please help transistor amplifier

Averagesupernova

You went about it backwards. R1 and R2 are of course significant in order to get the proper base voltage but they are also important to determine input impedance. So their values need to be kept in mind for this. A low output impedance will reflect way back to a lower input impedance. The amplifier Jony posted has a Zin of about 1000 ohms. Can you tell why a single stage amplifier with the gain and Zout that you want cannot have a Zin of 50K?

michael1978

ooo man i dont know how to deterimine zin and zout ;-) how do you select zin and zout, i am just a begginer, is zin R1 and R2 and zout rc and rl, if is yes, how you determine how you select calculate

Averagesupernova

Collector resistor is Zout.
Zin is R1||R2||(beta*Re). In the schematic that Jony posted Re becomes Re1||Re2.

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

michael1978

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

michael1978

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
thnx for answer

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....

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 ?

michael1978

yes thank you

michael1978

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

michael1978

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.

Averagesupernova

Michael, are you asking why Zin and Zout are important to know?

Jony130

This one is correct of course.

michael1978

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

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]

michael1978

thank you very much

Jony130

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

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