by michael1978
Tags: amplifier, transistor
P: 3,844
 Quote by michael1978 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.
P: 133
 Quote by yungman 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 dont explain good and you dont 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
P: 3,844
 Quote by michael1978 yes i know, because i buy some books here but pfff they dont explain good and you dont 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.
P: 133
 Quote by yungman 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 dont have so much time, you download books from internet? do you know any site?
P: 133
 Quote by Jony130 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 +
P: 3,844
 Quote by michael1978 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.
P: 3,844
 Quote by michael1978 yes i understand, i with try first with this book, and to look to open my card, because i dont 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.
P: 133
 Quote by yungman 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
HW Helper
P: 4,716
 Quote by michael1978 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 - 12.5 - 11.5) = 12.13 ⋍ 12Ω.
P: 133
 Quote by 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 dont 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
 P: 133 [QUOTE=michael1978;4194864]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 dont 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
 P: 133 [QUOTE=michael1978;4196654]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 dont 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
HW Helper
P: 4,716
 Quote by michael1978 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.
HW Helper
P: 4,716
 Quote by michael1978 why dont 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
P: 133
 Quote by NascentOxygen 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 dont 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
P: 133
 Quote by NascentOxygen 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)
P: 133
 Quote by Jony130 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?,
P: 389
 Quote by michael1978 JONEY why i get 0.200 Vac?,
I don't know why. Maybe you made a mistake in simulation program.

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