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autir

BJT biasing formulas

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You don't answer this question that i asked before;
The voltage gain of this amp = RC/RE, where Re not bypassed.
What would this gain be if RE bypassed?

The voltage gain of a common-emitter transistor stage is RC/(Re+RE). If RE is bypassed or not present then the gain is RC/Re. The output will be very distorted without negative feedback provided by RE.

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IF i have a class a amp with o/p z = RC = 2K ohm and
there is a second stage connected to it
what the best Zin value of this second stage to not loaded the 1st one
is it true Zin2 >= Zo1

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Hi Walid,
The input impedance of the second stage is in parallel with the RC of the 1st stage, so an input impedance for the 2nd stage of 4k will load-down the 2k of the 1st stage so its gain will be reduced to 0.67. An input impedance for the 2nd stage of 10 times the RC value of the 1st stage results in a gain reduction to 0.91 which isn't noticable to be heard or seen.
If the impedance equals the resistance then the gain is reduced to exactly half.

It is tricky to calculate the gain reduction caused by loading-down when the value of the coupling capacitor is too small for the frequency being analysed. As the frequency gets lower, the increasing reactance of the coupling cap increases its loss from coupling (highpass filter), but also decreases its attenuation of the 1st stage from loading-down. ;D

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I read carefully autir's design and audioguru's reply and my comment is:

If I disregard the low values of R3&R4 for a moment I can say that autir do a good job in organizing this subject and make it more direct.

I'll continued autir's design after these lines in the .pdf file (age 2):
By Kirchoff ’ s second law,the voltage drop across R2 will be VBB=VBE+IC * R4 ==>
VBB=0.87 Volt .

then,

according to audioguru's theorem: Idiv= around (10 * IB) ==>
Idiv = 10 *25uA =0.25mA
so, R1+R2 = 9V/0.25mA = 36K  .....(1)
VB = 0.87 V (from autir) ==>  0.87 = 9R2/(R1+R2) .....(2)
solving these two equations we get:
R1 = 32.52 K  and R2 = 3.48 K , you must round them to a practicle Values.

Now, autir choose Av=20 and IC=600*25uA = 15mA. I think it is an o/p stage because of this high gain and current, isn't so?
Values of R3&R4 are not good as audioguru said and autir may choose 2k and 100ohm.
to do this with keeping Av=20 and Ic=15mA, VB must increased:
VE= 100*15mA=1.5V, so VB = 1.5 + 0.66 = 2.16V, use this value instead of 0.87 in eq(2) above you get:
R2=8.64K and R1 = 27.36K and you still keeping the 10:1 divider and your initial conditions.

Note: I'm not trying to teach you, simply i repeat what u teach me, to be more rightful, i repeat what audioguru teach me, hope u correct me if there is some mistake.
thank you all.

note (2): Audioguru, you teach your dog somethings wrong. 

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yes autir, Arab food has these "side-effects" but appear after 40 years, so be aware to not have any dog in your house.
I hope u enjoy with arab food and thank u for these feelings.
I'm from Gaza-Palestine 

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Hi Walid,
1) When you increased the value of R3 and R4 to 2k and 100 ohms, you should have also reduced their current. 15mA through 100 ohms produces 1.5V like you calculated, but 15mA through 2k is 30V. The battery is only 9V. Therefore the transistor is saturated with its emitter and collector voltages near 0.43V. So the base voltage on your circuit is too high without a bypassed RE added, which would reduce the current.
2) My wifey and I produce those "side-effects" (turn-ons) on each other.
3) My dog doesn't understand these things. I play with my dog and my dog gets excited. I play with my wifey and she gets excited. They are understood to be the same by my dog. ;D

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I didn't "decide" to use a 10:1 ratio of divider current to base current. The -50%, +100% range of current gain and the temperature change of current gain "dictated" to use that ratio so that the transistor's DC operating point doesn't change too much with different transistors having the same part number and also doesn't change too much with temperature change.

I don't think that Thevenin's theory should be used to select biasing resistors because the ratio of the actual resistors is determined by the circuit's emitter voltage percentage of the supply voltage, which is determined by its RC/RE ratio. ;D

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Hi

lets do it again, I'll design a new one close to autir's:

Decisions:
Vcc=9v, 2N3904 transistor of hfe=230 typically and VBE=0.65v
Av=20, VCE=4.5V, IC=10mA

Calculations:
IB=10m/230=43.5uA, Idiv=10*IB=0.435mA

Rc=20RE, VCC-VCE=IC*(RC+RE) ====> RC=428.6ohm and RE=21.5ohm

VB=VBE+IC*RE  =  0.864V
R1+R2=VCC/Idiv,  VB=VCC*R2/(R1+R2) ====> R1=18.71K, R2=2K.

please correct me or tell me it is agood work

yours,
walid.

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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

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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

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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

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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 ...

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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

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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]

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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

post-2833-14279142888884_thumb.jpg

post-2833-14279142889_thumb.jpg

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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

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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.

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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

post-2833-14279143242289_thumb.jpg

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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:

post-1706-14279143242645_thumb.png

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