please help transistor amplifier

michael1978

AND value of r1 and r2 is r1=22k, and r2=4.7K and power supply 10V

Jony130

Yes for Vcc = 10V and hfe = 150

michael1978

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


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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

Jony130

I don't understand the question?

michael1978

you show me first this example
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---
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

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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

michael1978

yes i ask you, i want to know

Jony130

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

michael1978

i ask you for gain av 50 without re, and you show me example
---
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

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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

Jony130

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 = Rc||RL/(re + Re1)

michael1978

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

Jony130

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)

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michael1978

this is a bit difficult to me, and how much is ib to your example, can you explain in similary way because i find difficult, sorry joney i try to understand but is difficult, ah value of ib is 1 is correct

michael1978

joney please can you explain this, i know you explain to me but i dont understond, this 11 and the value of ib HOW YOU CALCULATE 11AND IB PLEASE joney you help a lot to design amplifier. i am simple an beginner

------------------
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

michael1978

??????????????????????????????

Jony130

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

michael1978

joney i know i think

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....

Jony130

Simply you forget about base current which loads the voltage divider in your calculations.



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

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