
#73
Nov2312, 02:46 PM

P: 133

  OK Av = Rc/re = gm*Rc = Ic/26mV*Rc = (Ic*Rc)/26mV 50[V/V] = (Ic*Rc)/26mV from this (Ic*Rc) = 1.3 I choose Rc = 1K So Ic = 1.3/1K = 1.3mA But as you can see this approach is not very good because give as low voltage swing. We need to use Re1 or use a different type of a circuit. But lest as add Re1 resistor. Now Av = Rc/(re+Re1) And for Rc = 1K we can use this Ic = 0.5Vcc/Rc = 5V/1K = 5mA So re = 26mV/5mA = 5.2Ω And Re1 = Rc/Av  re = 1K/50  5.2Ω = 20  5.2Ω = 14Ω This circuit is also not very good because low Re1 means that Hfe spread and temperature change will have significant impact on bias point stability. This is why I use Ve = 1V and add Re2 to reduce the voltage gain to 50[V/V]  after you show the second example Also we can split Re resistor with a capacitor. And use Re1 = 14Ω and Re2 = Ve/Ic  re1 = 1V/5mA  14Ω ≈ 180Ω And now we can select R1 and R2 Vb = Ve + Vbe = Ic*(Re1+Re2)+ Vbe = 5mA* (180 + 14) + 0.65V = 1.62V If we assume Hfe = 150 Ib = Ic/hfe = 5mA/150 = 34μA R1 = (Vcc  Vb)/( 11*Ib) = 22KΩ R2 = Vb/(10*Ib) = 4.7KΩ  is this the same like in example 1 and example 2 i mean this second example, and the first example without R2 has the same value value of r1 and r2 is r1=22k, and r2=4.7K and power supply 10V , and is this correct R1 = (Vcc  Vb)/( 11*Ib) = 22KΩ of this R1 = (Vcc  Vb)/( 10*Ib) = 22KΩ do you understand you show 1 example in 2 party, first without re2 second with re2, so this one example, do you understand 



#74
Nov2312, 02:50 PM

P: 133





#75
Nov2312, 02:55 PM

P: 389

This two equation holds only if Ve > 0.6V
R1 = (Vcc  Vb)/( 11*Ib) R2 = Vb/(10*Ib) 



#76
Nov2312, 03:02 PM

P: 133

 OK Av = Rc/re = gm*Rc = Ic/26mV*Rc = (Ic*Rc)/26mV 50[V/V] = (Ic*Rc)/26mV from this (Ic*Rc) = 1.3 I choose Rc = 1K So Ic = 1.3/1K = 1.3mA But as you can see this approach is not very good because give as low voltage swing. We need to use Re1 or use a different type of a circuit. But lest as add Re1 resistor. Now Av = Rc/(re+Re1) And for Rc = 1K we can use this Ic = 0.5Vcc/Rc = 5V/1K = 5mA So re = 26mV/5mA = 5.2Ω And Re1 = Rc/Av  re = 1K/50  5.2Ω = 20  5.2Ω = 14Ω This circuit is also not very good because low Re1 means that Hfe spread and temperature change will have significant impact on bias point stability. This is why I use Ve = 1V and add Re2 to reduce the voltage gain to 50[V/V]  Also we can split Re resistor with a capacitor. And use Re1 = 14Ω and Re2 = Ve/Ic  re1 = 1V/5mA  14Ω ≈ 180Ω And now we can select R1 and R2 Vb = Ve + Vbe = Ic*(Re1+Re2)+ Vbe = 5mA* (180 + 14) + 0.65V = 1.62V If we assume Hfe = 150 Ib = Ic/hfe = 5mA/150 = 34μA R1 = (Vcc  Vb)/( 11*Ib) = 22KΩ R2 = Vb/(10*Ib) = 4.7KΩ  ah joney i make you tired, is possibe to write me in one party not seperate all this here, like in first example with re2, you explain me very good, but this is not in one party , i can understand better PLEASE 



#77
Nov2312, 03:29 PM

P: 389

OK
For this diagram the voltage gain is equal to Av ≈ Rc/(re+Re1) Rc = 0.1*RL = 0.1*10K = 1K And for Rc = 1K we can use this Ic = 0.5Vcc/Rc = 5V/1K = 5mA So re = 26mV/5mA = 5.2Ω And Re1 = Rc/Av  re = 1K/50  5.2Ω = 20  5.2Ω = 14Ω This circuit is not very good because low Re1 means that Hfe spread and temperature change will have significant impact on the bias point stability. This is why I use Ve = 1V for good bias stability. To overcome this problem of a low value for Re1 = 14Ω and Ve = 70mV (Ve = Re1 *Ic = 5mA*14Ω = 70mV). I add anther Re resistor in series with Re1. Now we have Ve = Ic * Re1+Re2 and we want Ve = 1V so Re2 = Ve/Ic  Re1 = 1V/5mA  14Ω ≈ 180Ω And now we can select R1 and R2 Vb = Ve + Vbe = Ic*(Re1+Re2)+ Vbe = 5mA* (180 + 14) + 0.65V = 1.62V If we assume Hfe = 150 Ib = Ic/hfe = 5mA/150 = 34μA R1 = (Vcc  Vb)/( 11*Ib) = 22KΩ R2 = Vb/(10*Ib) = 4.7KΩ And I use C2 capacitor to "remove" Re2 from gain equation for AC signal. Av = RcRL/(re + Re1) 



#78
Nov2312, 03:34 PM

P: 133

JONEY THANK YOU, i understand everything, just this one no R1 = (Vcc  Vb)/( 11*Ib) = 22KΩ, where come 11, how come this 11 can you explain me thnx 



#79
Nov2312, 04:00 PM

P: 389

First think first.
The one of the methods to become independent of hfe changes is to select voltage divider resistor (R1 and R2) in such a way that base current don't load our voltage divider to much. The maximum base current is equal to Ib_max = Ic/Hfe_min where Hfe_min is a minimum value for Hfe in data sheet. And we select voltage divider current so that I_divader > Ib_max In our example voltage divider current is I2 current . But as you can see the current that flows through R1 resistor is equal to divider current plus base current. Additional if we assume that divider current is ten times greater than base current. I2 = 10 * Ib the R1 resistor current I1 becomes I1 = I2 + Ib = 10*Ib + Ib = 11*Ib And this is why R1 = (Vcc  Vb) / (10*Ib + Ib) = (Vcc  Vb)/(11*IB) and R2 = Vb/(10*Ib) 



#80
Nov2312, 04:22 PM

P: 133





#81
Nov2412, 04:22 AM

P: 133

 ib is 0,034 johy this i know R1 = (Vcc  Vb)/( 11*Ib) = 22KΩ R1 = (Vcc  Vb) / (10*Ib + Ib) = (Vcc  Vb)/(11*IB) (10*Ib + Ib) IS (10*IB(0.034)+0.034)=(0.34*IB) but 11 i dont know can you explain me please CAN YOU ANSWER ME 



#82
Nov2412, 03:17 PM

P: 133





#83
Nov2412, 06:12 PM

P: 389

Do you understand first kirchhoffs law ?
Do you know how loaded voltage divider work? 



#84
Nov2512, 06:58 AM

P: 133

and i do like i see in some book voltage divider like this for example we need base current of 50MICRO AND base voltage of 1.7V so now for example if is 10vcc r2 1.7v base/50micro is come 34ohm r1 must have 8.3V/50MICRO is come 166ohm so 10VCC/200=5MICRO so the current in serie is the same, maybe i do somewhere mistake i am curios to know your calculation how you get 10 and 11 in voltage divider, this is something new to me, but can you show another with example so i can understand, you show also the first example with gain of 50 do you rembember also there i see R2 = Vb / ( 5 * Ib) = 30K R1 = ( Vcc  Vb) / ( 6 * Ib) = 150KΩ wher how you get 5 and 6, and second example r2 10 and r1 11, can you show me example just one the second i find self, i know you can do it, just with example in some easy way please joney i whait yesterday all day message from you, because i want to learn it, and i learn good how to select desired voltage and i am happy, but this no R1 = (Vcc  Vb)/( 11*Ib) = 22KΩ R2 = Vb/(10*Ib) = 4.7KΩ valu of 11 and 10 joney plase show me example in easy way so i can understand thnx for reply.... 



#85
Nov2512, 08:09 AM

P: 389

So to reduce this loading effect on output voltage we need increase the current that is flow through voltage divider or output current should be a small fraction of the voltage divider current. So when we design voltage divider we assume that voltage divider current is larger then the load current. Normally Divider current is 5 to 30 times larger then the load current. So if we want 1.7V and load current is equal to 50μA and we choose voltage divider current 10 times larger then load current we have this R2 = 1.7V/(10*Iload) = 1.7V/(10*50μ) = 1.5V/500μA = 3.4K And from I Kirchhoff's Law we see that R1 resistor current is equal: I_R1 = 10*Iload + Iload = 11*Iload = 550μA So R1 = (10V  1.7V)/(550μ) = 15K 



#86
Nov2512, 11:04 AM

P: 133

R2 = 1.7V/(10*Iload) = 1.7V/((this is 10 time larger)10*50μ(base current)) = 1.5V/500μA = 3.4K this i understand but this i dont understand I_R1 = 10*Iload + Iload = 11*Iload = 550μA 10*Iload(50μA)+Iload(50μA)=11(how we get 11 now)*Iload(50μA)=550μA i dont understand that 11 how we get? 10*Iload(50μA)+Iload(50μA)=0.55 not 11 sorry i dont know where i make mistake, where come this 11, i dont know how to calculate, i know 11*Iload is 550μA thnx for answer 



#87
Nov2512, 11:21 AM

P: 389

Solve this pure math problem
B = 10A + A Also from KCL we see that R1 must provide current for R2 resistor and also the current for the load. So if Iload = 50μA and we pick voltage divider 10 times load current we have this: I_R1 = Iload + 10 times load current = 11 times load current = 11*50μA = 550μA 



#88
Nov2512, 11:41 AM

P: 133





#89
Nov2512, 11:46 AM

P: 133

Can you tell me how to calculate with calculator joney




#90
Nov2512, 11:48 AM

P: 389




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