Help with a common Emitter Amplifier

Thread Starter

janner

Joined Nov 15, 2009
6
Hi Guys,

This is my first post on here so first of all "Hello!"
Right down to business. I have to design a two stage common emitter amplifier with a gain of 30, I have been given loads of notes but Im needing someone to point me in the right direction with it. Firstly im trying to just design a single stage amplifier get it working then basically just join two together. I know that then I can get the gain by just using 1/beta in the feedback.

I have been told that I can choose the supply and then choose the Ic. I used a supply of 15V and then an Ic of 4mA. The formulas I have been given to get the rest of the circuit are as follows
1, Choose Ic I used: 4mA
2, Make Ve 10% of Vcc I calculated 1.5V
3, Re = Ve/Ic I calculated 375 ohms
4, Rc = Vcc/2Ic I calculated 1.875K ohms
5, R2 = 10 x Re I calculated 3.75K ohms
6, R1 = (Vcc-(Ve+0.7))x R2 I calculated 21.8K ohms
Ve+0.7

Ok so I have done all my calculations and then built the circuit on multisim. The circuit inverts the output but does not ampilfy it? :confused: The output voltage is extately the same magnitude as the input :confused:
I have attached a copy of my circuit with the wave forms on it but if any other info is needed please just leave a message.

This does seem a monster first post but any help anyone could give me would be much appreciated. Thanks :)
 

hobbyist

Joined Aug 10, 2008
892
Hi Guys,

6, R1 = (Vcc-(Ve+0.7))x R2 I calculated 21.8K ohms
Ve+0.7
I must say this calculation is a new one to me.

But if it works that's good. (the equation that is)

VB / R2 = ID

and

(VCC - VB) / ID = R1

So some how I'll need to rearange these equations to see if I can get that #6, equation.

Be an interesting algebra problem.
 

Thread Starter

janner

Joined Nov 15, 2009
6
Aye the equation is divided by (Ve+0.7) it just came out abit wrong when typing sorry.

I got it to work I just removed the RL resistor I had in the circuit. so thanks anyways guys.
 

hobbyist

Joined Aug 10, 2008
892
(VE + 0.7v.) = VB

so (VB / R2) = divider Curent.

The R1 value is {(VCC - VB) / divider current.}

Are you sure you got your R1 value right?

Maybe I need to rearange and substitute the equations to get what they gave you for R1 equation.
 

Thread Starter

janner

Joined Nov 15, 2009
6
no sorry the calculation for R1 is

R1=

(Vcc-(Ve+0.7)) xR2
Ve+0.7

so I got =

(15-(1.5+0.7)) x 3750 = 21.8KΩ
1.5+0.7

Like i said I have the circuit amplifiying with a gain of 5 but I want to make it much bigger so I need to add in a Capacitor (Ce) so Im off to do more reading on how to do this beacuse as soon as I put one in the circuit my output goes very wrong!
 

Audioguru

Joined Dec 20, 2007
11,248
The gain of the transistor without a load is slightly less than Rc/Re.
Increase the gain by increasing Rc or reducing Re then re-bias it.

You can also increase the gain without re-biasing it by connecting a resistor in series with a capacitor from the emitter to ground. This reduces the effective value of Re at AC.
 
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