# Emitter bias feedback with REE?

Discussion in 'General Electronics Discussion' started by 24Volts, Sep 7, 2012.

1. ### 24Volts

164
0
Mar 21, 2010
Hello,

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

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?

Any help / feedback is very appreciated!

Thank you!

regards
r

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Last edited: Sep 7, 2012
2. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,251
2,703
Jan 21, 2010
look at the range of hfe for that transistor.

What value did you use when doing your calculations and what are the limits based on the range of hfe?

3. ### 24Volts

164
0
Mar 21, 2010
Hi Steve,

It seems that the Hfe of this transistor is between 35 and 300. Here is the spec sheet:

However, I measured the Hfe for this particular transistor and it is 156!

But what I don't understand is the difference between the calculated Ie and the measured Ie?

See the attachment T1-EBF-B which show my calculations!!!

There is a difference of 300 ua.. is this tolerable??

regards
r

4. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,251
2,703
Jan 21, 2010
Do what I suggest and factor in a beta of 300.

Did you measure the beta at the same Ic as this circuit?

Is your input voltage exactly 10V?

Is your Vbe exactly 0.7V? (you won't be able to measure this accurately)

You seem to have calculated Rb when you could have measured it. Perhaps you should use the measured value of Rb and calculate the figure that has most uncertainty?

5. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,251
2,703
Jan 21, 2010
When you try another bias method, you'll find that you can get much better calculated results because the actual results are less dependant on the most highly variable aspects of transistor's specifications.

6. ### 24Volts

164
0
Mar 21, 2010
Hi Steve,

You mean:

Rb = ß[vcc-vbe / Ie - REE - RE]
Rb = 300[(10 - 0.616/ 1.01ma) - 25.7 -492]
Rb = 2632016 ohms

then:
Ie = (Vcc-Vbe) / (Rb/ß) + REE + RE
Ie = (10-0.616) / (2632016 ohms/300) + 25.7 +492
Ie = 1.01 ma

I can never get it as per calculations LOL.... Now with a beta of 300 I measure:

Ie = 0.683 ua ???

No, I took the transistor and plugged it in my VOM and set it to HFE and took the reading from my VOM.

Yes ... exactly 10VDC

While the circuit was on, I measured the voltage drop between base and emitter! This gave me a 0.616VDC drop.

I have the exact Rb of 2,632016 ohms (using some resistors and a pot)... so we can say that I measured it since this is its official value now.

But still, it is a mystery as to why, we can't draw a transistor circuit on paper (without measuring anything) and calculate the values of Rb, Re, and Rc using the mathematical models AND THEN when we build it, the Ie measures as calculated????

I mean why is it when we build these circuits we have to measure here and measure there and then tweak it until we get the precise Ie??? I may as well put a potentiometer in series with RE and turn it until I get exactly 1.01 ma... Having said this then why bother with all those formula when all that's required is to do is wing it????

A little discouraged here.

Thanks for your much appreciated help!
r

Last edited: Sep 8, 2012
7. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,251
2,703
Jan 21, 2010
I think you're failing to get it (or I am)

You do calculations with (say) an expected hfe of 80 and calculate values. You then try it out in practice and the measured currents are not as you expect.

What you do now is you go an calculate the currents that you would expect, using the circuit values as they are now, but for the limits of the range of (in this case) hfe.

You will find you get a range of values for that current. If all works out well, the actual current will fall between these limits.

If you want exactly 1.01mA from this circuit you can calculate Rb for the range of hfe values and then understand that the actual value you need to use will fall somewhere in that range. So you try values (or use a pot) so that you can get arbitrarily close to the desired value.

Getting discouraged by this circuit? Sure. As you're finding out, the biasing is very dependant on the characteristics of the transistor. You'll find that it is also dependant on temperature (hold the transistor between your fingers and the collector current will change).

I imagine the next circuit you come across will have a voltage divider to bias the base. When you look at that, you will notice a difference.

Transistors are never exactly as per the specs. You'll find the specs will give minimum values for some things, ranges for others, and maximal values for still more. Whilst two transistors with the same part number are likely to be similar, they won't be exact. The mathematical models don't allow for that (how can they?)

Part of the art of electronics is in designing circuits where these variations don't matter.

In your case, the exact collector current is probably not too important. What is important is that the transistor is somewhere between cutoff and saturation so that a change in base current can cause the collector current to increase and decrease.

8. ### 24Volts

164
0
Mar 21, 2010
Hi steve,

Man oh man! did I learn this the hard way!! LOL!

r

9. ### john monks

693
1
Mar 9, 2012
24V, where did you get the formula (26mv/Ie)?

10. ### 24Volts

164
0
Mar 21, 2010
Last edited: Sep 10, 2012  