common emitter amp  - Page 1
Forum section: electronics / beginners

author: NASK
author_rank: Contributor
timestamp: August 09, 2017, 04:33:24 am
content: I want to design common emitter amp having with 100 gain. my calculation is attached but design not amplify that much. therefore could i know what wrong in my calculation ?

author: ElektronikLabor
author_rank: Regular Contributor
timestamp: August 09, 2017, 06:17:39 am
content: A gain of 100 is pretty high; most npn Transistors won't do that.Try several stages of amplification; e.g. first stage with gain 20 and second one with gain 5.If you want only one stage; you can try out a darlington transistor, which has a much higher current gain.

author: Zero999
author_rank: Super Contributor
timestamp: August 09, 2017, 08:16:07 am
content: Quote from: ElektronikLabor on August 09, 2017, 06:17:39 amI think you've confused voltage and current gain.BJTs can be configured with a voltage gain, well in excess of their current gain. A Darlington transistor may have a higher current gain, than a single transistor, but it actually has a lower transconductance, therefore will give a lower voltage gain, when configured as a common emitter amplifier.Quote from: NASK on August 09, 2017, 04:33:24 amI can hardly read that schematic. It's so fuzzy and small.Try setting the DC voltage gain to something more manageable, so the voltage across the collector resistor can be stabilised to half the supply voltage. Add a bypass capacitor and resistor across the emitter resistor, to increase the AC gain to the desired level.See attached example.

author: NASK
author_rank: Contributor
timestamp: August 09, 2017, 09:52:27 am
content: thank you for the answers!!Hero999, your attached example circuit Icq equals almost 0.46mA. why did you use that much of small current , can I use 5mA for Icq. and your gain is almost 200, if start from Av=Rc/Re' equation can I achieve this given solution.

author: Zero999
author_rank: Super Contributor
timestamp: August 09, 2017, 10:37:35 am
content: Quote from: NASK on August 09, 2017, 09:52:27 amNo particular reason. You can increase the current to 5mA if you like and reduce all the resistor values to compensate.QuoteYou forgot to take the fact that the BJT doesn't have a limitless transconductance. In this case, the easiest way to account for this is imagining an internal emitter resistor, with a value of 25mV/Ic.Re' = 25mV/Ic = 25mV/0.46mA = 54RRe = Re' + R5 = 93R ;We can ignore R2, as it's much larger than R5.Av = 10000/93 = 108If you download LTSpice (it's free and runs under WINE if you use Linux/Mac OS) you can simulate the .asc file attached to my previous post.

author: phliar
author_rank: Contributor
timestamp: August 10, 2017, 12:41:59 am
content: The best video on designing a BJT common-emitter amp is the one from The Signal Path:

author: orolo
author_rank: Frequent Contributor
timestamp: August 10, 2017, 07:41:55 pm
content: The computations look a bit fuzzy, but it seems you usted a grounded emitter transistor with 500 ohms collector resistor. That kind of amplifier is very unpredictable and temperature dependent (might easily saturate with a slight temperature increase) besides having low input impedance. Of course, if you choose to use emitter degeneration, for a gain of -100 you need a x100 colector resistor, which will be unpractical. For example, for a 100 ohms emitter resistor, you'd need a 10k collector one, who would steal 50 volts quiescent voltage! For these situations you have two choices: either cascade two moderate gain amplifiers, as suggested, or use an active load: instead of the collector resistor, use a pnp or jfet current source. In the last case you'll have gain to spare, so with a bit of feedback you'll achieve both correct gain and high input impedance.In any case, more that one transistor is required.

author: NASK
author_rank: Contributor
timestamp: August 17, 2017, 01:22:36 pm
content: I designed circuit emittor bypassing capasitor value is almost 2uF accoding this calculationXc=1k/10 = 100Rc=1/2?fXc= 1/(2?*1khz*100R)=1.5uFwhen I use calculated value, amplification getting low , but I use 100uf it getting correct amplification. how to overcome this issue.

author: Zero999
author_rank: Super Contributor
timestamp: August 17, 2017, 02:17:15 pm
content: Quote from: NASK on August 17, 2017, 01:22:36 pmWhat do you mean by low? What's the predicted and actual amplitude? How does it compare with what you get at 5kHz?

author: NASK
author_rank: Contributor
timestamp: August 18, 2017, 02:10:56 am
content: Using 100uF capacitors 5mv signal amplify 1.06 V @ 1KHz, When using calculated capacitor’s value 2uF, 5mv amplify 350 mV @ 1KHz. Required value for amplification is 1V.Using 100uF capacitors 5mv signal amplify 1.06 V at 5KHz.attached pdf shows results and diagram

author: Zero999
author_rank: Super Contributor
timestamp: August 18, 2017, 08:17:57 am
content: Quote from: NASK on August 17, 2017, 01:22:36 pmWhere did you get that formula from?Try this:RE' = 0.025*IEC = 1/(2pi*FC*RE')Where FCis the 3dB point, when the voltage is 0.707 of the maximum. If you don't want any attenuation than make FCmuch lower, than the frequency of interest.With an FCof 500Hz and the component values you've used, I get around 8.8µF for C.

author: IanMacdonald
author_rank: Frequent Contributor
timestamp: August 18, 2017, 08:44:19 am
content: There will be two problems with achieving a gain of 100 this way. BJTs are essentially current amps, and driving the base with a voltage, with little or no series resistance, will give rise to nonlinearity due to the b-e diode law conduction. That, and the gain will vary quite a bit across samples so no use for a production run if it must be 100.Better to use a simple discrete op-amp. A two transistor complementary arrangement should be adequate. Only slightly greater component count and far better results. The feedback potential divider then determines the gain to within a few percent regardless of hfe tolerances.The gain here wil be too high, increase the 100R.

author: Zero999
author_rank: Super Contributor
timestamp: August 18, 2017, 09:50:22 am
content: Quote from: IanMacdonald on August 18, 2017, 08:44:19 amI must point out that the Hfe does not determine the voltage gain of the circuit. It's proportional to the transconductance (which is temperature and current dependant) and the collector resistor.I doubt this is intended to be a useful circuit. It's most likely a college assignment.

author: BigBoss
author_rank: Regular Contributor
timestamp: August 18, 2017, 02:54:22 pm
content: Here are the results of your Common Emitter Degenerated Amplifier.I have used MathCAD to calculate and "iterate" the equations.Ref: "Analysis and Design of Analog Integrated Circuits. Paul R. Gray, Paul J. Hurst, Stephen H. Lewis, Robert G. Meyer.John Wiles and Sons, 4th Edition" Page 197-200

author: firewalker
author_rank: Super Contributor
timestamp: August 18, 2017, 10:10:49 pm
content: For the math you could also check this:https://alexkaltsas.wordpress.com/2015/09/03/designing-a-transistor-amplifier-part-a/Alexander.