by michael1978
Tags: amplifier, transistor
P: 121
 Quote by yungman That's the best book....bar none!!!! Everything I posted is in the Malvino!!! I gone very far as an engineer using nothing more than Malvino. Yes, I got into BJT IC design with just the knowledge of Malvino, it's that good!!! Of cause I learn a whole lot more since then, but it's good enough to get you started as an engineer. You better read the part about Vt again, it's all there.
can i ask you something wich symbols is this || for exaple rc||rl
is not rc divide rl, i thought || is divide / but is not
can you tell how you calculate with this symbol ||
thnx
 P: 256 If you have a resistors connect in parallel you use this two parallel lines "||" to express it R1||R2 = (R1*R2)/(R1+R2)
 P: 3,797 I use // as parallel. And Jony130 is correct.
P: 121
 Quote by yungman I use // as parallel. And Jony130 is correct.
§thank you ver much
 P: 3,797 You're welcome, keep digging into Malvino, it's worth your time. I hold this book is such high regard I did try to order a newer edition last year, but I ordered the experimental manual by mistake. I lost my original book long long time ago....since the early 80s. I want to keep a copy in my library collection. Another part Malvino is very good is the op-amp. I still use what I learn from the book. Of cause you need more on Bode Plot stability design, but this really give you a strong start. I designed many closed loop feedback system, I use the Bode Plot starting from what I learn from Malvino.
P: 121
 Quote by yungman You're welcome, keep digging into Malvino, it's worth your time.
thanx man
 P: 121 may i ask you something FIRST is this formula OK for example rc=(3.600X10000)/(3600+10000)=2650 desired voltage 0.5, 2650/0.5=5300 so re=5300 voltage gain is 2650/5300=0.5 is this correct formula? if is this is correct formula why my simulator is 43mV not 50mv where is wrong my formula of my simulator? can you exaplian me please thanx SECOND how if you connect one re in serie with r'e, how decrase voltage gain, can you explain in similary ways please
 P: 256 Why you built a amplifier with gain smaller then one?
 P: 121 i am trying to learn like in book, transistor ce class a, and i do experments with software, but like you see my result are not correct, i do somewhere mistake
P: 121
 Quote by michael1978 thanx man
jjjjj
P: 121
 Quote by yungman I use // as parallel. And Jony130 is correct.
FIRST
is this formula OK
for example rc=(3.600X10000)/(3600+10000)=2650
desired voltage 0.5, 2650/0.5=5300 so re=5300
voltage gain is 2650/5300=0.5
is this correct formula? if is this is correct formula why my simulator is 43mV not 50mv
where is wrong my formula of my simulator? can you exaplian me please thanx

SECOND how if you connect one re in serie with r'e, how decrase voltage gain, can you explain in similary ways please
 P: 256 Show as a diagram and all components values. And tell as how long you have been learn electronics?
P: 121
 Quote by Jony130 Show as a diagram and all components values. And tell as how long you have been learn electronics?
long enough,
rc=3.600
rl=10000
rc=(3.600X10000)/(3600+10000)=2650
re bypass 5300, only 1 resistor in parallel with capacitor
voltage divider
r1=10000
r2 = 2.200
power supply 10v
input voltage 2mv at frequnecy 20K
 P: 256 If I understand you correctly your circuit look like this: and Vcc = 10V If so the collector current is equal to Ic ≈ Ve/RE ≈ 1.1V/5.3KΩ ≈ 0.2mA and Vc = Vcc - Ic *Rc = 10V - 0.2mA * 3.6K = 10V - 0.72V = 9.28V . And the voltage gain Av = (Rc||RL)/re = 2.65KΩ/130Ω = 20[V/V] And Vout = 20* Vin = 20 * 2mV = 40mV But your amplifier Q point (bias point) was not chosen properly. See this post http://www.physicsforums.com/showthr...69#post4058469 Where re = 26mV/Ic Ve = (Vcc* R2/(R1+R2)) - Vbe And next time try use Engineering notation instead of 3.600 Attached Thumbnails
P: 121
 Quote by Jony130 If I understand you correctly your circuit look like this: and Vcc = 10V If so the collector current is equal to Ic ≈ Ve/Re ≈ 1.1V/5.3KΩ ≈ 0.2mA and Vc = Vcc - Ic *Rc = 10V - 0.2mA * 3.6K = 10V - 0.72V = 9.28V . And the voltage gain Av = (Rc||RL)/re = 2.65KΩ/130Ω = 20[V/V] And Vout = 20* Vin = 20 * 2mV = 40mV But your amplifier Q point (bias point) was not chosen properly. See this post http://www.physicsforums.com/showthr...69#post4058469 Where re = 26mV/Ic Ve = (Vcc* R2/(R1+R2)) - Vbe And next time try use Engineering notation instead of 3.600

sorry i am a beginner
my circuits look like that,

but when you start to build an amplifier, wich are the first steps to take
and how to get a desired voltage,
i so my amplifier is bad, but is was good the re was 1k but me i changet to experiment to get desired voltage

what do you think, do you have time to build one amplifier with desired gain, can you show step by step and to explain in similary way please
P: 121
 Quote by Jony130 If I understand you correctly your circuit look like this: and Vcc = 10V If so the collector current is equal to Ic ≈ Ve/RE ≈ 1.1V/5.3KΩ ≈ 0.2mA and Vc = Vcc - Ic *Rc = 10V - 0.2mA * 3.6K = 10V - 0.72V = 9.28V . And the voltage gain Av = (Rc||RL)/re = 2.65KΩ/130Ω = 20[V/V] And Vout = 20* Vin = 20 * 2mV = 40mV But your amplifier Q point (bias point) was not chosen properly. See this post http://www.physicsforums.com/showthr...69#post4058469 Where re = 26mV/Ic Ve = (Vcc* R2/(R1+R2)) - Vbe And next time try use Engineering notation instead of 3.600
and how you get 130Ω? Av = (Rc||RL)/re = 2.65KΩ/130Ω = 20[V/V] PLEASE CAN YOU TELL ME YOUR CALCULATION
 P: 121 i am not an enigner, i am total beginner, i try to leran self
 P: 256 If you want to learn you need first understand how this amplifier work. Also you need to understand how BJT work? And I don't ask about how BJT really work from physical point of view.

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