Common Emitter Amplifier

Hi,
I'm looking to build a common emitter amplifier (biased using voltage dividers). I am assuming I need to set the Q point so that the input wave does not shift my transistor into the active or cutoff region. I was just wondering a couple of things:

1. What information on the datasheet, will tell me the Q-point (or the different regions) of the transistor, are they the same for every transistor of different?
2. Once I have this information, do I alter the resistor values to suit it or do something else?

My amplifier circuit is based on the one found on this page:
https://www.electronics-tutorials.ws/amplifier/amp_2.html

The Q point is determined by things like, the collector load, which is shunted by the input impedance of the next circuit and the expected level of the output signal and the supply voltage. For small signals where the above factors do not really count, the collector current would be set to that recommended for the minimum noise. To get maximum linearity the emitter resistor would not be bypassed, so the stage gain would be Rc/Re. For high frequency use (>200KHZ?) the collector load should be a low value so as to reduce the capacitance effects of the next stage.
Frank

Resistanceisfutile said:

1. What information on the datasheet, will tell me the Q-point (or the different regions) of the transistor, are they the same for every transistor of different?
2. Once I have this information, do I alter the resistor values to suit it or do something else?

My amplifier circuit is based on the one found on this page:
https://www.electronics-tutorials.ws/amplifier/amp_2.html

Click to expand...

What is your problem?
Do you know that you can SELECT a suitable Q-point?
Have you some selection CRITERIA?
Which resistors do you intend to "alter"? Are you referring to design values of the cited link?

Resistanceisfutile said:

Hi,
I'm looking to build a common emitter amplifier (biased using voltage dividers). I am assuming I need to set the Q point so that the input wave does not shift my transistor into the active or cutoff region. I was just wondering a couple of things:

1. What information on the datasheet, will tell me the Q-point (or the different regions) of the transistor, are they the same for every transistor of different?
2. Once I have this information, do I alter the resistor values to suit it or do something else?

My amplifier circuit is based on the one found on this page:
https://www.electronics-tutorials.ws/amplifier/amp_2.html

Click to expand...

The tutorial explains the simple transistor circuit very well. It shows an example circuit and lists its currents. You can use Ohm's Law to simply calculate the Q-point. You can simply calculate the maximum output voltage swing between saturation (it is not called active) and cutoff.

You select a transistor part number then its datasheet shows the range of beta called hFE. You select the power supply voltage and the collector resistor value. You select how much voltage drop will be across the emitter resistor which is 1V in the example circuit (then calculate its value) .

Audioguru said:

The tutorial explains the simple transistor circuit very well.

Click to expand...

For my opinion, the tutorial does NOT explain the circuit "very well". Therefore, the questioner should also read my comments as contained at the end of the text.

just a suggestion

the tutorial looks ok,
follow it and there should not be a problem.
also Vgain = hoe in parallel with RL * hfe.
now if you change the transistor for a different type
the Q pint will change, and the gain will change.
why? because VBE, and hfe and FT, varies between transistor types.

as the current through the collector is = (V(at divider)-VBE/RE), neglecting IB which is very small,

the Q point will change as VQ=Vsupply- (IC*RL)-VRE

note VRE = (IB+IC)RE : IB= IC/hfe

and if you want low distortion you got to to select the transistor with the best linear curve.
otherwise use a high voltage transistor, and a high supply voltage

I agree that the tutorial failed to explain the requirement for RE.

The author of the tutorial was taught the same as me that a transistor is current controlled. An input signal voltage swing causes the transistor's base current to swing that causes the transistors collector and emitter current to swing with amplification. The voltage gain is Rc/(RE + re) where RE is not bypassed by a capacitor and the Rc is the collector resistor parallel with any load resistance. The input impedance is hfe x (RE + re) where RE is not bypassed by a capacitor. gm was not used for transistors, it was used for Fets that are voltage controlled.

Some people might wrongly think that hFE (beta) is voltage gain. No, they have nothing to do with each other.
hFE (beta) is DC current gain. hfe is AC current gain. The resistor values and amount of collector current in the circuit determine the voltage gain.

The voltage gain of a transistor is exponential, so the gain is low at low collector current and the gain is high at high collector current. Then there is severe distortion when the output level is fairly high and there is no AC negative feedback.

I simulated the same transistor in the same circuit and changed the bias (R2 value) so that the collector current was high, then low. Then the Q-point was symmetrical but the input signal level was increased (but the output was not clipping:

Audioguru said:

The voltage gain of a transistor is exponential, so the gain is low at low collector current and the gain is high at high collector current.

Click to expand...

I rather would expect that the voltage gain linearly depends on the DC collector current (for constant Rc).
Remember the transconductance is gm=Ic/Vt.

Audioguru said:

I agree that the tutorial failed to explain the requirement for RE.

The author of the tutorial was taught the same as me that a transistor is current controlled. An input signal voltage swing causes the transistor's base current to swing that causes the transistors collector and emitter current to swing with amplification. The voltage gain is Rc/(RE + re) where RE is not bypassed by a capacitor and the Rc is the collector resistor parallel with any load resistance. The input impedance is hfe x (RE + re) where RE is not bypassed by a capacitor. gm was not used for transistors, it was used for Fets that are voltage controlled.

Click to expand...

just a suggestion

the reason for RE is to bias the transistor against temperature changes affecting the "Q" point. it is normally by passed with a capacitor of large value.
leaving this resistor un-bypassed can cause
the circuit to oscillate or peak because of any capacitance between collector and emitter, the DC voltage developed across RE is lost to the output voltage swing. and yeah if left un-bypass gain approx RL/RE

re is so small compare to RL why bother
with re in this simple circuit.
hfe varies with frequency,transistor type
dc bias, temperature, only lots of negative
feed back will give a stable gain, and for this you require a large loop gain. (more than one transistor).

THINKER_KAM_MAN said:

re is so small compare to RL why bother
.

Click to expand...

Do you know the physical background (meaning) of re? If yes - you should know that you are not allowed to neglect re if compared with RL.

just a suggestion

yeah i know what its is,
the tutorial covers it
the feed back ratio is small when RE is decoupled

THINKER_KAM_MAN said:

just a suggestion

yeah i know what its is,
the tutorial covers it
the feed back ratio is small when RE is decoupled

Click to expand...

I am afraid you are mixing re and RE. That is exact the problem I always expect using this symbol re for a quantity which in reality is NOT resistive.

just a suggestion

is the input Z of a common base configuration
not = 25/Ic then i think i will have to read the article again

The voltage gain of a common-emitter transistor is RC (total)/(re + RE). In my simulation with highest current the RE is bypassed and the average collector voltage is 1.6V so the average collector current is (10V - 1.6V)/10k= 0.84mA. Re is 26mV/0.84mA= 31.0 ohms.
The load is 100k so in parallel with RC causes the total RC to be 9.09k. Then the voltage gain is 9090 ohms/31.0 ohms= 293 which is close (?) to the simulated gain of 260.
If RE is not bypassed then the voltage gain is 9090 ohms/(31 ohms + 1k ohms)= 8.8 times.

@Audioguru

gm was not used for transistors, it was used for Fets that are voltage controlled.

Click to expand...

gm is used for only FETs not for BJTs?????

@LvW

Do you know the physical background (meaning) of re? If yes - you should know that you are not allowed to neglect re if compared with RL.

Click to expand...

What is the physical background of re? Why can't we neglect it even it is very small in magnitude compared with RL?

Eshal said:

@Audioguru
gm is used for only FETs not for BJTs?????

Click to expand...

No - gm is used for BJT`s as well. More than that - I think, it should be used only!
The transconductance gm is the most important gain parameter because it connects the input with the output: It is identical to the slope of the input-output relation Ic=f(Vbe) which has an exponential form.
Hence, the gain (common emitter) is: G=-gm*RL(1+gm*Re) with Re being the emitter feedback resistor and RL the effective resistance connected to the collector.
I know that the quantity re=1/gm is used in some textbooks [example:G=-gm*RL(1+gm*Re)=-RL/(1/gm+Re)=-RL/(re+Re)]
But - for my opinion - this is a very unfortunate approach which can lead to misunderstandings/misinterpretations.
More than that - it is simply false because re is NOT a resistive element. It is the inverse of gm and, thus, has the unit V/A. That`s all.

Eshal said:

What is the physical background of re? Why can't we neglect it even it is very small in magnitude compared with RL?

Click to expand...

I think, this has been answered above. The inverse of the transconductance (called by you re) cannot be neglected against RL because it appears together with RL not in a summation (addition). See the given gain formulas

If Re (1/gm) is neglected then the gain of a single transistor (with its emitter resistor bypassed) is infinity! Impossible.

Audioguru said:

If Re (1/gm) is neglected then the gain of a single transistor (with its emitter resistor bypassed) is infinity! Impossible.

Click to expand...

Audioguru - from the mathematical point of view you do not "neglect" re but you set re=0.
To "neglect" something means to compare it in a sum or difference with another value. And that`s not possible in a formula like Gain=-gm*RL (=RL/re).

The biggest uncertainty value for the CE design is β. If we assume you want the largest swing for a known input max, the design is straightforward to choose R values for the Min/max range of β. The supply voltage must be at least 3 to 4V above the Vpp swing for low distortion . The distortion starts when Vc swings close to 3V within the V+ rail on the higher side whereas distortion on the low side can get within 0.5V to1V for Vce as this knee is much sharper.

The emitter DC voltage regulates the base current which regulates the Q point, but for high gain designs the Re can be chosen quite small such that the DC Q point of the emitter Ve is at least twice the input voltage swing and doesn't need to be 1V when the gain is >100 for a 12V supply, rather it only needs to be 0.1~0.2V for a 50mVpp input.

Since the Q point affects the symmetry of the large swing, with 1% resistors, the Q point is entirely determined by the base current, and range of values of β, thus the parameters to look for are Max/Min ratio of β rather than the average value of β. Companies like ROHM do a great job of binning these average values to keep this ratio low and appending the part number with a letter for the rank of current gain. The bin ranges for Max/min are often 1.5. for some which may range from B=50 to 500. So pay close attention to this ratio. of β max/min

In the original example Ib~5uA based on the H bias values, β=100, so Vc~6V approx. But this changes dramatically for β=50 to 200. for a ratio of 4:1. If you cannot choose a better ratio, then a negative collector to base feedback approach is required or else reduce the maximum output swing for this stage to allow for Q point swings.

As I stated before low Collector current distortion starts where Vc gets near 3V bwlow V+ and 0.5V above the emitter, so the ideal Q point should be (12-3-0.5)=8.2 then /2 = 4.2V which gives a 8.V max swing. under ideal fixed β conditions. Never try to use this full linear range even for 1.5 ratio of β if you want good linearity or add negative feedback for DC bias with other design modifications and changes for reduce AC gain sensitivity with β. THis is done by added a smaller Re so that gain is fixed by Rc/Re ratio. rather than dependent on internal rE which is modulated by emitter current.