
#37
Nov1912, 02:47 PM

P: 133

i know a little bit, i know a biasing, i know to find a amplifier voltage, i speak about ce amplifier, but i dont know how to get desired voltage, that is my big problem, like you find 40mv 



#38
Nov1912, 03:23 PM

P: 2,450

First things first. Can you tell us what the approximate DC voltages are at various points in the circuit and why? If you can then you are well on your way to understanding it. I realize Jony has already posted things of this nature but I feel you may not know how to arrive at this. You say you know biasing but I feel you are coming up short in that area. Do you know how to select the collector resistor to get the collector voltage where you want it based on a few other things in the circuit? And, what are those things and why they affect collector voltage? For that matter do you know where the collector voltage should be? Nothing wrong with not knowing this stuff. Not trying to make you out to look dumb or anyting. We're here to help. Where would you like to start specifically?




#39
Nov1912, 05:33 PM

P: 133

but he show me how to find av, but not how to buid amplifier with desired voltage, like here they say formula is rc//rl/re, i know this formula from the book, but desired voltage he dont tell me, i can build one amplifier but not with desired vooltage, for example i want av about 50 and input 2mv = 100mv, i dont know how to do it, for that i ask help here, in similary way 



#40
Nov1912, 05:55 PM

P: 2,450

You need to know the load impedance if before you go much farther, that is assuming you are driving a load. I will assume you don't know what the DC collector voltage should be. To start with we will set it at about half the supply voltage, so 5 volts. What do you know to do from here? I don't plan on designing the whole amp for you. Doesn't matter to me if it is not homework, I will treat it as so and make you learn it.




#41
Nov1912, 06:09 PM

HW Helper
P: 4,708





#42
Nov2012, 07:14 AM

P: 133

do you see vcc 10V collector 10 IC 10ma Vce midpoint Bdc 100 now what to do, how to find desired gain 



#43
Nov2012, 07:19 AM

P: 133

do not exist some formula, how i have to do with calculator can you tell me please thnx 



#44
Nov2012, 09:32 AM

P: 2,450

Thats not exactly what I had in mind but it is a start. I think if you actually knew how to get those resistor values and voltages you would not need to ask how to figure the gain. Did you plug some numbers into a simulator or something to get this? I mentioned to you that we would put the collector voltage at 5 volts. You have it at 6 volts which is close enough. AC voltage gain of this amplifier is 4 UNLOADED. How can you not know that? You have previously posted that you know the formula for gain which is (RL RC)/RE. Zout of this amplifier is 400 ohms.
 Next. Zin. Originally you wanted a Zin of 50K. Do you see how that is next to impossible with transistor betas typically around 100 using components you have selected? 



#45
Nov2012, 11:07 AM

P: 133





#46
Nov2012, 12:35 PM

P: 133

can you explain me of design one amplifier with gain of 50, rc 1k, power supply 12 and the other select you please in similary way, like you explain in first example 



#47
Nov2012, 12:54 PM

P: 2,450

Let me ask you this michael: How did you select the resistors in the schematic that you posted in post #42?




#48
Nov2012, 01:20 PM

P: 389

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 + (Re1Re2) ) ( re + (Re1Re2) ) = (Rc RL) / 70 = 909Ω/50 = 18Ω re = 26mV/Ic = 26mV/4mA = 6.5Ω 18Ω = (re + (Re1Re2)) = ( 6.5Ω + (220Re2) ) Re1Re2 = 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. 



#49
Nov2012, 04:50 PM

P: 133

gain rc//rl/re, but i see there in resistor in serie with capacitor how come now gain RC//RL/RE2+RE1 am i correct, if not can you tell me i dont get gain of 50 



#50
Nov2012, 04:52 PM

P: 133





#51
Nov2012, 05:00 PM

P: 133





#52
Nov2012, 06:29 PM

P: 2,450





#53
Nov2012, 06:38 PM

P: 133





#54
Nov2012, 07:01 PM

P: 2,450

Collector resistor is Zout.
Zin is R1R2(beta*Re). In the schematic that Jony posted Re becomes Re1Re2. 


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