# single stage voltage divider amplifier

#### simplynew

Joined Nov 5, 2010
6
i am a first year student in electrical and computer engineering and my project is to design and build a single stage voltage divider amplifier with the following design specs:
1. voltage gain = 50
2. lower cut off frequency must be less than 100Hz
3. it must have maximum symmetrical swing
4. supply voltage = 15V
5. small signal transistor (2N3904/2N2222) is to be used

and i cannot get my values of Icq, etc by using the rule-of-thumbs

Last edited:

#### simplynew

Joined Nov 5, 2010
6
i can determine Icq by rule-of-thumbs, but without that i am not sure how to choose my value of Icq

#### simplynew

Joined Nov 5, 2010
6
i was thinking to get my Icq value from a datasheet of hfe against Ic and just choose a midway value for hfe for the specified transistor ( i am using the 2N3904).
is this a good way to go about choosing a value for Icq?

#### hobbyist

Joined Aug 10, 2008
890
You could use the data sheet,

looking at the parameters your given, you can choose whatever current you want within transistor parameters, so if you combine the two together, you can choose a collector current that would be in the range where hfe is at it's most nominal value, then from there begin designing the amp stage.

According to the data sheet for a 2N3904, 10mA. is given as a nominal value.

#### simplynew

Joined Nov 5, 2010
6
Ok so here is one of my attempts:

Icq = 10mA
Vcc = 15V
A (Voltgae Gain) = 50
Vbe = 0.6 V (Im not sure how to choose my Vbe because it is in a range)
Vc = Vcc/2 = 7.5 V (because then you get maximum symmetrical swing about Vc)

After all this is determined, I proceeded to calculate Rc and Re by Kirchhoffs Voltage Law:
Vcc = IcRc + Vce + IeRe
Vcc = Vce + Ic(Rc + Re) {Assuming that IC ~ IE }
Now I get
Ic = (Vcc-Vce)/(Rc+RE) (1)
Icq = Vcc/((2*A*Re)) {where A = Volatage gain} (2)

→ From equation (2) :
10mA = 15/(2*50*Re)
Therefore , Re = 15 Ω

Now,
Volatge Gain, A = - Rc/Re (3)
So, 50 = Rc/15
Rc = 750Ω

Ve = Veq = Icq * Re
= 10 mA * 15Ω
= 0.15V

In general Vbe = 0.60V
Therefore, Vbe = 0.60V + 0.15 V
Vbe = 0.75 V

I am still not sure how i am supposed to go about choosing a value for Vbe, do i again look at the data sheet?

Sp from here i have to now calculate R1 and R2 but this is where i am stuck, i am not sure how to do this without using the rule-of-thumb.

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#### Audioguru

Joined Dec 20, 2007
11,249

#### hgmjr

Joined Jan 28, 2005
9,029
Greetings Simplynew,

Don't be put off by AG's comment. It is always a good idea to get as many suggestions as you can when you have a question.

hgmjr

#### simplynew

Joined Nov 5, 2010
6
Another attempt:
Designing a CE Voltage divider circuit using Vcc = 15V
Choose Icq = 1mA for good current gain and frequency response and without causing undue heating.
Vbe = 0.7Volts.

For good quiescent current stability, Vre = 1/10 Vcc = 1.5 volts.
Hence, Re = 1.5/1mA = 1.5kΩ

For maximum symmetrical swing, Rc = (15-1.5)/(1.5*1mA) = 9kΩ

Now, Vbq = Vre +Vbe = 2.2Volts

Current through R1 and R2, Ir = 1/10 Icq = 0.1 mA.
This gives R2 = 2.2Volts/0.1mA=22kΩ
And R1 = (15-2.2)/0.1mA=128kΩ

I can design my circuit this way, but i am not sure why Vre has to be 1/10 Vcc
And why Ir has to be 1/10 Icq
My teacher said this way is using the rule-of-thumb and I need to calculate my resistor values in a different way.

#### hgmjr

Joined Jan 28, 2005
9,029
Can you sketch your circuit and then take a picture of it and then attach it to a post here.

hgmjr

#### simplynew

Joined Nov 5, 2010
6
i'm guessing that i will have to use a bypass capacitor at the emitter because if i do a AC analysis my gain calculation is too high so i will need to split Re into two resistors in series and bypass one of them.

ignore my values for capacitors in my design as i have not yet done an AC analysis, i'm just doing DC analysis right now to obtain my resistor values.

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#### hgmjr

Joined Jan 28, 2005
9,029
i'm guessing that i will have to use a bypass capacitor at the emitter because if i do a AC analysis my gain calculation is too high so i will need to split Re into two resistors in series and bypass one of them.

ignore my values for capacitors in my design as i have not yet done an AC analysis, i'm just doing DC analysis right now to obtain my resistor values.
On first inspection, your Vc is centered fairly well between ground and the 15V power supply. However, your voltage gain is 6 which is well below your target of 50.

Re is going to need to be reduced and Rc increased. In concert with these changes you will need to resize your base bias resistors.

hgmjr

#### shteii01

Joined Feb 19, 2010
4,644
You do not get to choose Vbe. Vbe depends on the transistor, you do not need to choose it. Generally Vbe is anywhere from .6 to .75-.8 volts, it should be in the data sheet somewhere.

#### Audioguru

Joined Dec 20, 2007
11,249
It will be wrong to decrease Re and increase Rc to increase the voltage gain because then the Iq of the transistor will change a lot when the temperature changes and when different transistors (even with the same part number) are used. You are correct to split Re into a bypassed resistor in series with an unbypassed resistor, then the AC gain will be 50 and the DC gain will be 6 for good stability.

But the voltage gain of a transistor also depends on the internal emitter resistance of a transistor that depends on emitter current. Didn't your teacher teach this simple formula or isn't it in your text book?

The datasheet has a detailed graph of the Vbe for a "typical" transistor. Use it to calculate the base bias resistors.

#### hgmjr

Joined Jan 28, 2005
9,029
i am a first year student in electrical and computer engineering and my project is to design and build a single stage voltage divider amplifier with the following design specs:
1. voltage gain = 50
2. lower cut off frequency must be less than 100Hz
3. it must have maximum symmetrical swing
4. supply voltage = 15V
5. small signal transistor (2N3904/2N2222) is to be used

and i cannot get my values of Icq, etc by using the rule-of-thumbs
Were you provided with an output impedance target?

What is the upper end of the frequency range your amplifier should be designed to pass?

hgmjr

#### hobbyist

Joined Aug 10, 2008
890
Ok so here is one of my attempts:

Icq = 10mA
Vcc = 15V
A (Voltgae Gain) = 50
Vbe = 0.6 V (Im not sure how to choose my Vbe because it is in a range)
Vc = Vcc/2 = 7.5 V (because then you get maximum symmetrical swing about Vc)

After all this is determined, I proceeded to calculate Rc and Re by Kirchhoffs Voltage Law:
Vcc = IcRc + Vce + IeRe
Vcc = Vce + Ic(Rc + Re) {Assuming that IC ~ IE }
Now I get
Ic = (Vcc-Vce)/(Rc+RE) (1)
Icq = Vcc/((2*A*Re)) {where A = Volatage gain} (2)

→ From equation (2) :
10mA = 15/(2*50*Re)
Therefore , Re = 15 Ω

Now,
Volatge Gain, A = - Rc/Re (3)
So, 50 = Rc/15
Rc = 750Ω

Ve = Veq = Icq * Re
= 10 mA * 15Ω
= 0.15V

In general Vbe = 0.60V
Therefore, Vbe = 0.60V + 0.15 V
Vbe = 0.75 V

I am still not sure how i am supposed to go about choosing a value for Vbe, do i again look at the data sheet?

Sp from here i have to now calculate R1 and R2 but this is where i am stuck, i am not sure how to do this without using the rule-of-thumb.

Getting back to your first question.

Sp from here i have to now calculate R1 and R2 but this is where i am stuck, i am not sure how to do this without using the rule-of-thumb.

These are not rules of thumb, as much as calculations, in determining these values.

To solve for R1 and R2, follow these steps carefully.

1). Assume 1/2 of VCC is across RC. therefore [IC = Vcc / 2*RC]
2). IE ~= IC so [VRE = (IC * RE)]
3). Vbe ~= 0.7v. and [VB = (VRE + Vbe)]

Now this next step can be approached in several different ways.
For the base to ground resistor. (R2, in your schematic)

4a). R2 = 10 to 20 times RE.
5a). ID = divider current so [ID = (VB / R2)]
6a). R1 = [ (Vcc - VB) / ID]

Another approach for R2.

4b). IBq = (ICq / HFE) DC values.
5b). ID = (10 * IBq)
6b). R2 = (VB / ID)
7b). R1 = [(Vcc - VB) / ID]

another approach for R2 would be to make it a value that would meet input impedance requirements.
But that gets much more involved.

#### Suraj91

Joined Nov 20, 2010
2
we are in the same class and we have the same problem. who are u? lol

#### Suraj91

Joined Nov 20, 2010
2
Getting back to your first question.

Sp from here i have to now calculate R1 and R2 but this is where i am stuck, i am not sure how to do this without using the rule-of-thumb.

These are not rules of thumb, as much as calculations, in determining these values.

To solve for R1 and R2, follow these steps carefully.

1). Assume 1/2 of VCC is across RC. therefore [IC = Vcc / 2*RC]
2). IE ~= IC so [VRE = (IC * RE)]
3). Vbe ~= 0.7v. and [VB = (VRE + Vbe)]

Now this next step can be approached in several different ways.
For the base to ground resistor. (R2, in your schematic)

4a). R2 = 10 to 20 times RE.
5a). ID = divider current so [ID = (VB / R2)]
6a). R1 = [ (Vcc - VB) / ID]

Another approach for R2.

4b). IBq = (ICq / HFE) DC values.
5b). ID = (10 * IBq)
6b). R2 = (VB / ID)
7b). R1 = [(Vcc - VB) / ID]

another approach for R2 would be to make it a value that would meet input impedance requirements.
But that gets much more involved.
what is the proof for (the apparent rule of thumb" we will need the proof in order for the circuit to be accepted, even if it is very tedious. an explanation of why we used 1/10vcc will not be accepted. we need hard proof. which we do not have. the same goes for 1/10 ic