Please help transistor amplifier

[NascentOxygen]

why don't become plus + in place from minus - like this re1||re2 (re1*re2)/(re1+re2)

This expression has re2 on both sides. We need to extract it to one side, and if you do the algebra you will see how the - sign comes about.

re2 = ( 220 * 11.5 ) / (220 - 11.5) = 12.13 ⋍ 12

 

[michael1978]

In my post, the -12.5 is crossed out. It's replaced by -11.5.

yes i know but i ask somthing else
look this
re2 = ( 220 * 11.5 ) / (220 - 11.5) = 12.13 ⋍ 12Ω.
why don't become plus + in place from minus - like this re1||re2 (re1*re2)/(re1+re2)
re2 = ( 220 * 11.5 ) / (220 - 11.5) = 12.13 ⋍ 12
, of this is other formula
can you answer me

 

[michael1978]

This expression has re2 on both sides. We need to extract it to one side, and if you do the algebra you will see how the - sign comes about.

re2 = ( 220 * 11.5 ) / (220 - 11.5) = 12.13 ⋍ 12

yes you right i am learning a little bit math , from the book basic math for electronics

is not the same like this equation

re1||re2 (re1*re2)/(re1+re2)

 

[michael1978]

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 + (Re1||Re2) )

( re + (Re1||Re2) ) = (Rc|| RL) / 70 = 909Ω/50 = 18Ω

re = 26mV/Ic = 26mV/4mA = 6.5Ω

18Ω = (re + (Re1||Re2)) = ( 6.5Ω + (220||Re2) )

Re1||Re2 = 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.

JONEY why i get 0.200 Vac?,

 

[Jony130]

JONEY why i get 0.200 Vac?,

I don't know why. Maybe you made a mistake in simulation program.

 

[michael1978]

I don't know why. Maybe you made a mistake in simulation program.

yes Jony, i do it one more time and is now 0,68 Vac now smaller, of i have to change the emitter capacitor , because the value of capacitor i make 1000u, of transistor , which transistor you think to use do you know some type?
because input is 2mv*50gain=100MV
thnx for reply

 

[NascentOxygen]

The BC109C is a popular high gain general purpose transistor.

 

[michael1978]

The BC109C is a popular high gain general purpose transistor.

i will change now

i don't have that type of transistor

 

[michael1978]

look my circuit, i don't know where i make mistake, normal gain have to be 50*2Mv=100mV Peak

 

[NascentOxygen]

It's not a very good design. For bias point stability (against V_B and ß variations) you should allow a couple of volts across the emitter resistor; you have just 0.4V.

For maximum symmetrical output swing, you should allow approx as much voltage across R_C as V_CE. But you have ~1.8V and 7.8V, resp.

At what frequency are you measuring the ac gain?

 

[Jony130]

No the design is good. Simply the BJT that michael1978 use in his simulation has a very low current gain.
Ie = 0.4V/220Ω = 1.8mA

And Ib = IR4 - IR5

IR4 = 1V/30K = 33.4μA

IR5 = (10V - 1V)/150kΩ = 9V/150KΩ = 60μA

So Ib = 60μA - 33.4μA = 26.6μA

And Hfe = β = (Ie/Ib - 1) = 66.6

And this circuit was design for Hfe > 420

 

[michael1978]

It's not a very good design. For bias point stability (against V_B and ß variations) you should allow a couple of volts across the emitter resistor; you have just 0.4V.

For maximum symmetrical output swing, you should allow approx as much voltage across R_C as V_CE. But you have ~1.8V and 7.8V, resp.

At what frequency are you measuring the ac gain?

at 20kh
thnx for reply

 

[michael1978]

No the design is good. Simply the BJT that michael1978 use in his simulation has a very low current gain.
Ie = 0.4V/220Ω = 1.8mA

And Ib = IR4 - IR5

IR4 = 1V/30K = 33.4μA

IR5 = (10V - 1V)/150kΩ = 9V/150KΩ = 60μA

So Ib = 60μA - 33.4μA = 26.6μA

And Hfe = β = (Ie/Ib - 1) = 66.6

And this circuit was design for Hfe > 420

so result is correct 66.6, for ib like this is formula for current Ir5-Ir4? is not the same for ir4 and ir5 they are in serie ir4 and ir5
but Jony you told Ve to be 1v why is now Ve 0.4V?
thnx for reply

 

[michael1978]

so result is correct 66.6, or voltage gain is 66.6, if so 66.6*2Mv=133mv again the result is not corrcect
for ib like this is formula for current Ir5-Ir4? is not the same for ir4 and ir5 they are in serie ir4 and ir5
but Jony you told Ve to be 1v why is now Ve 0.4V?
thnx for reply

thnx for reply
]so result is correct 66.6, or voltage gain is 66.6, if so 66.6*2Mv=133mv again the result is not corrcect
for ib like this is formula for current Ir5-Ir4? is not the same for ir4 and ir5 they are in serie ir4 and ir5
but Jony you told Ve to be 1v why is now Ve 0.4V?
thnx for reply[/QUOTE]
thnx for reply

 

[michael1978]

thnx for reply
]so result is correct 66.6, or voltage gain is 66.6, if so 66.6*2Mv=133mv again the result is not corrcect
for ib like this is formula for current Ir5-Ir4? is not the same for ir4 and ir5 they are in serie ir4 and ir5
but Jony you told Ve to be 1v why is now Ve 0.4V?
thnx for reply

thnx for reply[/QUOTE]

hi Jony, can you answer me please

 

[Jony130]

thnx for reply
]so result is correct 66.6, or voltage gain is 66.6, if so 66.6*2Mv=133mv again the result is not corrcect
for ib like this is formula for current Ir5-Ir4? is not the same for ir4 and ir5 they are in serie ir4 and ir5
but Jony you told Ve to be 1v why is now Ve 0.4V?
thnx for reply

66 is a BJT current gain, not a amplifier voltage gain.
So in order to make this circuit work properly. You need change the BJT that you use in your simulation program. You need to find the BJT with Hfe > 300 in your simulation.

 

[michael1978]

66 is a BJT current gain, not a amplifier voltage gain.
So in order to make this circuit work properly. You need change the BJT that you use in your simulation program. You need to find the BJT with Hfe > 300 in your simulation.

ah 66 is current gain, so you mean to change type of transistor, but is possible for you to tell me any type of transistor which work, because there are a lot of type of transistor, and somebody he tell me one type of transistor but that transistor don't exist in my list of transistors
and yes Jony if you can tell me, i try to find self soms but i can't find it, that is the problem
thnx for reply

 

[Jony130]

BC548C
What simulation program do yo use ?

 

[michael1978]

BC548C
What simulation program do yo use ?

o joney i don't have, i have bc548A but not BC548C, i use B2 Spice A/V

 

[michael1978]

hey Jony i try with bc548A, and i get voltage gain of 84mv is this correct?

 

[Jony130]

Use BC547C

 

[michael1978]

Use BC547C

yes Joney i have that type i change, and now voltage gain is 94mv is correct now?

 

[Jony130]

Well first you need to learn how to use the simulation program.
I use B2 Spice V5 and as a BJT use BC547C. And AC sweep analysis show that voltage gain is equal to 50.1V/V
For 20mV at input I get 1V at output.

 

[michael1978]

Well first you need to learn how to use the simulation program.
I use B2 Spice V5 and as a BJT use BC547C. And AC sweep analysis show that voltage gain is equal to 50.1V/V
For 20mV at input I get 1V at output.

but this was disegn for 10v battery not 12v

 

[Jony130]

Well for Vcc = 10V the voltage gain is equal to 46V/V (0.9V at output for 20mV at input) because we don't included RL in our calculations when we solve for Re2.

 

[michael1978]

Well for Vcc = 10V the voltage gain is equal to 46V/V (0.9V at output for 20mV at input) because we don't included RL in our calculations when we solve for Re2.

my circuit look like this

 

[Jony130]

If so, your voltage gain is equal to 46V/V

 

[michael1978]

my circuit look like this , the voltage gain is correct almost the same like yours, but metter gain is not correct, WHY GAIN METTER NOT CORRECT

 

[michael1978]

If so, your voltage gain is equal to 46V/V

yes almost 2Mv*46=92MV my 94MV

 

[Jony130]

So everything is OK

 

[michael1978]

So everything is OK

yes now, but can you tell me jony how to find in transistore HFE
because when you design this circuit

first you say
First we need select BJT I choose BC546C with typical hfe = 520 and Hfe_min = 420
and after
Ib = Ic/Hfe_min = 4mA/420 ≈ 10μA (base current)

how me to find me Hfe, you know which software i use
thnx for help

 

[michael1978]

happy christmas to everybody,

jony can you help me for Hfe, i have to look in datasheet of i can look also in B2 spice?

 

[Jony130]

You can use your simulation program to find Hfe or datasheet it's up to you.

 

[michael1978]

You can use your simulation program to find Hfe or datasheet it's up to you.

so i have self to look in datasheet for transistor, because my simulator he tell me maximum of hfe but not a minimum of hfe

thnx for reply

 

[NascentOxygen]

You should run your simulation a number of times, testing that it works satisfactorily with hfe_max and hfe_min, and (for multi-transistor circuits) all combinations.