BJT biasing formulas

[Alun]

I was going to tell walid about your new computer but I thought that I should ljust et you speak for yourself. ;D

I imagine it depends on where you live,  PCs both here in the UK and Canada are quite cheap but I don't now about Gaza-Palestine, I think they might be just a bit harder to get hold of over there.

Hi Alun

I miss u ....

Thanks, don't worry I will return to the origional subject, just not now as I'm not in the mood to do calculations, I'll see how I feel tommorow.

 

[Alun]

Alright Walid,
I've converted the document to pdf for you, I opened it in OpenOffice 2 and it read perfectly then I exported it to a .pdf. I find it quite ironic  how it won't read in Microsoft's Word 97 but it reads perfectly in Sun's OpenOffice.

class_A_preamp_v.0.1.pdf

 

[audioguru2]

Thanks for translating the document, Alun.

Hi Walid,
I wish your schematic wasn't a negative pic.

1) Your R1 and R2 divider doesn't have enough current. R1 can't supply the base current without even having R2. Your error was caused because you selected a Thevenin parallel combination of R1 and R2 to supply a current equal to the base voltage, but the divider's current must be much more than the base current. I use a ratio of 10 times for Idiv/Ib.

2) Your choice of an old BC548C transistor limits your purchase since few manufacturers still make it and few other transistors have such a high curremt gain. I used BC548C transistors with Philips about 40 years ago. They invented it but no longer make it.

3) You choice of 15mA for the emitter and collector currents is very high and cause a calculation for R3 and R4 values that are not common. I would have used 2k for R3 and 100 ohms for R4 or 20k for R3 and 1k for R4 if the load impedance is high enough.
Also, The MCC datasheet has few spec's at 15 mA but has full spec's at 2mA.

4) Your Vbe value must be accurate since the emitter voltage is so low. My MCC datasheet shows typically 0.74V with a collector current of 15mA. The difference of 80mV from your 0.66V causes the emitter and collector currents and to err by 5.6mA which is more than 37%. The collector voltage will also err by more than 37%.
Also, the Vbe spec has a range of Vbe that isn't spec'd at the high collector current of 15mA. The range is from 0.55V to 0.7V at a collector current of 2mA.
Therefore a bypassed emitter resistor must be added to raise the emitter (and base) voltage so that the range of Vbe causing a change of the emitter and collector currents is insignificant. ;D

 

[autir]

Alun thank you for translating the .doc file.

Hi Walid,
I wish your schematic wasn't a negative pic.

Feel free to call me Autir  ;D

1) Your R1 and R2 divider doesn't have enough current. R1 can't supply the base current without even having R2. Your error was caused because you selected a Thevenin parallel combination of R1 and R2 to supply a current equal to the base voltage, but the divider's current must be much more than the base current. I use a ratio of 10 times for Idiv/Ib.

You are absolutely right. I have verified it by running simulations having used several values for R1 and R2. I use Thevenin in a wrong way.

2) Your choice of an old BC548C transistor limits your purchase since few manufacturers still make it and few other transistors have such a high curremt gain. I used BC548C transistors with Philips about 40 years ago. They invented it but no longer make it.

In the past you, among others, have advised me to stick with the BC546-550 family. Why were these issues not pointed out?
BC548s are cheap and readily available where I live and appear a lot in scematics on the Web. I assumed that it was a fair choice.

3) You choice of 15mA for the emitter and collector currents is very high and cause a calculation for R3 and R4 values that are not common. I would have used 2k for R3 and 100 ohms for R4 or 20k for R3 and 1k for R4 if the load impedance is high enough.
Also, The MCC datasheet has few spec's at 15 mA but has full spec's at 2mA.

4) Your Vbe value must be accurate since the emitter voltage is so low. My MCC datasheet shows typically 0.74V with a collector current of 15mA. The difference of 80mV from your 0.66V causes the emitter and collector currents and to err by 5.6mA which is more than 37%. The collector voltage will also err by more than 37%.
Also, the Vbe spec has a range of Vbe that isn't spec'd at the high collector current of 15mA. The range is from 0.55V to 0.7V at a collector current of 2mA.
Therefore a bypassed emitter resistor must be added to raise the emitter (and base) voltage so that the range of Vbe causing a change of the emitter and collector currents is insignificant. ;D

You are absolutely right.
Thank you very much! 

 

[audioguru2]

Feel free to call me Autir

Sorry, Autir.
I was distracted by trying to run away from Walid who might try to kiss me or somethin'.
I was also distracted by thinking that maybe Walid is a pretty girl and about meeting her. ;D

Since I haven't used a BC548C transistor for about 40 years, I didn't know they aren't made by many companies any more until I looked for its datasheet.

 

[Staigen1]

Hey Audioguru, did you use BC548C 40 years ago?

//Staigen

 

[audioguru2]

Hey Audioguru, did you use BC548C 40 years ago?

Hi Staigen,
I started with the BC108 in a metal case which has the same chip as a newer BC548. Then came the BC148 in a strange boxy plastic case and serrations in the pins to lock it to a pcb. Then came the BC548 in a TO-92 plastic case. I also used PNP BC178, BC158 and BC558 ones. The last number being a "7" resulted in a higher voltage rating and a "9" resulted in them being selected for low noise. I still have some of all of them and they still work. The BC179, BC159 and BC559 PNP ones have an extremely low noise. ;D 

 

[Staigen1]

Hehe, yeah, the BC548 wasn't born at 1965, but i belive BC108 was(BC107, BC108, BC109)

//Staigen

 

[audioguru2]

Hi Staigen,
I had a great time working as a nubee engineer for Philips, it was both good and not so good. I got to see and play with all their new stuff like their compact cassette recorder/player and LEDs.

A friend in accounting wanted me to fix his electronic organ. I never saw one before but fixed it in a few minutes after figuring out how to get it apart. I was rewarded with a big bottle of fine cognac.

One Saturday morning I got a telephone call from the president of Philips Canada. He wondered if I would be interested in fixing a European-built Philips colour TV he gave to a friend, before the ballgame that day. I never knew the guy, never looked inside the TV before, didn't have much test equipment nor had a schematic. I worked in Car Radio Engineering, not in their TV production division!
I thought, "Oh goody. If I pull this off I will be rewarded with a big promotion, a big raise or a big gift."
The TV was completely dead.
Lucky #1: I poked around inside and spotted a cracked rectifier diode.
Lucky #2: I had exactly the same rectifier in my tool kit.
Lucky #3: I replaced it and the TV worked! ;D
I asked the maid to inform the owner when he returns that the TV has been fixed courtesy of the president of Philips Canada. I left a message for the president that the TV was fixed.
Later, I got nothing, not even thanks. The president didn't return my calls. What happened? Did my rectifier also fail during the game? I dunno to this day. :'(

 

[walid1]

Hi guys

I learned English through reading Electronic books and articles, so I can talk in this subject. I other sides of English language I'm very weak and depend totally on my dectionary.
I want to tell Audioguru something about his new PC like WAW but the language limit my feeling.
I want to tell Alun that i hope him to be very well tomorrow
I want to tell Alun that I thank him for all the efforts changing .doc to .pdf for me
If it is in arabic i can tell you a good words but I hope you understand me and I'm sure you do.

Alun:"What operating system are you running"
Walid: windows95

Alun:"How much RAM do you have?"
Walid: 48MB, it was 16MB and later I add 32MB

anyway i live with this PC tell i buy anew one pentium 4
thank you Alun.

A final note: I don't know till now what the difference between Autir and autir, please tell me to be aware.
For kissing, In Arab socities, it is a habit to kiss one man another and this as a welcome, anyway i still learn from my best friends.
Tell me audioguru what should I say to tell you that I respect u instead of what i said.

I'm going to study the class A amp of autir to see what we can do with it.

 

[walid1]

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?
thanks.

Staigen gains 289 posts by sending 289 Hehe....hoho haha hehe isn't funny!

 

[audioguru2]

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.

 

[walid1]

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

 

[audioguru2]

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

 

[walid1]

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. 

 

[walid1]

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 

 

[audioguru2]

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

 

[autir]

@Walid:
I don't believe that there is anything wrong with my circuit apart from the wrong application of the Thevenin theorem (that is, regarding calculations. The choice of

 

[audioguru2]

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

 

[walid1]

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.