# Emitter bias feedback with REE?

Discussion in 'The Projects Forum' started by rougie, Sep 6, 2012.

1. ### rougie Thread Starter Active Member

Dec 11, 2006
410
2
Hello,

Today I have been experimenting with emitter bias feedback method for biasing a transistor. Please locate in the following link:

where it describes the 1st attempt of calculations as:

β = 100 and Ic = 1.01ma

and

β = 300 and Ic = 2.76ma

in the "emitter bias" section.

When I redid the circuit and calculated it with the real values as shown in T1-EBF-A.png attachment, the measured Ie (1.875ma) was not equal to the calculated Ie (1.54ma)?

Therefore I thought of recalculating it while applying REE calculations. As you can see at attachment T1-EBF-B.png, the calculations resulted as being further erroneous at 1.01 ma???

In summary, the calculations are somewhat close but still off from the actual measurements. Please note, that in my schematic I made VBB the same source as VCC.

Does anyone see where I went wrong?

Any help / feedback is still very appreciated!

Thank you!

regards
r

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2. ### rougie Thread Starter Active Member

Dec 11, 2006
410
2
I apologize for this question, as I figured it out!

Regards

Last edited: Sep 6, 2012
3. ### rougie Thread Starter Active Member

Dec 11, 2006
410
2
Hello ...

I seem to have gotten much closer than I did in the initial post.

When I did the circuit and calculated it with the typical values as shown in attachment T2-EBF-A.png (left side), the Ie calculated value was 0.101 ma but when measured with the actual component values, Ie was1.31ma.

I redid this circuit calculations taking into account REE as shown in T2-EBF-B.png and the same discrepancy occured.

Is it normal to have such a difference between theoretical and practical measurements?

Thank you!

regards
r

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4. ### Audioguru AAC Fanatic!

Dec 20, 2007
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Beta is not a fixed number. For a transistor part number it is a range of numbers maybe from 30 to 900. Did you guess that your transistor has a beta of 100? It is probably not.
Beta also changes when the temperature changes.

5. ### rougie Thread Starter Active Member

Dec 11, 2006
410
2
Audioguru, the 100 has been taken from:

and all I did was redo their calculations. But then in reference to the right side of the T1-EFB-A.png file, I measured it and it is officially 156... please view for yourself at attachment below.

And then I redid the calculations with the correct beta, Vbe, and measured values of the resistors!

Isn't that the way we calculate the circuit? Look at both attachments since the last attachment shows the calcs taking REE into account... but I am off by approximately 300 ua??

r

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6. ### Audioguru AAC Fanatic!

Dec 20, 2007
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I doubt that your meter measures hFE at the same collector current as used in the article. The hFE is different for each transistor (even if they have the same part number), is different at different currents and is different at different temperatures.

Your circuit should use negative feedback to minimize the differences.

7. ### rougie Thread Starter Active Member

Dec 11, 2006
410
2
Yes, I know every transistor has a different beta based on many factors.... But what I want to know then Audioguru, how am I supposed to calculate Ie of the circuit shown in my attachments above.

Can you please fill in the blank for 'B':

Ie = (Vcc - Vbe) / (Rb/B) + REE +RE
Ie = (10 - 0.616)/ (1.4M/ B) + 25.7 + 492

B *must* have a number right?

SO, FOR THIS TRANSISTOR AT THIS TEMPERATURE AT THIS CURRENT AT MY OFFICE RIGHT NOW:

Calculating the unknown B using the known measured values with:
Vcc-IbRb-Vbe-IeRe = 0

and using:
B = Ic/Ib

Solving for B gives me 205...
and the next pn2222 transistor i will use might be 145, but the same RE will make it so that Ie won't vary much...

I don't know what else to say!

thanks anyways
r

Last edited: Sep 6, 2012
8. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,168
1,188
First of all REE = re = Vt/Ie = 26mV/Ie but this resistance is a AC signal resistance (dynamice resistance). So for DC re = 0Ω.
As for the emitter current.

Vcc - Ib*Rb- Vbe - Ie*Re = 0 (1)

and

Ie = Ib + Ic = Ib + β*Ib = Ib *(β+1) (2)
Or

Ib = Ie/(β+1) (1)

Ie = (Vcc - Vbe)/( Rb/(β+1) + Re )

As for β measurement. Simply measure voltage across Rb resistor and divide this VRb voltage by Rb

Ib = VRb/Rb

Measure voltage across Re resistor and Ie = Ve/Re

And finally β = (Ie/Ib - 1)

9. ### rougie Thread Starter Active Member

Dec 11, 2006
410
2
Hello Jonny130,

In reference to the following link:

Within the *DC* emitter bias configuration analysis, it is quoted that:

Based on the above quote, I really thought that REE was applicable to my DC emitter bais configuration calculations!!!

But you are telling me that:
For DC REE = re = 0Ω.

Just wondering here.... correct me if I am wrong... you seem to be saying one thing but there saying another?

Okay, I didn't think of it this way!! I will try this measurement tomorrow and see if I get better accuracy for Ie calculated vs Ie measured.

r

10. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Well for me re is a small-signall resistance between base and emitter when we looking into emitter with base short to ground (short for AC signal via capacitor) re = dVbe/dIe.
And by a small-signal I mean that BJT is already properly bias and work active region. So for me it is a mistake to mixing a small-signal parameters with DC current one.

11. ### Jony130 AAC Fanatic!

Feb 17, 2009
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When we designs the circuit we don't have to know exact value for Hfe.
All we need to know is a minimum value for Hfe from datasheet.
And instead of one RB resistor we use two resistor in voltage divider connection.

Now let as find Ib current. From the II Kirchhoff's law we can write

Vcc = I1*R1 + I2*R2
(1)

I1 = Ib + I2 (2)

I2*R2 = Vbe + Ie*Re (3)

Ib = Ie/(β+1)
(4)

And we can solve this for Ib.

$Ib = \Large\frac{R2Vcc - Vbe(R1+R2)}{(\beta +1)Re(R1+R2)+(R1R2))}$

And if we want to design this circuit we start selecting Ve voltage.
Ve voltage should be large or equal Vbe to improve temperature stability.
So for Ve = 1V we have Ie = 1V/Re
Voltage at base will be equal to Vb = Ve + Vbe ≈ 1.65V.
Now we can pick voltage divider resistors.
One of the way to become independent of hfe changes is to select voltage divider resistor in such a way that base current don't load our voltage divider to much. The maximum base current is equal to Ib_max = Ic/Hfe_min where Hfe_min is a minimum value for Hfe in datasheet.
And we select voltage divider current so that I_divader >> Ib_max.
I_divider greater than 30....10 * Ib_max is enough safety margin.
I_divider = I2

R1 = (Vcc - Vb)(I2 + Ib) =
(Vcc - Vb)/(31*Ib) (Vcc - Vb)/(30*Ib)

R2 = Vb/(I2) = Vb/(30*Ib)

If we meet this conditions I_divider >> Ib then Ie ≈ ((Vcc*R2/(R1+R1)) - Vbe )/Re
And now Ie is almost totally independent of Hfe.

Of course today no one except hobbyist and students use and design such a simple amplifier.

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12. ### rougie Thread Starter Active Member

Dec 11, 2006
410
2
Hello Jony130,

you meant this right:
.....................v
R1 = (Vcc - Vb)/(I2 + Ib) = (Vcc - Vb)/(31*Ib)≈ (Vcc - Vb)/(30*Ib)

Now its clear! I don't understand why no tutorials/videos/articles mention this.

In another forum, another fellow said the same thing you said except he suggests to pick say a value of 80 HFE ... and go from there. I don't know if this would make any sense though!

In any case, Jony130, what you have just explained here, should be mentioned in the "biasing calculations" section of the tutorial. In particular this phrase:

This would of cleared up a lot of gray zones I had about transistors from the start!

Is it okay if one wants to use the emitter and collector bias configurations instead for designing circuits?

A sincere thanks for your transparency !

regards
r

Last edited: Sep 8, 2012
13. ### Audioguru AAC Fanatic!

Dec 20, 2007
9,418
903
A transistor with simple one-resistor collector bias has a very low input impedance due to the negative feedback from the collector to the base. Also its voltage gain is affected by the output impedance of the signal source.

A transistor with voltage divider bias does not have these problems.

14. ### rougie Thread Starter Active Member

Dec 11, 2006
410
2
So does this mean that if I would be designing with transistors for the sole purpose of signal amplification signals, I should just focus on the voltage divider bias configuration?

r

15. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Yes, you shoudl focus on the voltage divider.

But also you shoudl know that we can remove negative feedback for AC signals from the collector to the base by splitting RF resistor and adding to the circuit one capacitor.
And then the input impedance is equal RB1||(β+1)re

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16. ### rougie Thread Starter Active Member

Dec 11, 2006
410
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Okay, a few last things....

in the following formula:

Ie = (Vcc - Vbe)/( Rb/(β+1) + Re )

in the β+1, can we omit the "+1". In the tutorial, they leave it out as !!!

Ie = (Vcc - Vbe)/((Rb/β)+1 + Re)

The thing is, it wouldn't make much difference in the calculations if we leave it or not...!

Also on another topic,

I eventually, require to have a small ramping voltage from 0 to 3.3 VDC at the base of a transistor. The output is a varying current flowing from the collector to the emitter of 0 to approximately 80ma.

To do this, I can simply use the emitter bias feedback configuration ... right! I wouldn't really need a voltage divider at the base since voltage at base is provided from the output of a DAC!

Regards
r

Last edited: Sep 8, 2012
17. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,168
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Hmm

Rb/β = 1000/100 = 10

Rb/(β+1) = 1000/101 = 9.9009

(Rb/β)+1 = (1000/100) + 1 = 10 + 1 = 11

Rb/β = 100/100 = 1

Rb/(β+1) = 100/101 = 0.99009

Well yes if you want to design a voltage to current converter from 0A to 80mA you don't need a voltage divider.
All you need is a Re resistor Re ≈ 0.8V/80mA = 10Ω.
Also don't forget about power dissipation Ptot = Ic * Vce in BJT. 80mA it is a lot of current for small BJT which can lead to destruction of a BJT.

18. ### rougie Thread Starter Active Member

Dec 11, 2006
410
2
oooops ... Sorry Jony130, I made a mistake on the way I asked the question.... Here I go again:

in the following formula:

Ie = (Vcc - Vbe)/(Rb/(β+1) + Re )

in the β+1, can we omit the "+1". In the tutorial, they leave it out as:

Ie = (Vcc - Vbe)/((Rb/β) + Re)

The thing is, it wouldn't make much difference in the calculations if we leave it or not.

======================================================

So, if Rb = 100 and β = 100 and Re = 10
Just focusing on the second part of the equation ... this part >>> "(Rb/(β+1) + Re )"

Then using the top one (as you use) we have:

(Rb/(β+1) ) + Re
(100/(100+1)) + 10 = 10.999999999

However using the formula that is shown in the turorials of this site, we have:
(Rb/β) + Re
(100/100) + 10 = 11.0

So there is not much difference... so when I use the above formula to do my calculations can I assume that I can simply use:
Ie = (Vcc - Vbe)/((Rb/β) + Re)

As for my voltage to current circuit, according to spec (If I read this right) PN2222 can have an Ic max of 600ma... I should be okay!

Now I might be going a little ahead of myself here, but the thing is, the configurations that I have been reviewing this past week ex:

base bias,
collector bias,
emitter bias and
voltage divider bias

I was able to calculate Rb by the formulas in the tutorial and by the formulas you gave me earlier. But all this time I always had a constant voltage drop across Rb. However in this new circuit the base input will ramp from 0 - 3.3VDC....More specifically, the voltage will continuously ramp from 0-3.3VDC four (4) times a second! On a ramping voltage like this... should I calculate Rb at Voltage base in /2? For example:

VBin = 3.3/2 = 1.65VDC

Rb = β(((Vbb - Vbe)/Ie)- RE)
Rb = β(((1.65 - 0.7)/Ie)- RE)

bla... bla

Like I say I am going a little ahead of myself.... and I didn't do much transistor examples with AC signal voltage at the base yet.

r

Last edited: Sep 9, 2012
19. ### Ron H AAC Fanatic!

Apr 14, 2005
7,049
659
Your voltage-controlled current source (or sink) will be very nonlinear, because the base-emitter voltage will chsnge a lot over the full scale current change. Not to mention the fact that you will get no current out when the input is less than ≈0.5V, unless you add an offset to the input.
Voltage-controlled current sources usually include an op amp and a transistor in a feedback loop. Post here if you want more info on this.

20. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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I agree. This aspect of the book is based on some curious reasoning which would be difficult to defend.