by michael1978
Tags: amplifier, transistor
P: 410
 Quote by michael1978 so this is the steps if i need to build an amplifier thnx 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
Simply voltage gain of a CE amplifier is always equal to

Av = ( resistance seen from collector toward the output) / (re + (resistance seen from emitter to gnd)

So we have

Av = (Rc||RL)/ ( re + Re1||Re2) = 909Ω/( 6.5Ω + 220Ω||12Ω ) ≈ 909/18.5Ω = 49.13[V/V]

Zin = R1||R2||( Hfe+1 *(re+Re1||Re2) ) = 25KΩ||( 421 * 18.5Ω) = 25K||7.8KΩ = 5.9KΩ

And
Zout ≈ Rc ≈ 1KΩ
P: 133
 Quote by Jony130 Simply voltage gain of a CE amplifier is always equal to Av = ( resistance seen from collector toward the output) / (re + (resistance seen from emitter to gnd) So we have Av = (Rc||RL)/ ( re + Re1||Re2) = 909Ω/( 6.5Ω + 220Ω||12Ω ) ≈ 909/18.5Ω = 49.13[V/V] Zin = R1||R2||( Hfe+1 *(re+Re1||Re2) ) = 25KΩ||( 421 * 18.5Ω) = 25K||7.8KΩ = 5.9KΩ And Zout ≈ Rc ≈ 1KΩ
thank you,
you show me a good example
but i have one more question, if somebody want to desigin a amplifier, they do it like your example, of another design, i mean for ce claas a amplifier
one more time thnx for good example
P: 133
 Quote by Averagesupernova Collector resistor is Zout. Zin is R1||R2||(beta*Re). In the schematic that Jony posted Re becomes Re1||Re2.
yes he show me a good example, but is so much important zin and zout in amplifier,
did you show jony example, this steps i have to take all time if i design an amplifier
P: 133
 Quote by michael1978 so this is the steps if i need to build an amplifier thnx 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
jony i start to look at you amplifier,
but i dont understand this ( re + (re1||re2) ) = (rc|| rl) / 70 = 909Ω/50 = 18Ω

where did you take 70? 909?50? can you explain me please....
P: 410
 Quote by michael1978 jony i start to look at you amplifier, but i dont understand this ( re + (re1||re2) ) = (rc|| rl) / 70 = 909Ω/50 = 18Ω where did you take 70? 909?50? can you explain me please....
There is a error instead-of 70 it should be 50. The voltage gain we want is equal to 50[V/V].
OE amplifier voltage gain is equal to
Av = Rc/re so to find desired re value I use this equation:

re = Rc/Av

And in my example the gain I want is equal to 50[V/V] and Rc = 1K; RL = 10K
So form this

Rc||RL = 909Ω

re = Rc||RL/Av = (1K||10K)/50 = 909Ω/50 = 18Ω

It is clear now ?
P: 133
 Quote by Jony130 There is a error instead-of 70 it should be 50. The voltage gain we want is equal to 50[V/V]. OE amplifier voltage gain is equal to Av = Rc/re so to find desired re value I use this equation: re = Rc/Av And in my example the gain I want is equal to 50[V/V] and Rc = 1K; RL = 10K So form this Rc||RL = 909Ω re = Rc||RL/Av = (1K||10K)/50 = 909Ω/50 = 18Ω It is clear now ?

yes thank you
P: 133
 Quote by Averagesupernova Collector resistor is Zout. Zin is R1||R2||(beta*Re). In the schematic that Jony posted Re becomes Re1||Re2.
is so important to know resitance zin and zout, i am a begginer, how do you select zin zout is easy zout=RC but zin? you select, but zin how do you select
P: 133
 Quote by michael1978 yes thank you
oo joney sorry i have to ask you this
is this correct re2 = ( 220 * 11.5 ) / (220 - 12.5) = 12.1 = 12Ω.
of this one re2 = ( 220 * 11.5 ) / (220 - 11.5) = 12.1 = 12Ω.
 P: 2,537 Michael, are you asking why Zin and Zout are important to know?
P: 410
 Quote by michael1978 of this one re2 = ( 220 * 11.5 ) / (220 - 11.5) = 12.1 = 12Ω.
This one is correct of course.
P: 133
 Quote by Jony130 This one is correct of course.
jony can i ask you something, is possible for you to make one another amplifier but without Re2, with gain av 50, i know you maket very good the one, i understand all, but how to make without re2, you can select self the value but desired voltage to be 50, can you do it the same as in first example , if you want thank you PLEASE
 P: 410 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]
P: 133
 Quote by jony130 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]
thank you very much
 P: 410 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 Attached Thumbnails
P: 133
 Quote by Jony130 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]
AND value of r1 and r2 is r1=22k, and r2=4.7K and power supply 10V
P: 410
 Quote by michael1978 AND value of r1 and r2 is r1=22k, and r2=4.7K and power supply 10V
Yes for Vcc = 10V and hfe = 150
P: 133
 Quote by Jony130 Yes for Vcc = 10V and hfe = 150
THIS is the same example without re2, but you add in serie re2
i mean this laste circuit, 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 this is correct R1 = (Vcc - Vb)/( 11*Ib) = 22KΩ

of this R1 = (Vcc - Vb)/( 10*Ib) = 22KΩ

---
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]

thnx
 P: 410 I don't understand the question?

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