HELP Single Stage BJT Common Emitter Amplifier resistor calculation

[NiCeBoY]

Hello.. can anyone help me out to calculate the value of resistors R1, R2, Rc and RE please?

The values should be such as the out gain is between 10dB to 20dB

The transistor used in this circuit is bc547 and the load resistor is a fixed one of 10k.
Below is the circuit:
**broken link removed**

Thanks

 

[Hero999]

Do your own homework.

 

[silvarblade]

like hero said you get a reply by making some effort on your own!

 

[Willbe]

gain is between 10dB to 20dB

The voltage or power or current gain?

I will say that:

the circuit output resistance looking into the collector should be >5x smaller than the load resistor and

the resistance looking into the circuit input should be >5x greater than the impedance of your input coupling capacitor at the freq. of interest and

you should first pick an equivalent circuit for your transistor, something with at least one ideal diode in it and

the Vbe at turn-on for a silicon transistor is ~0.5 v and it is fully on at ~1 v and

for temp insensitivity RE should be >5x the internal Re of the transistor [which value depends on the emitter current according to an inverse relationship containing a 26 in the numerator] and

it would help if you were conversant with Leon Charles Thevenin and

the current through R1 and R2 should be >5x than the current into the base and

if you think in terms of KΩ rather than Ω it will probably go easier and

using resistors of 10% or 20% tolerance are probably good enough since your gain spec has quite a wide interval.
What I'm getting at here is that you should think in terms of 1.0, 1.2, 1.5 or even more widely spaced resistance values, rather than 1.0, 1.1, 1.2, 1.5, and

Speaking of your own effort, for extra credit [and to redeem your reputation on this forum], three questions:
.....Are the paragraphs in this post, and these questions, arranged in any particular order?
If so, what is the order? Why this order?
.....Resistor values are evenly spaced on a
a) linear scale?
b) log scale?
.....What's with the 5x stuff?

Why so long-winded? I need practice in tech writing in case I ever get another engineering job. . .

Nice schematic!

 

[NiCeBoY]

Hello,
Its voltage GAIN.
I should use resistors of the E12 series...

which the values can be found here :
https://www.st-andrews.ac.uk/~www_pa/Scots_Guide/info/comp/passive/resistor/e12/e12.html

Thanks man..

 

[Hero999]

For low gains the gain is roughly equal to Rc/Re.

For higher gains the HFE of the transistor becomes a limiting factor and needs to be taken into account.

The formulae for taking the transistor HFE into account can be found on Wikipedia.
https://en.wikipedia.org/wiki/Negative_feedback_amplifier

Rc needs to be lower than RL by an order of a magnitude.

The cut off frequency can be calculated with the following formula and needs to be below the lowest frequency of interest.
[latex]F_C = \frac{1}{2\pi CR}[/latex]

 

[audioguru]

I don't think the school-kid has any understanding of a transistor amplifier stage.
Maybe the teacher is no good.
Maybe the school-kid missed too many classes.

 

[NiCeBoY]

lol no am present on every class man...

the formula FC=1/2(pie)CR
i have never heard of it before...

 

[Nigel Goodwin]

lol no am present on every class man...

the formula FC=1/2(pie)CR
i have never heard of it before...

It's one of the VERY basic formulas related to electronics, you should have done that long before trying to design amplifier stages.

 

[NiCeBoY]

what does FC stands for?

 

[ericgibbs]

what does FC stands for?

The formula for calculating the Impedance of a Capacitor is:-

Xc = 1/ ( 2 * Π * f * C) in Ohms

That expanded is: Impedance of a Cap = 1 ( 2 * pi * frequency * capacitor value in F)

example: if a capacitor of 4µF is connected to a frequency of 50Hz then,

Xc= 1/( 6.28 * 50 * 4^-6) = 796Ω

Do you follow that.?

 

[on1aag]

Hi NiCeBoY,

I've got a question for you:

Did your teacher(?) give you these resistor values ?

on1aag.

 

[NiCeBoY]

Oh Xc Xl Z and all i know lol...
sorry

No i have to calculate the resistor values..


I want to make it compatible with approximate calculation which means

BRE should be greater or equal to 10.R2

The gain should be between 10 - 20 Db.

And Vcc should be 15V

The input should be a small value like 200mV or 500 mV @ 10Khz so that when being amplified its does not exceed the Vcc value or else it won't work...

 

[Willbe]

Everything you need has been posted. The ball is in your court.

Think of the circuit as a box that has to meet input and output specs. This is the most abstract level of this design; it is a top down approach. The very last thing you do with this method is to specify actual component values of those things "inside the box".

You need a max voltage gain, Vout/Vin, of 10^(20/20) so you know how many volts rms out you should be seeing with 500 mV rms in.
You also should know that the DC/AC equivalent circuit for a power supply is an ideal voltage source in series with zero impedance.

So, top-down,
Do you need attenuation or gain?
Do you just need level shifting and no gain at all [if you don't need either use a piece of wire]?
Do you just need current gain?
Do you just need voltage gain?
Do you need power gain ['cause if you don't and the impedance levels of source and load are OK, you could just use a transformer, unless the boss says not to]?

Run some calculations with pencil & paper.
You don't even need a calculator, just figure order-of-magnitude resistor values, like 1, 2, 5, 10, 20 or even 1, 3, 10, 30, all evenly spaced on a log scale. 1, 10, 100 is too coarse for this application.
Do the math in your head, 'cause computers crash more often than your head does.

If it doesn't work on paper first it won't work in the real world.

With 10 µF at 60 Hz you have a 270 Ω source impedance so you know how high, minimum, the base bias network values should be.
E.g., 1.2 k in parallel with 12 k is ~1 k, and this paralleled value should be >5 x 270 Ω, so you could raise this somewhat.

1 k and 10 k in parallel is about ~1 k, and 1 k is about 1/10 th of 270 Ω + 10 k, so maybe the possible maximum output impedance of this circuit is OK [imagine the transistor, C to E, as just a switch, full on and then full off, and what resistance value does the load resistor see when looking back into the circuit when this happens?]

0.5 v/270 Ω = ~ 2 mA rms max input current able to be sourced by the input signal, and
the current into the load resistor should be 0.5 mA rms max,
so you may not need current gain at all, so how are you going to convert the current gain [that you don't need] of a transistor [a current controlled current source] into the voltage gain [that you do need]?

If you can thoroughly understand the operation of this circuit, so that you know it cold backwards and forwards, you'll be in good shape for the problems you will later face.

It's not so much that you figure out these resistor values, it's that you learn how to reason through a circuit like this, that you learn to reason like a good engineer, logically and methodically, knowing when you can round off and when you can't.

It's you and your wits against the laws of physics; may the best side win.

 

[Hero999]

You don't even need a calculator, just figure order-of-magnitude resistor values, like 1, 2, 5, 10, 20

https://en.wikipedia.org/wiki/Preferred_number#1-2-5_series

 

[Willbe]

https://en.wikipedia.org/wiki/Preferred_number#1-2-5_series

Today I have learned something; I've heard of Renard but this article goes way deeper.

 

[NiCeBoY]

Hello,
Finally i calculated these values for R1 & R2 & Rc & RE...

Now the problem i am getting is that my calculated values and measured values is not the same..
Here is my circuit :

**broken link removed**

Using Beta = 110 as on the datasheet :
and also using approximate analysis..
My calculated values are as follows:

Vb=1.597V
Vce= 5.13V
Ic~=IE=0.897mA
Vc=6.03V
Ib=8.08 micro.Amps
Input inpedance = 4791.96 Ohms
Output impedance = 5000 Ohms
Vo=4.44V

Can anyone check my circuit and let me know if its in good working condition?

Thanks in advance...

 

[silvarblade]

well niceboy sorry to say this but you have done some errors in your calculation:

First of all you have calculated the the dc op point vb not the one with the i/p signal.
I/p impedance is correct, s far as the o/p impedance is 5k(you are correct)


you high pass filter has F3db=3.325hz

the simulator which ever you have used is fairly accurate but i would advise you to use ltspice very helpful to understand all the details.

here are some of the values that i cal dc op point not simulated it yet.

vb=1.5969(you are correct)

ve=? for ve taking vbe=.6 then Ve=Vb-Vbe=.996v(simulator is nearly correct)

Ic=(dont know how to get the similiar symbol)Ie=Ve/Re=.996Amps(simulator is nearly correct)

Vc=Vcc-IcRc=5.04v(you have connected Vc incorrectly in your figure)

Gain=Rc/Re=10

so approximating the o/p voltage i think
your o/p voltage is a bit high but you have hugely overcalculated in your Vo calculations.

 

[Hero999]

An output impedance of 5k is too high to drive a 10k load, it'll be fine to drive 100k though, if you want to drive a 10k load the output impedance should be 1k maximum but I'd recommend even lower.

 

[NiCeBoY]

Vc=Vcc-IcRc=5.04v(you have connected Vc incorrectly in your figure)

Where is the fault plz?