Designing an amplifier using discrete components

Thread Starter

Muhammad Humayun Iqbal

Joined Nov 11, 2016
4
Hello people,

I am asked to design a multistage amplifier which should have a gain of 100. i.e if i provide 10mV, it should amplify it to 1V.
And i am supposed to add 8ohm Speaker as a Load Resistor.
I am bound to use simple transistors, resistors and capacitors. The instructor only mentioned that i will have to use 2 common emmitter stages following a darlington pair to obtain such amount of amplification.
I don't have any idea about what values should i take for resistors and the capacitors separating multi stages.
Kindly anyone guide me with the calculations required to find out the values.

Thank You.
 

WBahn

Joined Mar 31, 2012
29,979
An assignment like this is to evaluate YOUR ability to apply what you have learned over the span of the course. If you truly have no idea where to even start with this assignment, perhaps you need to repeat the course.

Or, you could make an attempt at working your assignment and we can provide some guidance based on YOUR efforts, but we aren't going to tell you want to do step by step.
 

Thread Starter

Muhammad Humayun Iqbal

Joined Nov 11, 2016
4
An assignment like this is to evaluate YOUR ability to apply what you have learned over the span of the course. If you truly have no idea where to even start with this assignment, perhaps you need to repeat the course.

Or, you could make an attempt at working your assignment and we can provide some guidance based on YOUR efforts, but we aren't going to tell you want to do step by step.
well, i am afraid. I might have to repeat this course. :(

You go this amplifier configuration. Two common emitter. In the second stage Darlington. The scheme is covered by the negative feedback. This feedback stabilizes the gain and reduces distortion.
I had it in my mind already. Pretty much did the same. I am just having trouble in setting the resistors' value.

Is C1 connected to R4/R6? Your pic seems to show the wire from C1 jumping directly to Q1 base.
Yes, C1 is connected to R4/R6.
 

wayneh

Joined Sep 9, 2010
17,496
I'm no expert but the emitter resistor in stage 1 is just 10Ω and there is no base resistor on Q1. This means the input impedance is very low and might overwhelm the source?

I also wonder why C3 is smaller than C6 and C2. Seems to me it should be the other way around.
 

WBahn

Joined Mar 31, 2012
29,979
well, i am afraid. I might have to repeat this course. :(
That won't be the end of the world. Look at it as an opportunity to really learn this stuff. You've been through it once so you have an idea of what you need to learn and what you didn't learn. That makes it a lot easier to really focus on learning the stuff that is important this time around.
 

Ylli

Joined Nov 13, 2015
1,086
Let's start with the first stage. Currently, you have the base of Q1 biased via R4 & R6. This voltage divider put about 12 * (50/80) = 7.5 volts on the base. Using a nominal 0.7 volts base - emitter drop, that puts 6.8 volts on the emitter. With that 10 ohm emitter resistor, you have biased this stage for a current of 680 mA.

Do you really want that much (no you don't). With this much collector current, and a 3K collector load resistor, the transistor will be fully saturated over all practical input levels.

That 3k in the collector is probably OK for now. Rethink the emitter resistor and the base biasing resistors and then perhaps we can move on to the second stage.
 

Bordodynov

Joined May 20, 2015
3,177
Hi.
Your circuit is not suitable (with nominal values of resistors). This scheme does not provide the necessary output amplitude. I will give you advice on the processing of your circuit.
1. The last stage (Darlington), turn on the circuit with common collector (emitter follower).
2. The load switch into the emitter.
3. Set the input voltage divider to load at least 2 volts.
 

absf

Joined Dec 29, 2010
1,968
Hello @Mohammad,

if you reread post #2, #9 and #12 especially #12 again carefully. You'll find that the hints there is really useful for your design. ;)

Allen
 
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