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# BJT biasing formulas

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Hi audioguru
u told me before that IE= 0.55mA was wrong and IE actually = 0.45mA
please, make another figure like the first one marked "s transistor.png " with the wright numbers.

I made a minor mistake near the beginning of my calculations so IE is still 0.55mA but the collector voltage is roughly 4.5V instead of the desired 5.5V. I tried changing the 330k bias resistor to 360k, changing the 47k bias resistor to 43k and other combinations when I realised that when the base current is included in IE and the numbers aren't rounded off, then the actual collector voltage will be near 5.0V which will be fine.
I will leave re-calculation for you to do if you want since 5% resistors will change the voltage about 0.5V anyway. ;D

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Hi
If Zin=R1//R2//(hfe*(re+R4)), where re=25mV/IE, then
re = 25/0.55=45.45
Zin = 330k//47k//(230*(45.45+1000)
Zin = 330k//47k//240.45k = 35.13K as audioguru said 34.9K

Augioguru, why u think this slight difference. I tryed all combinations of 26 instead of 25 and 0.45mA instead of 0.55mA and the results are more difference.

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Forget about only less than a 1% difference, the tolerance of the resistors and coupling cap will be 5% won't they?
Years ago I was taught 26mV, but new transistors' datasheets show it is wrong.

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Dear Audioguru,
This may be the last question to complete this circle of questions, but i expect u also want to finish this.
you said this statement: "A 10:1 ratio in a voltage divider doesn't allow the divided voltage to change much when a low gain transistor causes a 5:1 ratio or a high gain transistor causes a 20:1 ratio.
The resistors could also be 360k for R1 with 51k for R2."
1)I understand from this that you are usually take a 10:1 ratio but you may go slightly up or down. Can I take this as a rule in my own design?
2) You said also:"doesn't allow the divided voltage to change much when..."
Are you mean by this "divided voltage" that the VB which now = 1.2V doesn't change much by the applied AC signal, and WHY?
3) if R1=330k and R2=47K, then, the divided voltage VB = 10 * R2/(R1+R2)= 1.25v and ID = 10/(47+330)=26.5uA and the ratio is 11, where
if R1=360k and R2=51K, then, the divided voltage VB = 10 * R2/(R1+R2)= 1.24v and ID = 10/(51+360)=24.3uA and the ratio is 10.
and from all that we can concude again that the raio must be around 10.
I love you Audioguru I feel that i'm very close to the truth.

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you said this statement: "A 10:1 ratio in a voltage divider doesn't allow the divided voltage to change much when a low gain transistor causes a 5:1 ratio or a high gain transistor causes a 20:1 ratio.
The resistors could also be 360k for R1 with 51k for R2."
1)I understand from this that you are usually take a 10:1 ratio but you may go slightly up or down. Can I take this as a rule in my own design?

My rule always works, the rule in your tutorial fails when the transistor's gain is low like in power transistors. Take your pick which rule to use.

2) You said also:"doesn't allow the divided voltage to change much when..."
Are you mean by this "divided voltage" that the VB which now = 1.2V doesn't change much by the applied AC signal, and WHY?

You don't understand that the bias voltage divider is for the transistor's DC operating point. AC won't be affected if the coupling cap is big enough and the transistor has its DC operating point correct.
When the transistor's gain is low at half, its bias current is doubled which reduces the base voltage due to the extra current subtracting from the current in the lower divider resistor and adding in the upper divider resistor. Since the base voltage is lower, then the emitter voltage and current are also lower and therefore the reduced current in the collector resistor increases its voltage.
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I love you Audioguru

Ehmmm.... guys?!?  ;D
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Ehmmm.... guys?!?
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I have designed a class A preamp from scratch and written the procedure in a Word document. I would feel very obliged to all that can read it and point out any mistakes/ misconceptions etc.

class_A_preamp_v.0.1.zip

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The voltage gain of this amp = RC/RE, where Re not bypassed.
What would this gain be if RE bypassed?
thanks.

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Hi Autir
I downloded this ZIP file and when opening it in WORD97, the document looks like this:

EMBED Equation.3
14,28 Ohm.
R3=20
EMBED Equation.3
EMBED Equation.3
EMBED Equation.3
285 Ohm.
By Kirchoff's second law, the voltage drop across R2 will be VBB=VBE+IC
EMBED Equation.3

thanx.

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This is exactly why I strongly discourage the use of the word document format as it isn't a proper oficial industry supported standard.

autir,
Exchanging documents in word format is generally a bad idea as they can transmit macro viruses, often not read reliably on other computers, and can be hard to open on non-Windows machines.

Try converting it in a differant format like .rtf or .pdf

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Hi Alun

I miss u ....
I this have petium 1 133MHz PC since march 1997, it is 1300\$ that days, today it is less than 100\$.
I can't buy new one pentium 4, not only for money but because this PC satisfies all my requirements with less problems.
Audioguru, have 486 PC
So i agree with u that AUTIR replace his file with .pdf
thank you.

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walid,
What operating system are you running? if it's Windows 2000 SP 4 or later then MS Word 2003 view should work.

How much RAM do you have? OpenOffice 2 requires 128MB and is more modern than Word 97 or go for OpenOffice 1.1.5 if you've only go 64MB and if you've don't even have that then why not give AbiWord a go? All of the aforementioned are at least 99% MS compatable, this only things that normally cause problems are macros and some complex layout formatting and be aware also that macros can spread viruses so it's not big loss them not being supported.

Most people don't give much thought to what productivity software they use, they normally just stick with MS Office as it's what "everybody uses" and it's what they're are used and they are blissfully unarware of the alternatives which are often cheaper and better in some cases. I think this is a shame as there's lots of money to be saved and they should choose productivity software carefully as it's what they're going to be using most of the time.

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The hard drive on my 486/100MHz pc broke on Dec 26/04. On Dec 27/04 I bought a Pentium4/2.93GHz name-brand pc and 19" LCD monitor on sale with many manufacurers' rebates. On Dec 28/04 I got most of my money returned to me because I spotted an ad with a better price and my store guaranteed to beat any price. They were shocked to see that their own ad on the internet had a much better price! When the rebate cheques didn't arrive, I complained and got them 2 days later in the mail. I gave the very nice "photo" printer it came with to my daughter but I could have sold it. The amplified speakers it came with are so good that I haven't bothered to connect the pc to my stereo only 3m away. The CD and dual-layer DVD burners work well and are very fast.
I bought this very good pc nearly for nothing! I was lucky to have good timing.

The accurate voltage gain calculation of a common-emitter transistor stage is RC/(Re+RE). Re is inside the transistor and usually limits the max voltage gain available (with RE bypassed) at a certain supply voltage and with a collector resistor. The voltage gain can be increased with a high supply voltage because then a higher value can be used for the collector resistor. The voltage gain can be increased a lot by using a current-source or bootstrapped collector resistor to replace the collector resistor and an emitter-follower to buffer its very high impedance. Audio amplifier and opamp ICs use these methods.

My MS Office XP Professional program messes-up when I open docs with schematics so I didn't bother trying to see yours.

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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.
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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

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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

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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!  :D
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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.
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Hey Audioguru, did you use BC548C 40 years ago?

//Staigen

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

//Staigen

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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."
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. :'(

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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.

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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!

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