Optimizing a BJT Amplifier Design for Specific Parameters

[Bourbon daddy]

The outcome is: design a single stage amplifier with the following parameters:

Rin=1KOhms
Rout=1KOhms
Voltage Gain=15
corner frequency=300Hz
specified Transistor is BC109BP
AC supply, 0.5V
DC supply 20V
supply frequency, not specified, I have assumed 1K

We have not been given guidance on this, so I am muddling through a few blogs, particularly;

http://ampdesigns.tripod.com/engg_stuff.html

So far my results have been far from satisfactory in virtually every sense.

The circuit that I have based my own on can be found on this link:

http://ampdesigns.tripod.com/engg_stuff.html

(I can't seem to upload images of my own circuit, any pointers on this subject matter greatly appreciated).

A few of my calculations:

Icq=Icpeak+Icmin=0.03+0.005=0.035A

Selection of Re

Vre=10% VCC=2V

Re=Vre/Icq=2/0.035=57Ohms

RB is set at 1KOhm

R1 and R2

Vr2=Vbe+Vre=2+0.7=2.7V
Vr1=VCC-Vr2=17.3

R1/R2=Vr1/Vr2=6.407

Rb=1K

Rb=R1 in parallel with R2, keeping a 6.407 ratio between them

R1=8K
R2=1.24K

Rb=1.07KOhms

Xce=Re/10

c=1/2PI*(1k)(5.7)

=28uF

Xcc=Rc + Rload

c=1/2PI*(1k)(2K)

=80nF

Xcb=R1*R2/R1+R2...to note, I have used the frequency of 300Hz in this Calc as I believe . it sets the corner freq.

c=1/2PI*(300)(1K)

=530nF


I don't know what aspect of the design process is incorrect, but the output waveform seems to indicate that the capacitors are incorrectly sized to begin with, I am also not achieving a voltage gain of 15.

If anyone can point me in the right direction I will be very grateful.

Vin=0.6 Vout=3.863

Gain=approx. 6

If there is a way of uploading multisim files I will upload the circuit in its current form.

Here is a word document with some circuit images on.

https://www.physicsforums.com/attachments/68396

 

[rude man]

You need to show a diagram. Attach a permitted file such as .pdf, .doc or .jpg.

If your attachment was a WORD file then just change the name to xxxxx.doc.

 

[Bourbon daddy]

View attachment Physics forum.pdf

Here is an image attached as a pdf, hopefully you are able to open it.

 

[Bourbon daddy]

View attachment amplifier circuit.pdf

First pdf was not very good quality, this is includes some formulae that I have used.

 

[rude man]

What is U3? What is the thing connected between the base and emitter of Q1? The thing shunting R5? And what is th capacitor doing ending in mid-air next to R3?

Pretty hard-to-read schematic ...

 

[Bourbon daddy]

View attachment Physics forum2.pdf

Apologies if it wasnt clear enough, U3 is an ammeter, thing connected between b and e was a voltmeter, same as R5... emitter cap was removed as I had removed it to check output signal waveform.

All peripheral equipment is now removed, so hopefully you will find it clearer.

What I seem to be struggling with is that I keep reading different mathematical laws to calculate the various components, none of them actually seem to work. In the last image, the component values are quite different, as I keep experimenting.

I have just read that G=(R2llRload)/r3

Hence the change to R3, (now set to 33.3Ohms)

I am getting closer to achieving a gain of 15, but the output waveform is heavily effected by the caps, and the corner frequency is someway off 300Hz.

 

[rude man]

Ooh, much better!
Ok now, to summarize:
Rin=1KOhms
Rout=1KOhms
Voltage Gain=15
corner frequency=300Hz : is this amp supposed to amplify frequencies below or above the corner frequency? Or do you want a passband around 300 Hz?

specified Transistor is BC109BP
AC supply, 0.5V: what's this, the input voltage? there is no such thing as an ac power supply for this circuit.

DC supply 20V

Are you limited to a 1-transistor design? If not, look up what a Darlington configuration is.

 

[Bourbon daddy]

According to my limited knowledge and my brief teachings, this is what I am working with;

Rin=1KOhms

For this, my belief is;

Rin=(R1llR4)llR3

I know that considering the values in the attached image my Rin would not equal 1K. I have been experimenting with different values and explanations of what Rin actually is. (Rin=R1llR3)

I seem to have this idea that I should be keeping a ratio of approx. 5.5 between R1 and R4. I can't remember why :/

Rout=1KOhms

Rout=R2


Voltage Gain=15

To set the voltage gain, first I tried;

G=R2/R3=1k/66.6

Then I tried;

G=(R2llRload)/R3=1k/33.3

corner frequency=300Hz : is this amp supposed to amplify frequencies below or above the corner frequency? Or do you want a passband around 300 Hz?

My guess is either I am looking for a passband around 300Hz, either that or at 300Hz, dB should be approx -3dB.

Either way I am interested in frequencies above the 300Hz range.

To calculate this, I think I am interested in Cap in.

My understanding is the reactance of this component is equal to R1llR4, knowing that, we get

c=1/2pi*(300)(R1llR4)

If R1llR4 is equal to 1K

c=530nF

(again, different on the image due to me trying out different things)


specified Transistor is BC109BP
AC supply, 0.5V: what's this, the input voltage? there is no such thing as an ac power supply for this circuit.

I have probably been a bit amateur with my terminology, I know a transistor does not require an external power source.

0.5V is the input AC voltage, the peak of this, I assume is what I am amplifying. This voltage is generated by the function generator XFG1.

DC supply 20V

Are you limited to a 1-transistor design? If not, look up what a Darlington configuration is.

For the purpose of this task I am limited to this configuration, however I do need to analyse the Darlington pair for another task.

Finally, thank you for your time and responses so far, I do appreciate it. I am seeing my tutor at lunchtime today, so hopefully I can discuss my problems with him, but I would rather have better results than what I am already getting.

 

[rude man]

My opinion is that you will not be able to design a 1-transistor circuit of robust design (like being able to unplug the transistor & replace with one of same type) with a gain of 15V/V and an output impedance of 1K ohms. You do need a 33 ohm emitter resistor if you want a gain of 15 with a 1K load as you show. That significantly loads down the input bias pair R1 & R4 and it's beta-sensitive. That screws up your bias point. Quiz: what should the dc voltage at the emitter be? The input impedance is very beta-sensitive also. Transistor circuits should not be beta-sensitive.

Judging by your diagram you're looking for a high-pass amplifier (input frequency > 300 Hz). Your load coupling capacitor should be lower than what it is (why?). Your input capacitor should be large enough to pass > 300 Hz though it could be larger, depending on whether you want a 20 dB/decade or 40 dB/decade rolloff at 300 Hz.

 

[Bourbon daddy]

Ok, first of all, I have basically started again, I have recalculated all component values (prior to going through your points) and what I get is;

a gain of approx. 16... currently I am happy with that, however, the output waveform is heavily distorted in the positive cycle, which suggests to me that the biasing is incorrect.

you are correct in assuming it is a high pass filter and because of his I have actually increased the cap in value, my calculations are as follows;

Cap in xc=R1llR4

c=1/(2pi(300)(1695K)
=313nF

The way I see it is, the role of this capacitor is to allow all frequencies above 300 to pass uninterrupted. All DC voltages will also be blocked.

Due to capacitors C1 and C2 I would expect to see no DC voltage at the emitter.

I am seeing a corner frequency of close to 300Hz, close enough for me to be happy.

My concern now is the heavily distorted output waveform, I need to go to class in a bit, but I will post pictures later on.

 

[rude man]

you are correct in assuming it is a high pass filter and because of his I have actually increased the cap in value, my calculations are as follows;

Cap in xc=R1llR4

c=1/(2pi(300)(1695K)
=313nF

The way I see it is, the role of this capacitor is to allow all frequencies above 300 to pass uninterrupted. All DC voltages will also be blocked.

Correct. But C1 is a second high-pass filter. What value do you intend for it? And question: is it a good design if the cutoff frequency varies with the load resistance?

Due to capacitors C1 and C2 I would expect to see no DC voltage at the emitter.

Alas, there you'd be wrong. What is the voltage at the base? What can you say about a transistor's Vbe unless it's totally shut off?

I am seeing a corner frequency of close to 300Hz, close enough for me to be happy.

My concern now is the heavily distorted output waveform, I need to go to class in a bit, but I will post pictures later on.

As I said, the design is not going to give you clean waveforms unless you start to think 2 transistors OR a higher output impedance OR a lower gain.

 

[Bourbon daddy]

"Correct. But C1 is a second high-pass filter. What value do you intend for it? And question: is it a good design if the cutoff frequency varies with the load resistance?"

C1 in my understanding,

A) Blocks DC voltage
B) Sets the cut-off frequency

I have only concerned my self with the corner frequency, as set by C2. This Capacitor I have neglected.

to calculate the value of the capacitor,

C=1/2Pi(f)(R2llRload) with the frequency being 1KHz

You would require this capacitor to maintain its integrity over a range of load resistances I would imagine. Basic circuit theory tells me that if it wasnt you could be met with a scenario of cut off frequency increasing, reactance reducing, this could eventually result in an effective short across the cap, rendering it unsuitable.


"Alas, there you'd be wrong. What is the voltage at the base? What can you say about a transistor's Vbe unless it's totally shut off?"

Haha... Of course I am wrong... I am kicking myself over this, but its a new subject and I am still getting to grips with it.

Of course you would have DC voltage at the emitter, As for the Vbe, in operational mode, I have been taught that you would expect to see a value of 0.6/0.7V.

I think that Ve is approx. 10% of Vcc.

To calculate, you could work out;

Vcc-Vc-Vce=Ve

Voltage at the base can be calculated using

Vb=Vcc*(R2/R1+R2)

In my scenario I should be measuring 3V, but in reality I am actually getting roughly 0.3V


"As I said, the design is not going to give you clean waveforms unless you start to think 2 transistors OR a higher output impedance OR a lower gain."

After reworking my circuit I have managed to capture a good waveform that is showing heavy clipping on the negative cycle (phase shifted 180Deg so really I suppose it is the positive cycle).
It is a good clear image that will give me a lot to talk about in my assignment, and although I have not been able to successfully design an appropriate amp, I can evaluate and analyse my results and compare them with the theoretical results.

I am getting a gain of around 16, but obviously only on half of the waveform.

 

[rude man]

If your base voltage is only 0.3V then obviously the transistor is cut off fort small input voltages. Put in about 10 mV ac at say 1 KHz and see what you get at the output. the answer is zero!

I will drop this thread now but if you get a chance to go Darlington or some other set of specs we can talk again.