BJT biasing formulas

[audioguru2]

Hi Wald,
Your calculations are correct but you have the same problems as before:
1) The dictated collector current requires odd resistor values. Select the RC and RE resistors first. A standard 2k for RC and 100 ohms for RE will result in a gain of 20 and a collector current near 2mA.
2) The collector current of 10mA is rather high which will quickly drain a little 9V battery and cause a low input impedance.
3) The lack of adding a bypassed emitter resistor to raise the emitter and base voltages means that the DC operating point will change too much with different transistors.
You calculated using a Vbe of 0.65V. The datasheet shows a typical Vbe of 0.72V at a 10mA collector current and it could be from 0.65V to 0.85V.
Calculate again using your resistors and a Vbe of 0.85V, and you will see that your transistor's collector voltage will be very high and the transistor is nearly cutoff! :'(

Design the circuit again with RC= 2K, RE= 100 ohms and a bypassed additional RE= 560 ohms to raise the emitter and base voltages about 1V. Then it will have a voltage gain of 20, a fairly low current, a fairly high input impedance and work well with any 2N3904 transistor with a current gain from 100 to 400 and a Vbe from 0.55V to 0.75V (at a collector current of nearly 2mA). ;D

 

[walid1]

Hi audioguru,
What did you mean by the following:
1) Odd resistor values
2) Select the RC and RE resistors first


AND why u audioguru tend to make VE very small like 2mA*100ohm=0.2V. I see many circuits in that the VE is greater than 1 volt.

WHEN IC is big (10mA), we use a transformer power supply rather than batt.

audioguru: "3) The lack of adding a bypassed emitter resistor to raise the emitter and base voltages means that the DC operating point will change too much with different transistors."

Walid: (1) u all the time want VE very small and now you want to raise it
      (2) u now introduce a new aspect (a bypassed emitter resistor) we never use in the last design. I think u like to dispersed my mind.
      (3) Please make good work for your friends and design it as you want and show it to us. I beleive u never do so, u want to always ask u......
      (4) Why now, the DC operating point will change too much with different transistors. Why it does not change in your first design.
I wait.... but please, we now study the autir' first design without any bypass

 

[audioguru2]

What did you mean by the following:
1) Odd resistor values
2) Select the RC and RE resistors first

You can't buy nor find resistors with the "odd (unusual) values you calculated. You need a voltage gain of 20 so choose resistors with common values like 2k and 100 ohms.

AND why u audioguru tend to make VE very small like 2mA*100ohm=0.2V. I see many circuits in that the VE is greater than 1 volt.

I don't make nor want the VE to be low. The voltage gain requirement of 20 dictates that the voltage across RE is 1/20th of the voltage across RC. With the voltage gain of 10 in the other circuit, the VE is twice as high.

WHEN IC is big (10mA), we use a transformer power supply rather than batt.

Look at a list of common 5% resistor values and the only 20:1 ratio is 2:0.1, such as 200 ohms to 10 ohms which makes the current 20mA, or 2k to 100 ohms which makes the current 2mA, or 20k to 1k which makes the current 0.2mA, etc.
When the current is high then the input impedance is low and the input coupling capacitor must be bigger.
The low input impedance will load-down a high impedance source like an electret mic.

audioguru: "3) The lack of adding a bypassed emitter resistor to raise the emitter and base voltages means that the DC operating point will change too much with different transistors."

Walid: (1) u all the time want VE very small and now you want to raise it

 

[walid1]

It is ok, when we come to design such amp, and when choosing Av we must take other things into account at the same moment, that things are the IC value and the VBE value.
example: u want Av=10, then u have many values of (RC/RE) which = 10, like 10K/1K, 4.7K/470, 1k/100 .. etc

The difference in choosing one of these ratios depends on What Ic you want.
Lets assume that your VCC=9V, and you choose 10k/1k ratio, then the saturated current Isat=9/(10K+1K)= 0.8mA, the best choise of IC=Isat/2=0.4mA to be at the center.
Taking into account VCE=VCC/2=4.5V, then only 4.5v remain and it is divided between RC and RE according to there ratio, that is VC=10VE.

If VE = 1K*0.4mA=0.4V then V(RC)=4V

With the same VCC=9V and RC/Re=4.7K/470, then Ic= Vcc/(2*[RC+RE])= 0.87mA and VE= 0.87m*470=0.41V and so on..
In sammary,
First: decide Av
second: put several values of RC/RE = Av
Third: Choose the RC/RE ratio that give you the desired Ic, then complete your calculations ...

 

[audioguru2]

Third: Choose the RC/RE ratio that give you the desired Ic

I know what you mean but it might be confusing for others. I would say "Since the gain is 10 then choose RC=10RE values that give you the desired IC." ;D

 

[walid1]

I understand from your reply that mine is true.

 

[walid1]

Hi
I back again to this very important subject
first thank you audioguru you do a good work to make me understand
i print this 5-pages subject after editting it in WORD and finally have 19 A4 pages
i take it with me every where reading them. it is hard to read directly from the computer screen, i need to underline the important and new aspects and compare bwn some replies
When I finish it I'll either put a detailed design or continue asking question and god help guru
walid

 

[walid1]

hi all
Hi audioguru... my big teacher
I'll show you here what guru taught me in the last 5 pages, then we'll
continue to reach a high level... hope so

WHAT I WANT
(1) design a preamp (voltage divider unbypassed)
(2) using the famous PN2222A transistor whose datasheets are attached below
(3) I need a voltage gain of 10 for low distortion.
(4) I have a 9v batt as a power supply (VCC).

CALCULATIONS and ANALYSIS

STEP 1

Av (the voltage gain) = RC/RE = 10, so RC=10 RE and V(RC) = 10 V(RE)
Before u choose RC and RE values u must consider the following:
a) the value of RC = o/p impedance (Zo)
b) If this stage have a load like a second stage follow it then the total
  Zo= RC//Zin2
C) Zin2 must be >= 10 RC to not load down the 1st stage and decrease its Av
d) IC = IE = Isat/2 , where Isat is the saturation current and we choose its
  half to be at the center fore symmetrical swing of the o/p signal
        Isat = VCC/[RC+RE]

 

[walid1]

e) When Ic is high (look at Fig VD01 below) hfe (beta) is low, Zin is low
  and VBE is high (Look at Fig VD02 to see how VBE changed with IC)
f) When Ic is high it will drain your batt quickly

NOW

I want (as u may noticed in the figures) that Ic = 2mA
2mA means Isat 4mA
4mA = 9/[RC+RE] 
[RC+RE] = 9/4m = 2250
RC= 10 RE ==>  11RE = 2250  ==> RE = 204.5 ohm take it 200 ohm (two 100 in series)
so, RC =10 RE = 2K ohm

NOTIC
your Zo = RC = 2K, so Zin2 must = 20k at least to not affect Av1

STEP 2

For symmetrical swing up and down, VCE = VCC/2 = 4.5V
Remain 4.5V from the 9V batt, these 4.5V divided between RC and RE according to their values
4.5/11 = 0.41v, so, V(RE) = 0.41V and V(RC) = 4.5 - 0.41 = 4.09V
The number 11 arise from the fact that RC contribute by 10 and RE contribute by 1, total=11

STEP 3

At Ic = 2mA, beta = 200
so, IB = 2m/200 = 10 uA
Id (the current pass through R1 and R2 with no transistor connected) = 10 IB = 100uA
Id = VCC/[R1+R2] = 9/[R1+R2] ==> [R1+R2] = 90 K
We calculate VE as = 0.41v and from the curve VBE = 0.64 SO, VB = 1.05V

I(R2) = Id - IB = 90 uA ==> R2 = VB/V(R2) = 1.05/90 u =11666.7 ohm take it 12 K ohm
R1 = 90 -12 = 78K , connect 68K in series with 10K

STEP 4

Zin = R1//R2//[hfe*(re+RE)]
re = 26 mV/IC = 13 ohm
so, Zin = 10.4K // 42.6K = 8.4K (fairly low i'll discuss this later)
the o/p voltage can swing from [VCC] up to [VC- Isat*RE] down, that is it
will swing from 9V to (4.91 - 4m*200)= 4.11v
this swinging ac signal is carried on a VC dc voltage
The max upper peak of this signal which represent the negative part of the
i/p ac signal, = VCC - VC = 9 - 4.91 = 4.09 V
The max lower peak of this signal which represent the positive part of the
i/p ac signal, = 4.11 V
Note the symmetric it is very close to 100%, thanks to Mr AUDIOGURU for every thing.


To increase Zin, u have to lower IC and recalculate the other values.

Later I'll continue my discussion about:
bypassing the whole RE
bypassing part of RE
choosing the caps value between stages
discussing a real designs and show if this is good or bad ...

I hope guru read this and put his golden comments
thank u all
WALID

View attachment 39162

View attachment 39163

 

[audioguru2]

Hi Walid,
Your calculations are correct and you have a transistor stage with a voltage gain of 10 and it has very low distortion. ;D

 

[walid1]

good discussion

 

[walid1]

Hi

I said before, at 12/6/2006 that:

Later I'll continue my discussion about:
bypassing the whole RE
bypassing part of RE
choosing the caps value between stages
discussing a real designs and show if this is good or bad ...

today I'll discuss with audioguru and you, the effect of adding a bypass cap to RE, its name is CE

Till now we can design a perfect voltage divider BJT transistor configuration without CE

When adding this CE, two circuit parameters are changes:
(1) The voltage gain Av now = RC/re and not as before adding CE it was RC/RE.
With CE, you get mor Av, and distortion takes place.
(2) The i/p impedance seen at the base, ZB, ZB = hfe*re, it is very smaller compared with the case without CE (ZB = hfe*(re+RE)).
I think that the total Zin (= RB1//RB2//ZB) will not affected much because of the paralle connection.

my question: are there any other parameter affected by adding CE, and
Why we disregard the distortion accompanying with the increased Av?

thank you

 

[audioguru2]

Hi Walid,
Negative feedback reduces a transistor's voltage gain and distortion and increases its bandwidth and input impedance. Where gain is more important than low distortion or wide bandwidth like in a child's toy walkie-talkie then a CE is added.

If a current source is used as the collector resistor or bootstrapping is used then the voltage gain is high and the distortion is low. But the output impedance is high so an emitter-follower is needed at the output. Transistors are cheap in IC opamps so that is what they do.

 

[walid1]

Hi guru
please look at the Fig below. It is thr first stage of your FM TX
I omit C2 to make it more simple to discuss

Negative feedback reduces a transistor's voltage gain and distortion and increases its bandwidth and input impedance.

Where is the Negative feedback in your circuit (below)

If a current source is used as the collector resistor or bootstrapping is used then the voltage gain is high and the distortion is low. But the output impedance is high so an emitter-follower is needed at the output.

please put some examples to a current source and bootstrapping and how they connected to collector

you guru add a CE to your Tx

thank you
View attachment 40151

 

[audioguru2]

Where is the Negative feedback in your circuit?

R5 provides negative feedback at lower audio frequencies. At 10kHz and above, C4 bypasses it for high voltage gain and high distortion but you can't hear the 20kHz to 30kHz distortion harmonics.

please put some examples to a current source and bootstrapping and how they connected to collector

Here they are:View attachment 40153

 

[walid1]

Hi guru

u enlarge the discussion and introduce a new aspects -at least foe me- and i can

 

[audioguru2]

total Z seen by the MIC (Zmic)= 10K//11.2K= 5.3K

Correct.

C1 (=330n) and Zmic are a high pass filter, its cut off freq (Fo)
Fo= 1/(2 pi C1 Zmic)= 91 Hz

No. The mic impedance is about 5k and it is in parallel with its 10k load/powering resistor so it has an effective series resistance of 3.3k. The 3.3k, the 330nF coupling capacitor and the 11.2k load resistance are all in series so the cutoff frequency is 1/(2 pi C1 [3.3k + 11.2K])= 33Hz.

Fo = 64 Hz, so this filter cuts all freqs less than 64Hz including mains hum.

No. This is a very simple filter. At the cutoff frequency the response is down -3dB which is a small amount. If the cutoff frequency is 500Hz then 50Hz would be reduced but not eliminated.

 

[walid1]

Good evening Guru
Thank u for fast and good reply

This is a very simple filter. At the cutoff frequency the response is down -3dB which is a small amount. If the cutoff frequency is 500Hz then 50Hz would be reduced but not eliminated. 

Now I'll not ask you about filters and the -3dB, I'll postpone this to another discussion, but I noticed that C1 and the other Rs are not intended as a filter.
You choosed C1 as the smallest possible cap  to couple the two parts of the circuit, and not els.

Only at high frequencies and if there is no load.

At high frequencies, to what range?
To a freq make Xc4 = R5, if so, to f=2258Hz a human voice.

At 15kHz, the reactance of the 150nF C4 capacitor 
is 1070 ohms in parallel with R5= 71.1 ohms.

Xc4 at 15KHz = 70.7 ohm in parallel with R5= 61.5 ohms, not so far.
I know this is not problem, and this occur whe somebody do fast calculation.

Therefore at 15kHz the voltage gain is about 0.707 x 10k/(71.1 + 86.70= 63.4.

Two questions:
(1) why u choose the 15KHz personally?
(2) I understand [10k/(71.1 + 86.70)] but not 0.707
I know that 1/sqr(2) = 0.707
but why multiplying the Av by it, is it a known formula?

At 300Hz, the reactance of ....

this statement answer the #1 question above, 300Hz is the lower limit of the human voice and the 15KHz is the upper limit.

At 300Hz, the reactance of the 150nF C4 capacitor is 53.5k so it doesn't affect the gain. Then the voltage gain is only 10k/(470 + 86.7)= 18. So the treble boost at 15kHz is about 3.5 times.

The idea is very clear and you provide me with a very good and new information, but to others who may read our discussion, I tell them GURU has some headache because he answring hunreds of questions every day, GOD save guru from any bad thing.

At 300Hz, the reactance of the 150nF C4 capacitor is 3.5k
3500//470 = 414 ohm
Then the voltage gain is only 10k/(414 + 86.7)= 20. So the treble boost at 15kHz is about 3.2 times. NO DIFFERENCE

GURU you are good man and "Electronics Lab - Community = GURU"
Thank u for every word you write to help me.

 

[audioguru2]

You choosed C1 as the smallest possible capView attachment 40156

 

[walid1]

Hi
I'll back to the first design; C_E without CE.
I think in the past that increasing the value of Ic have better property to get higher o/p power,
and I was prefering -according to this- to design circuits with higher Ic values
GURU explained that this was a mistake and from my discussions with him I found that the
disadvantages of this selection:
1-consumption of the battery rapidly
2- more Ic = less hfe
3-i/p impedance to less with more Ic This leads to download the previous stage
My questions :
(1) Are there any other disadvantages?
(2) when I have to choose higher values for Ic?