Help!! - Circuit Design: Voltage divider amplifier

[simplynew]

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


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


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?

 

[ericgibbs]

hi,
Look thru the Class A amplifier designs on this link.
class a amplifier design - Google Search

 

[simplynew]

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.

i uploaded my design layout.

 

[ericgibbs]

hi,
An Ic of 10mA is too high for a small signal class A amplifier, I would consider about 2mA.

1. Work out the transistor base input current required when the Gain of the transistor is 50.
2. To ensure that the Vbe remains steady, have about 10 times the value of the base current flowing the base resistor divider.
3. By having a lower Ic value of say 2mA, the base divider current will be lower, so the base resistors will have a higher value, so the input resistance can be higher.
4. The value of the input impedance is also affected by the Re resistor value.

If you can keep the input impedance to the base higher this means that you can use a lower value coupling capacitor at the input and still meet the 100Hz target.

Also when calculating the voltage swing at the collector, allow for the losses due the voltage across Re and the transistor.

Relook at your design.

 

[audioguru]

Your circuit will be a DC thermometer because the Vbe changes with temperature changes. Add a bypassed emitter resistor (maybe 68 ohms) in series with the 15 ohms emitter resistor so that the base voltage can be higher and will be affected less by temperature changes.

The base bias resistors should have a current that is about 10 times the amount of base current which is shown on a graph of "typical" current gain on the datasheet. Then a low gain transistor will operate almost the same as a high gain transistor.

The transistor has an internal emitter resistance so the emitter resistor value you use must be less than you have. Ask your teacher or look in your text book for the formula of the internal emitter resistance that depends on the emitter current.

 

[simplynew]

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.

 

[audioguru]

The Vre should be 1/10th Vcc because some transistors have a Vbe of 0.6V and others have 0.7V. A transistor that has a Vbe of 0.65V at room temperature will be 0.6V or less when hot and will be 0.7V or more when cold. The voltage across the emitter resistor reduces the effect of the different voltages of Vbe.

With an Rc of 9k and an Re of 1.5k then the voltage gain with no load is less than (9/1.5=) 6, not 50. Do you know the complete formula for the gain and how to increase the gain?