without the solution manual. You are spending almost a month on this and I am sure it's cover in only about 4 to 5 pages in Malvino. I am sure anything I said at this point, HAD been totally covered by Jony130. Read the book over and over and over until you understand. Or read this thread over and over. Write out everything Jony130 wrote step by step to verify that you are following, don't just read, write it out.

thanks for time, in malvino i have version 7, the book tell only how to desigin an amplifier, and joney he make quicky the amplifier, yes he spent a lot of time for me, the book which you buy in amazone they are used books not new but not expensiv
 
  • #143
no i try, i can't make amplifier without adding Re1, because &
10vcc
R2=170Ω=1.7V
R1=830Ω=8.3V
Ve=1v but i need one resistor for Ic
Ic≈Ie
how can i select current to Ie i need to add resistor, and after to select Vce and Vc

is that , that we need to add resistor to Ve to select Ie, otherwise you can't make an amplifier
but if you add for example to be Ie = 1mA, ic=5K*1mA = 5V after the gain going to change it can't be which i want
 
  • #144
michael1978 said:
no i try, i can't make amplifier without adding Re1, because &
10vcc
R2=170Ω=1.7V
R1=830Ω=8.3V
Ve=1v but i need one resistor for Ic
Ic≈Ie
how can i select current to Ie i need to add resistor, and after to select Vce and Vc

is that , that we need to add resistor to Ve to select Ie, otherwise you can't make an amplifier
but if you add for example to be Ie = 1mA, ic=5K*1mA = 5V after the gain going to change it can't be which i want

So you know it is 1.7V at the base and 1V at the emitter.

YES, you set Ie with the emitter resistor ( Re1+Re2). This is the KEY of DC biasing. You first set up the operating current Ie of the transistor, this is the first and foremost thing. Collector don't set the current of the BJT, collector current is only follow the emitter current. AGAIN, this is in Malvino.

1) In the diagram, Re1+Re2 is about 200 ohm. You put 1V at emitter, you put 1V across the Re1 and Re2. SO you force 5mA through the resistors...Which, means you set the Ie to 5mA. This is how you set up the DC bias current.

2) After setting the Ie, then you start looking at the collector. You know β is high, so you know the Ic≈5mA. If you have Vcc=10V, and you have 1K resistor at the collector, then you know 5mA through 1K is 5V. So the collector is at +5V.

3) AGAIN, gain is determined by the impedance at the collector divided by the impedance at the emitter. We gone over this over and over and over already. Read the old posts AGAIN.

These are all in Malvino. Please read that before you ask any more question. If you don't get it, read it, write out the numbers, work through the problems before you post.
 
  • #145
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

attachment.php?attachmentid=53770&stc=1&d=1355076807.png


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

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  • #146
Jony130 said:
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

attachment.php?attachmentid=53770&stc=1&d=1355076807.png


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

JONEY thank you for reply i am learning a lot from you, you are the beste;-) COMPLIMENT
 
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  • #147
yungman said:
So you know it is 1.7V at the base and 1V at the emitter.

YES, you set Ie with the emitter resistor ( Re1+Re2). This is the KEY of DC biasing. You first set up the operating current Ie of the transistor, this is the first and foremost thing. Collector don't set the current of the BJT, collector current is only follow the emitter current. AGAIN, this is in Malvino.

1) In the diagram, Re1+Re2 is about 200 ohm. You put 1V at emitter, you put 1V across the Re1 and Re2. SO you force 5mA through the resistors...Which, means you set the Ie to 5mA. This is how you set up the DC bias current.

2) After setting the Ie, then you start looking at the collector. You know β is high, so you know the Ic≈5mA. If you have Vcc=10V, and you have 1K resistor at the collector, then you know 5mA through 1K is 5V. So the collector is at +5V.

3) AGAIN, gain is determined by the impedance at the collector divided by the impedance at the emitter. We gone over this over and over and over already. Read the old posts AGAIN.

These are all in Malvino. Please read that before you ask any more question. If you don't get it, read it, write out the numbers, work through the problems before you post.
did you see the post 146 of Joney, maybe i am mistake but malvino he don't show you how to get desired voltage, i wil like to ask you why Jony, he put so big resistance of voltage divider i say for r1 22K and for R2 4.7k how he calculate,? me i put it R2=170, and R1=830, look how big difference, he show one example but me i put in other way,
 
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  • #148
michael1978 said:
did you see the post 146 of Joney, maybe i am mistake but he don't show you how to get desired voltage, i wil like to ask you why Jony, he put so big resistance of voltage divider i say for r1 22K and for R2 4.7k how he calculate,? me i put it R2=170, and R1=830, look how big difference, he show one example but me i put in other way,

We were talking about the diagram of post 114 all along until now.

The newer diagram is not as desirable. I believe it called self bias or something. It depend a lot on the beta of the transistor. Voltage divider bias is a better way to go.
 
  • #149
yungman said:
We were talking about the diagram of post 114 all along until now.
yes but i say to joney i going to make one amplifier, and he help answer me thanks from him, ok why Joney select so big resistance of R1 22k and R2 4.7k
and he get the sam base almost 1.7V, how i have to do it big resistance like joney
 
  • #150
michael1978 said:
yes but i say to joney i going to make one amplifier, and he help answer me thanks from him, ok why Joney select so big resistance of R1 22k and R2 4.7k
and he get the sam base almost 1.7V, how i have to do it big resistance like joney

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.
 
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  • #151
yungman said:
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
 
  • #152
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.
 
  • #153
yungman said:
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
 
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  • #154
michael1978 said:
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.
 
  • #155
yungman said:
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
 
  • #156
michael1978 said:
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
 
  • #157
michael1978 said:
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.
 
  • #158
yungman said:
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
 
  • #159
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!
 
  • #160
yungman said:
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,
 
  • #161
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.
 
  • #162
yungman said:
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
 
  • #163
michael1978 said:
you can find it in internet

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

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.
 
  • #164
yungman said:
I am old school, I want it to be in my hands!:rofl:

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
 
  • #165
michael1978 said:
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.
 
  • #166
yungman said:
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?
 
  • #167
Jony130 said:
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

attachment.php?attachmentid=53770&stc=1&d=1355076807.png


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 +
 
  • #168
michael1978 said:
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.
 
  • #169
michael1978 said:
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.
 
  • #170
yungman said:
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
 
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  • #171
michael1978 said:
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] - 11.5) = 12.13 ⋍ 12Ω.
 
  • #172
NascentOxygen said:
...

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
 
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  • #173
michael1978 said:
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
 
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  • #174
michael1978 said:
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
 
Last edited:
  • #175
michael1978 said:
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.
 
<h2>1. What is a transistor amplifier?</h2><p>A transistor amplifier is an electronic device that uses transistors to amplify an electrical signal. It is commonly used in electronic circuits to increase the strength of a weak signal.</p><h2>2. How does a transistor amplifier work?</h2><p>A transistor amplifier works by using a small current to control a larger current. This is achieved by applying a small input voltage to the base of the transistor, which then allows a larger current to flow from the collector to the emitter. This amplifies the signal.</p><h2>3. What are the benefits of using a transistor amplifier?</h2><p>Transistor amplifiers offer several benefits, including high gain, low distortion, and high input impedance. They are also small in size and require low power, making them ideal for use in portable electronic devices.</p><h2>4. What are the different types of transistor amplifiers?</h2><p>There are three main types of transistor amplifiers: common emitter, common base, and common collector. Each type has its own unique characteristics and is used for different applications.</p><h2>5. How do I choose the right transistor amplifier for my project?</h2><p>When choosing a transistor amplifier, it is important to consider factors such as the desired gain, frequency range, and input and output impedance. It is also important to select a transistor with a suitable power rating for your project.</p>

1. What is a transistor amplifier?

A transistor amplifier is an electronic device that uses transistors to amplify an electrical signal. It is commonly used in electronic circuits to increase the strength of a weak signal.

2. How does a transistor amplifier work?

A transistor amplifier works by using a small current to control a larger current. This is achieved by applying a small input voltage to the base of the transistor, which then allows a larger current to flow from the collector to the emitter. This amplifies the signal.

3. What are the benefits of using a transistor amplifier?

Transistor amplifiers offer several benefits, including high gain, low distortion, and high input impedance. They are also small in size and require low power, making them ideal for use in portable electronic devices.

4. What are the different types of transistor amplifiers?

There are three main types of transistor amplifiers: common emitter, common base, and common collector. Each type has its own unique characteristics and is used for different applications.

5. How do I choose the right transistor amplifier for my project?

When choosing a transistor amplifier, it is important to consider factors such as the desired gain, frequency range, and input and output impedance. It is also important to select a transistor with a suitable power rating for your project.

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