# A little overwhelmed with BJT amplifier design

Discussion in 'The Projects Forum' started by HunterDX77M, Dec 12, 2011.

1. ### HunterDX77M Thread Starter Active Member

Sep 28, 2011
104
2
I need to make a BJT voltage amplifier that gives me a gain of 25 (either in phase or out of phase). My restrictions are as follows:
-Transistor: 2N2222 with a $\beta$ of 150
-Transistor power dissipation is less than 50 mW
-Resistor only biasing
-single DC power supply
-Resistors have to be less than 500 kilo ohms
-Coupling and bypass capacitors are "large" (1 mF)
-Load resistance is 16.81 kilo ohms

I started off with a really simple design, that I think would work except that it has me using a resistor that is about 4 ohms, which makes no practical sense. Assume that the input AC signal is a "small" signal, and the small signal model can be approximated with the hybrid-$\pi$ model

See attached diagram:

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2. ### Audioguru Expert

Dec 20, 2007
10,508
1,168
A single transistor is not a power amplifier so it cannot drive a load as low as only 4 ohms and have voltage gain at the same time.

3. ### crutschow Expert

Mar 14, 2008
19,827
5,546
Start with a simple single transistor common emitter circuit with an emitter resistor to control the gain and go from there.

4. ### Adjuster Late Member

Dec 26, 2010
2,147
302
The requirement for power consumption to be less than 50mW seems odd, unless there is some additional requirement. This might be something like a minimum available output voltage level, or minimum output impedance. With no such stipulations, this power level should be extremely easy to meet. Are you sure you have not left something out?

Edit: Unless perhaps the load impedance is in ohms, rather than kilohms?

5. ### jimkeith Well-Known Member

Oct 26, 2011
541
100
4Ω seems to be in error somehow--I would consider this a false start.
This is doable, but with the 50mW requirement, you must work backwards.
What is your Vcc?
Assuming that quiescent Vce = Vcc/2, it is easy to calculate the collector pull-up resistor for Pc = 50mW. Both pull-up resistor and transistor will dissipate the same power, if biased in the center.
After Ic is determined, you can work on the biasing

6. ### thatoneguy AAC Fanatic!

Feb 19, 2009
6,357
727
Here is my SWAG at it, needs a rather high supply voltage, though.

It's for reference only, and may not fulfill all the requirements, but the concept is close, I think. Though I'm not much good at this.

7. ### Destroyer x New Member

Apr 7, 2011
6
0
It is a little bit stupid, but works fine...i am preparing a video and as soon it will be uploaded i will publish the link here... also i will know if links can be published here.

My Youtube Channel is here... the link:

regards,

Carlos

8. ### Adjuster Late Member

Dec 26, 2010
2,147
302

1. The base potential divider resistances are a bit high for good bias stability, and certainly exceed the requirement for all R < 500kΩ. R1 and R2 could do to be reduced by a factor of two, or a bit more to get down to convenient preferred values.
2. All capacitors should be "large", but here they are not: as can be seen by the nearly 90° phase angle between input and output, this circuit is affected by capacitor reactance, probably mainly in Ce.
3. The gain is too big: this can be conveniently be done by putting a resistor in series with Ce, as earlier recommended by Crutschow.

Apr 7, 2011
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10. ### Audioguru Expert

Dec 20, 2007
10,508
1,168
I wish the voice in the video spoke better Engrish.

11. ### Destroyer x New Member

Apr 7, 2011
6
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As we may have other countries in an International forum...some knowledge of other languages could help.

I am sorry folks.... my messages are result of a big effort to communicate and to help people...i have never studied English...was my effort and goodwill that made me learn a couple of words, and i hope the ones are interested in the message, will make some effort to understand and will forgive the bad quality in the English language.

I am very good in my own language... a second and third language is an effort to be friendly with you.

regards,

Carlos

Last edited: Dec 14, 2011
12. ### Audioguru Expert

Dec 20, 2007
10,508
1,168
The video used "trial and error" to guess at the design of the single transistor amplifier. The internal emitter resistance of the transistor (26/emitter current in mA) was not included which made the calculation for gain too low.

13. ### thatoneguy AAC Fanatic!

Feb 19, 2009
6,357
727
OP said caps needed to be 1uF (how I interpreted it).

Each resistor can only be < 500k, not the sum of all of them, and I only broke that rule once (though it could be changed). It is a generic solution for both threads on this going right now, but not a perfect solution for either.

14. ### Destroyer x New Member

Apr 7, 2011
6
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our friend...make it more correct, more precise and with better language.

regards,

Carlos

15. ### Adjuster Late Member

Dec 26, 2010
2,147
302
And I wish that he had not recommended having only an emitter resistance determined by considering the AC gain requirement. The result will be poor DC bias stability.

It would be much better to have a DC emitter resistance dropping at least an appreciable fraction of a volt, with the lower impedance required to get the right AC gain obtained either using series sections with one part bypassed, or using one emitter resistor for DC bias, with a resistor in series with the bypass capacitor.

16. ### Destroyer x New Member

Apr 7, 2011
6
0
This will present a more clear "scene" to our young friends.

I could see, in this same forum, another question not too much different, even the demands where almost the same, the transistor and the gain.

Seems these guys are from the same school.

regards,

Carlos

17. ### Adjuster Late Member

Dec 26, 2010
2,147
302
While I dislike the modern use of 1mF = 1millifarad = 1000μF, it is technically correct: many simulation programs will take it as such.

The emitter circuit impedance is low, and with Ce = 1μF, the amplifier would only have full gain at frequencies above a few kHz.

The lower bias chain resistances were as much to increase the chain current to improve the bias stability, as to keep the resistor rule.

18. ### thatoneguy AAC Fanatic!

Feb 19, 2009
6,357
727
All true, but what I was trying to get across was the relatively high value of the supply needed in order to get the power below 50mW.

19. ### Destroyer x New Member

Apr 7, 2011
6
0
Observe it is in series to the ground through the input impedance... you can see the input resistor value in series with this condenser makes a filter to ground... deviating high frequencies to the ground.

This is the reason why we use the smallest possible value in the input, not to loose low frequencies but also avoiding to filter high frequencies..... audio enters the transistor base, but also it is filtered...the huge condenser "eats" or drains high frequencies (and some mids and lows too) to the ground....but for high frequencies this condenser is alike a short because the capacitive reactance.

regards,

Carlos

20. ### Audioguru Expert

Dec 20, 2007
10,508
1,168
No.
The input capacitor passes all high frequencies and blocks DC and low frequencies because it forms a simple highpass filter. It is not a filter to ground.
It has no effect on high frequencies but an inductive electrolytic capacitor will reduce radio frequencies.