Hello all

I built a common emitter amplifier with these requirements: VCC=20V, Ic=10mA Av=2

What i did is calculating RC first of all with this formula VC=(VCC-RC•IC) where i chose the vc half of vcc, to have full dynamic so VC= 10V. and i found that RC=1k, now i chose RE in order to have 2 as gain, so RE=500 Ohm
After that i calculated VE as VE=RE•iC , since ic=ie and found that VE=5V.

At this point know that VB= VE+VBE
so now i calculated VB=R2/R1+R2•VCC so 5.7=R2/(R1+R2)•20 , now if we suppose that hfe is 100 and the IR2 current needa to be 10•IB we get that R2= 6.3kohm and R1 by doing the inverse formula we obtain R1=16k

By simulating it looks working fine, but i had a question now RE the way i calculated it isnt the best for stability of the circuit cus transistor changes its parameters also with its temperature, then how can i consider that too? thanks all.

 

I don't see anything wrong with your approach.

 

I don't see anything wrong with your approach.

View attachment 349196

but even at high temperatures it will still work fine? with that RE?

 

RE provides negative feedback. This provides temperature stability.

 

RE provides negative feedback. This provides temperature stability.

Yeah but what value i gotta choose? in my case i chose it in order to set the gain, but maybe there is another approach where u use like such formula or so: RE= VB-VBE+k(T-T0)/IC

 

You cannot have it both ways.
Rc was already restricted to 1 kΩ.
To get a gain of 2 you would have to tweak RE to 420 Ω.

The current gain of the transistor can vary from one transistor to another and with Ic.
The circuit performance will not change with beta.

 

You cannot have it both ways.
Rc was already restricted to 1 kΩ.
To get a gain of 2 you would have to tweak RE to 420 Ω.

The current gain of the transistor can vary from one transistor to another and with Ic.
The circuit performance will not change with beta.

Yeah i know, but maybe there is another approach to have the same gain with same ic current , but maybe better RE value for higher temperatures? its just my theory tho, i dont know, thats why im asking

 

maybe better RE value for higher temperatures?

The emitter resistor value has little effect on the temperature sensitivity for a circuit gain of 2.
The value of the collector and emitter resistors are selected for gain, and to get the desired frequency response, since higher values will give a lower maximum frequency response, due to the effects of parasitic circuit capacitances.

 

The emitter resistor value has little effect on the temperature sensitivity for a circuit gain of 2.
The value of the collector and emitter resistors are selected for gain, and to get the desired frequency response, since higher values will give a lower maximum frequency response, due to the effects of parasitic circuit capacitances.

Well in case i wanted a higher gain? how i would proceed?

 

Well in case i wanted a higher gain? how i would proceed?

The same as you calculated a gain of 2.

 

The same as you calculated a gain of 2.

Ye this is clear i mean, but if i had higher gain needed, so my transistor would be more sensible to temperature changes, in that case what u would do?

 

To get higher gain, you can bypass Re or part of Re with a capacitor. The DC operation is the same, but the capacitor reduces the Re for AC.

For instance, split the 500 Re into 100 and 400 and bypass the 400 with the appropriate sized capacitor and you get an AC gain of 10 with the same stability as your original calculation.

 

To get higher gain, you can bypass Re or part of Re with a capacitor. The DC operation is the same, but the capacitor reduces the Re for AC.

For instance, split the 500 Re into 100 and 400 and bypass the 400 with the appropriate sized capacitor and you get an AC gain of 10 with the same stability as your original calculation.

Oh in this way im adding a parallel resistor ( so RE is lower basically for AC component ) and in this way i get higher gain right ?? thats a smart way, but u need to see relation beetwen RE and the impedance of the cap that changes with frequency no?

 

See

 

The emitter resistor value has little effect on the temperature sensitivity for a circuit gain of 2.

Please, could you verify/explain this statement?
In contrary - I think the DC-stability of the circuit against temperature changes is proportional to the loop gain LG=-gm*Re .

 

See
View attachment 349311

View attachment 349312

View attachment 349313

View attachment 349315

i think he meant a cap in parallel to RE

 

Oh in this way im adding a parallel resistor ( so RE is lower basically for AC component ) and in this way i get higher gain right ?? thats a smart way, but u need to see relation beetwen RE and the impedance of the cap that changes with frequency no?

Yes, C must be calculated to have impedance small compared to R at the lowest frequency you want to amplify.

This is the standard way of controlling the gain of a CE AC amplifier, making it insensitive to transistor beta.

 

Post #12 was referring to raising the AC gain without compromising the DC stability.
You do this by keeping the emitter resistance the same but adding AC bypass to a portion of the resistance.

 

Post #12 was referring to raising the AC gain without compromising the DC stability.
You do this by keeping the emitter resistance the same but adding AC bypass to a portion of the resistance.

View attachment 349409

Exactly!

 

Please, could you verify/explain this statement?
In contrary - I think the DC-stability of the circuit against temperature changes is proportional to the loop gain LG=-gm*Re .

Okay, but the temperature change effect at a DC gain of two would not be significant over normal temperature changes.