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Please help transistor amplifier 
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#19
Nov1812, 01:57 PM

P: 133

is not rc divide rl, i thought  is divide / but is not can you tell how you calculate with this symbol  thnx 


#20
Nov1812, 02:01 PM

P: 409

If you have a resistors connect in parallel you use this two parallel lines "" to express it
R1R2 = (R1*R2)/(R1+R2) 


#21
Nov1812, 02:33 PM

P: 3,898

I use // as parallel. And Jony130 is correct.



#22
Nov1812, 02:52 PM

P: 133




#23
Nov1812, 02:53 PM

P: 3,898

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 opamp. 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. 


#24
Nov1812, 02:57 PM

P: 133




#25
Nov1912, 02:12 AM

P: 133

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 


#26
Nov1912, 10:26 AM

P: 409

Why you built a amplifier with gain smaller then one?



#27
Nov1912, 12:42 PM

P: 133

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



#28
Nov1912, 12:43 PM

P: 133




#29
Nov1912, 12:45 PM

P: 133

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 


#30
Nov1912, 12:48 PM

P: 409

Show as a diagram and all components values.
And tell as how long you have been learn electronics? 


#31
Nov1912, 01:05 PM

P: 133

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 


#32
Nov1912, 01:28 PM

P: 409

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 = (RcRL)/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 


#33
Nov1912, 01:45 PM

P: 133

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 


#34
Nov1912, 02:04 PM

P: 133




#35
Nov1912, 02:05 PM

P: 133

i am not an enigner, i am total beginner, i try to leran self



#36
Nov1912, 02:25 PM

P: 409

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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