# electronic amplifier design (2nd attempt)

#### rjnachshin

Joined Nov 10, 2015
12
Hi for my electronics class we were asked to design a circuit that meets the following requirements.
Zi greater than 47k
Zo has a max of 4.7k
the overall no load voltage gain is over 1000
the output voltage capacity is at least a 5v p-p with no visible distortion at 5v p-p input
Vcc = 15v

I designed the following circuit but I'm still having trouble picking the values for the capacitors and If my design meets the qualifications. I'm attaching the circuit. Im using a beta of 150 and a ro of 50k. Any feed back would be very helpful!!! If you have a better idea for how to approach the problem also please let me know.

Thank You

#### Attachments

• 69.1 KB Views: 20
Last edited:

#### Alec_t

Joined Sep 17, 2013
14,237
the output voltage capacity is at least a 5v p-p with no visible distortion at 5v p-p input
That implies a voltage gain of at least 1, not "over 1000". Are you missing an "m" somewhere?

#### marcf

Joined Dec 29, 2014
288
Yes, think you meant 5mVac for the input.

In order for 4.1k (not a standard value try 4.3k) to drop 7.5v in order to put the amp into its most linear operation, you need to have 1.8mA of collector current. With a current gain of 150 you need at least 1.8ma/150 (120uA) of base current. There is no way you are going to obtain this with a 9.48Meg (not a standard value) resistor. (15v/9.48M = 1.5uA) At 1.8ma, the 330 ohm resistor would drop a little over 0.6v. The input impedance of the amp is about 50K. (150*330ohms) in parallel with 900K (not a standard value).

This is pretty close to your goal of 47k.

#### JoeJester

Joined Apr 26, 2005
4,390
Pick a series, like E24 for your design.

#### Attachments

• 28.6 KB Views: 5

#### rjnachshin

Joined Nov 10, 2015
12
Yes, think you meant 5mVac for the input.

In order for 4.1k (not a standard value try 4.3k) to drop 7.5v in order to put the amp into its most linear operation, you need to have 1.8mA of collector current. With a current gain of 150 you need at least 1.8ma/150 (120uA) of base current. There is no way you are going to obtain this with a 9.48Meg (not a standard value) resistor. (15v/9.48M = 1.5uA) At 1.8ma, the 330 ohm resistor would drop a little over 0.6v. The input impedance of the amp is about 50K. (150*330ohms) in parallel with 900K (not a standard value).

This is pretty close to your goal of 47k.

#### rjnachshin

Joined Nov 10, 2015
12
I have made my third design and his one has met the required goals!...well i think it did from my simulation. I need to find the parts electrical values, and the secondary rating like like power rating for resistors and voltage rating for capacitors. Also i need reference designators which I'm not sure what that means also. But I'm attaching my new design and the output waveform

#### Attachments

• 31.6 KB Views: 11
• 13.9 KB Views: 12

#### Bernard

Joined Aug 7, 2008
5,784
It will be hard to spec. capacitor values with unknown frequency
For 100 Hz, might aim for 3 uF for input, which is odd value so best choice 4.7 uF to 5 uF. @ 25 V or better.

#### hobbyist

Joined Aug 10, 2008
892
Could you please show us the method you used to determine the values of your bias resistors.

#### JoeJester

Joined Apr 26, 2005
4,390
Output 5v peak to peak and a gain of 1000 means the input is 5V peak to peak / 1000 or 5 mV peak to peak.

Your diagram has a signal generator at 2.5 mV rms or 7.07 mV peak to peak.