# common emitter amp design

Discussion in 'Electronics Homework Help' started by KUMARA SHP, Aug 21, 2014.

1. ### KUMARA SHP

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Aug 1, 2014
I need to know how much should be the I current compare to the base current how can we determine it?

check my image

2. ### KUMARA SHP

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Aug 1, 2014
R1 & R2 are divider resistors , Rc is load resistor

3. ### KUMARA SHP

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Aug 1, 2014
I took I as 10 greater that base current, if I choose it 20 times greater than , what should be the better ?
when R1 & R2 values are grater will noise introduce?
what will be the better values for this amp

4. ### KrisBlueNZSadly passed away in 2015

8,393
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Nov 28, 2011
You posted this thread in the wrong section. The Off-Topic Members' Lounge is for off-topic subjects, i.e. subjects not directly about electronics. This is an on-topic subject! It also looks like homework so I have moved it to the homework section.

5. ### KUMARA SHP

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Aug 1, 2014
thanks .I am new to this , so that mistake will never happen

6. ### KUMARA SHP

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Aug 1, 2014
I need to find transistor with Hfe=350 , I didn't find anyone , help me can I change H fe value in proteus or multisim
I need to simulate circuit

7. ### Arouse1973Adam

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Dec 18, 2013
You have set yourself a gain of 33. This leaves you with a fixed ratio of RC and RE. A good rule of thumb is to choose RE so it drops 1V across it, or better still 10% of the supply voltage. But you won't be able to do this. So if RE is 1K then RC will be 33K.

The resistor values should be chosen by knowing the output resistance needed for the next stage. If this is of no concern just choose a value which will keep RE between 1K and say 2K2 whilst dropping half the supply across RC. Once you know the collector current, use the typical gain of the transistor to work out the base current. The gain of the transistor will vary with collector current. You should find this on the data sheet.

One this is done choose R1 and R2 to pass through them 10 times the base current. Then modify the values keeping the ratio the same so the base voltage is just on the verge of increasing the voltage on RE and the current through them is 10 times the base current. This voltage will be approx. 0.7 V + VRE.

Note this is a poor design, it should have an RE bypass capacitor to increase the gain at a.c. this allows the choice of better d.c stability component values.

This should get you started.

8. ### KrisBlueNZSadly passed away in 2015

8,393
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Nov 28, 2011
Good answer Adam. But the emitter decoupling capacitor is deliberately left out, because the desired circuit gain is 33 at AC. This also produces lower distortion because the emitter degeneration resistor becomes a source of negative feedback at AC, as well as at DC. Other than that, you've covered all the bases (pun intended)

9. ### Arouse1973Adam

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Dec 18, 2013
Thanks Kris, yes series feedback I think it's called isn't it? I didn't know that about a.c distortion, I assume your on about clipping and not purity of the waveform in normal use?
Thanks
Adam

10. ### KrisBlueNZSadly passed away in 2015

8,393
1,272
Nov 28, 2011
I haven't heard of it being called series feedback, but you could well be right. I just call it "emitter degeneration", but that's only a specific case of "series feedback".

No, I'm talking about accuracy of the output waveform. The base is driven from the input source, which is a voltage signal, but the base current is not linearly proportional to the base voltage, as you know. The collector current is proportional to the base current, and the collector voltage (the output of the stage) is proportional to the collector current, because the collector current is converted to a voltage by the collector load resistor.

So there is an inherent nonlinearity between the input and output voltages, because the base is current-driven (at least from the point of view of how the collector current is determined; I don't want to get into another discussion here about whether a transistor is actually voltage-driven or not - not the least because I don't actually know!)

Two ways of reducing this nonlinearity are to add a resistor in series with the input (which converts the input voltage into a current, roughly), or to add an emitter degeneration resistor. In either case the voltage gain of the stage is reduced, and the distortion is improved. The emitter degeneration resistor is preferred, because it actually implements negative feedback and it also helps stabilise the DC conditions.

Sometimes you will see two emitter degeneration resistors in series, with a capacitor across one of them. In this case, the un-bypassed one determines the AC negative feedback and improves the distortion, and the bypassed one provides negative feedback at DC only, to improve the circuit stability.

11. ### Arouse1973Adam

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Dec 18, 2013
Thanks for that Kris
Adam

12. ### KUMARA SHP

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Aug 1, 2014
thank for all comments got lot of knowledge
this was my school work
I designed it & simulate in multisim
my calculations showed me base current as 28 mA , but simulate showed me as 19mA can it happen
I neglected base resistance & other little stuff in h parameter model

13. ### Arouse1973Adam

5,170
1,092
Dec 18, 2013
Simulations, real life and calculations will all have variations but 9 mA out seems quite a bit. Can you show us your calculations.
Adam

14. ### KUMARA SHP

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Aug 1, 2014
not 9mA , 9uA in base current ;
sorry my above post has mistake
it should be 28uA & 19uA

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