# Multi-stage BJT amplifier design

Discussion in 'Homework Help' started by Sirius Black, Mar 15, 2012.

1. ### Sirius Black Thread Starter New Member

Mar 10, 2012
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I am to design an amplifier using 2N4401 BJTs. The specs include having a voltage gain of (+-)100. It needs to have input resistance above 1k and low output resistance. I decided to use a Common Emitter for the first stage to get the required gain. Then follow this with a common collector to achieve my low output resistance. I have attached my design so far, and keep in mind the capacitors are only for blocking DC, I have no restrictions on the bandwidth for which my amplifier should work. My values for R1, R2 are to keep a voltage of about .66V at Vb because I assumed Ic to be 1ma and the voltage drop across Re to be .06V. Vb = Vbe(.6)+.06. For R3 and R4 I wanted Vb to be half of Vcc(12V) so that I had a large amount of room for my output voltage swing.

My questions are:

-Does this design look like I'm on the right track to designing this amplfier, if not what am I doing wrong?
-To get my 100 gain, I approximated the gain on the CE to be -Rc/Re which should be 100, is this correct to assume?
-How can I simulate my circuit in LTspice, or a similar SPICE program? I don't know my Vs, but I assume it is ac because I must meet a design spec of at least enough room for my output voltage to swing (+-)2.5V. I also don't know my load resistance.

Thanks,
Sirius Black

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2. ### Audioguru New Member

Dec 20, 2007
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Your base voltage is way too low. The dastasheet shows that a "typical" 2N4401 has a base-emitter voltage from 0.4V to 0.8V depending on the temperature. Some transistors will be a little higher and others will be a little lower.

If you want your circuit to work with ANY 2N4401 and at ANY temperature then set the base voltage at around 1.3V to 2.5V then use the voltage across the added emitter resistor to automatically adjust the current in the transistor.
Use two emitter resistors in series. A low value one sets the AC voltage gain and has no bypass capacitor. A higher value emitter resistor that is bypassed sets the transistor's current.

So your transistor was biased with its base voltage too low so most 2N4401 transistors would have been cutoff.
You had the only emitter resistor bypassed so the transistor would have had as much gain as it can which might be 160.

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3. ### Sirius Black Thread Starter New Member

Mar 10, 2012
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Thanks for the response! That makes sense why I couldn't get the simulation to give me what I wanted. Could you explain how you found the two Emitter resistances? (R4 and R5)

4. ### Audioguru New Member

Dec 20, 2007
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896
I calculated the voltage divider using standard resistor values. The base voltage is 1.8V which is high enough for 1.19V to be across the emitter resistors to set the transistor's current.
The datasheet shows the base-emitter voltage is about 0.61V so the emitter voltage is about 1.19V and the emitter and collector current is 0.78mA. Then the collector voltage is 6.7V.

A transistor has a built-in emitter resistance. Its value is calculated as 0.026 divided by the emitter current in Amps. It affects the maximum voltage gain and the input resistance. It is about 33 ohms here. Then the total emitter resistance is 33 + 24= 57 ohms.

The 6.8k collector load has 100k, 120k and the input of the 2nd transistor in parallel with it so it is actually about 5.7k ohms.

Then the voltage gain= 5700/57= 100.

5. ### Sirius Black Thread Starter New Member

Mar 10, 2012
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0
It's amazing how much you know about this. Thanks for all your help. Now I need to do a DC analysis and a small signal analysis to find input resistance to the first stage and output resistance to the second stage. If I have any more questions I'll probably post them up here.

Thank you again!

6. ### MrChips Moderator

Oct 2, 2009
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That's why he is audioguru.