Fixed Bias Transistor help.

Discussion in 'General Electronics Chat' started by Thecomedian, Dec 12, 2013.

1. Thecomedian Thread Starter New Member

Oct 12, 2013
22
3
Attachment shows the circuit I was working with. I chose the resistances to get 1mA for C and ~90uA for B, which should be 100 gain. The resistances were also chosen to have a voltage drop down to 1 volt on B.

However, the voltage on C is only 30-40mV, which means the junction B-C isn't turned on. What am I missing?

• problem.jpg
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Apr 5, 2008
15,803
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Hello,

The RB to RC values are about 10 to 1.
This will drive the transistor into saturation.

Doing the calculation for Ib the value will be ( 10 - 0.70 ) / 100000 = 93 μA.
If β would be 100 the collector current would be 9.3 mA, that is not possible as the powersupply is to low or the collector resistor is to high.
( the calculated voltage over on the collector resistor would be 9.3 mA X 9300 Ohms = 86.49 Volts ).

The 30- 40 mV you see on the collector of the transistor is the saturation voltage.

Bertus

3. Thecomedian Thread Starter New Member

Oct 12, 2013
22
3
Ah, thanks.

http://www.oddmix.com/tech/px/transistor_amplifier_fix_bias_f1.gif

I copied this and actually get an amplification with a sine wave, albeit with some distortion. In this case the ratio was increased 200:1 from 10:1. I'm still trying to figure out what the rules are for biasing a transistor properly while also using the resistors to get the proper volt/current across B and C... I've been all over the next for months and there doesnt seem to be a simple explanation.

4. nigelwright7557 Senior Member

May 10, 2008
488
71
You can use 2 base resistors to get fixed bias mode but a emitter resistor is recommended for that.

I have seen the base resistor taken from the collector which will give negative feedback and set the bias for you.

You also need a coupling capacitor to keep the DC levels correct.

5. LvW Active Member

Jun 13, 2013
674
100
Perhaps your start is - sorry - completele false?
You have chosen "the resistances to get 1 mA" collector current?
How did you make this choice? (Hopefully not based on Ohms law).
From many discussions I know that some people beliefe the collector resistor Rc would set the current Ic ("to limit Ic").
Therefore, my comment above.

by the way: The correct procedure is simple and straight-forward:
1. Select a value for Ic,
2. Calculate the corresponding Ib and a suitable value for Rb (yes - in this case with Ohm`s law). by doing this, assume a dc voltage Vbe=0.65 Volts
3. Select a value for Rc that allows at least 3...4 Volts across the transistor (Vce).
________________
This should work -however, stability of the dc operating point (Ic, Vce) is rather poor. The alternative mentioned by nigelwright (base resistive divider) offers better stability.

6. Thecomedian Thread Starter New Member

Oct 12, 2013
22
3
Okay, I had calculated (Vcc - forwardVdrop(Vbe)) / (1/100,000) to try to get 10uA on Base, and came out with 860k resistance for Rb, but I also chose the Rc based on the same calculations and came up with the 9300 resistance. I changed the first calculation for Rb to (10-9) thinking of getting 1 volt at Base at 10uA, but that didn't quite work, obviously. So I tried to set 100k resistance and was stuck from there.

Using 860k, I can get a working distorted sine with Rc @ 9300. An almost undistorted one Rc @ 1k. what's odd to me is the 9300 has about 94mV on collector for DC operating point. At 1k, I get ~6V. I suppose this major difference has to do with the fact that the internal resistance of a transistor is quite low, so it forms a voltage divider with the 1k that's better for a half-supply operating point, while the 9300 is way too big compared to the transistor internal resistance. Would this be correct?

7. MikeML AAC Fanatic!

Oct 2, 2009
5,451
1,066
Mr Funnyman,

Here is a simple way of biasing a transistor. I am showing you how to analyze what is happening using LTSpice.

Note that I made only a simple change to your original circuit; namely that the bias resistor is connected to the collector of the transistor; not to the positive supply rail. This provides some negative feedback to the bias circuit and is useable.

It is not the best way (look-up stabilized bias) biasing a transistor, but it works reasonably well, as I will show:

First, I am using LTSpice to find what value R2 needs to be. In the first sim, I am doing a DC solution for a range of possible values for R2. Note the statement .step param bias 100K 2meg 100K. This computes (and let's us plot) the collector voltage V(c) and the collector current Ic(Q1) for stepped values of R2 between 100K and 2meg. Note that at a value of R2=1megΩ, the collector current and collector voltage is biased to a value where the transistor could be used as an amplifier, because the collector voltage can swing up and down from the 3V value.

Second, I wanted to see how the 2n3904 biased with the 1megΩ base resistor R2 would work as an amplifier. I had to put a coupling capacitor between the base and the signal source so that the source does not short the bias current to ground.

I made the input signal V(in) only 4mV, and swept it from 1Hz to 1megaHz in the middle sim (AC analysis). Note that V(c) is 0dbV, meaning it is 1Vpp, while the input voltage plotted on a db plot is -48db, which means that the gain is 250 (+48db). Note the effect of the input coupling capacitor below 100Hz.

Finally, in the third sim I show one of the limitation of this simple means of biasing. I am doing a Time Domain sim, with an input of a 4mVpp sine wave at 1000Hz, but I also vary the temperature from -50 to +50 deg C. Note that DC offset at V(c) is effected by temperature. Also notice that the gain is slightly effected (the peak to peak voltage changes).

This shows that this simple bias method is somewhat effected by temperature. It would also be effected by the β of the particular transistor. The more complicated "stabilized bias" method was devised to overcome these shortcomings.

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8. LvW Active Member

Jun 13, 2013
674
100
I had some problems to really understand what you were doing.
For example: How could you do "the same calculation" for Rc as for Rb?
As I have mentioned in my former posting:
* Rb can be calculated by Ohms law (as you did correctly using 9.3 Volts across Rb)
* However, the corresponding Ib belongs to a corresponding Ic - INDEPENDENT on the Rc value. The BJT acts as a voltage controlled CURRENT SOURCE and you have to select (SELECT!!!) a suitable Rc value that allows approximately Vcc/2 for Rc and the other half across the BJT (Vce).

Of course you also can perform a simulator-aided design, but I think it is not necessary for such a simple circuit. But it is true that connecting the resistor Rb (of course with another value for the same Ib) at the collector node improves the stability of the operational point. In this case, your calculation starts with the selected dc voltage at the collector node (instead of Vcc).

9. BobTPH Active Member

Jun 5, 2013
806
121
You have to also remember the beta varies from one device to another of the same type and varies with colllector current. This is why the emitter resistor and 2 resistor biasing method is better, it reduces the dependence on beta.

Bob

Last edited: Dec 13, 2013
10. shteii01 AAC Fanatic!

Feb 19, 2010
3,504
512
Rules? You are trying to figure out? Seriously?

There is a whole CHAPTER in my textbook on how to bias transistor.

You need to get a textbook in the library and do some reading.
Or.
You can continue stumbling in the dark.

11. ian field Distinguished Member

Oct 27, 2012
4,447
791
You'd be better off taking the top of the bias resistor to collector, this compensates gain variation and temperature changes.

If for some reason the collector current becomes higher than you designed for, collector voltage falls and so does the current through the bias resistor.

Philips used this biasing arrangement almost exclusively in their Electronics Engineer construction kits - for the most part it usually worked well.

To get a decent amount of gain; you can split the collector to base bias resistor in two and create a tapping point - make the top resistor about 5x the collector resistor and AC shunt the tapping point to GND with a capacitor.

If you take the base bias resistor straight up to Vcc, you'll probably find it next to impossible to set the operating point - and that's before the transistor gets up to any mischief!

12. LvW Active Member

Jun 13, 2013
674
100
As mentioned by shteii01, as far as BJT biasing is concerned - there is a lot of information, principles, formulas, rules, recommendation,...which fill a complete chapter of a textbook.

However, the goal of all these alternatives is
* to fix a dc operating point on the non-linear Ic=f(Vbe) characteristic which allows to swing around this bias point without introducing too much distortions;
* Thus, this bias point must be located in the quasi-linear region of this characteristic.

Therefore, afer selection of a suitable collector current you have to calculate the parts values for the selected biasing scheme.
Why different biasing principles?
There are some vague assumptions in your calculation - the most uncertain parameter is the current gain B (or beta) which is given in the data sheet with a tolerance as large as 200% (or even more). In addition, the current gain is rather sensitive to temperature changes.
Therefore, you should try to use a biasing principle which is relative insensitive to these uncertainties.
And this leads to the best solution:
Resistive voltage divider at the base node and a feedback resistor Re in the emitter path.
Rule of thumb: Re~0.1*Rc and Ir~(6..10)*Ib.
Ir is the current through the upper resistor of the base biasing resistor chain.

Remark: As an alternative solution, you can follow the guidelines as given by ian_field above.