your welcome

It looks like you got it now,

the examples I gave will help you get into the range of linearity, but there is a lot of design requirements, involved when more paramaters are given.


please post your results so that those on this board can check your work and give additional help.

 

As to your post #4,

you can solve for RB resistor after you choose a voltage value for RE, or a gain value resistor, for RE, so as to solve for VB, there you will have to assume a Beta value, or disclose a value through data books, but here you could see how an amp designed with one base resistor, will be Beta dependant, which is not a good practice for transisistor amp design.

That's why negative feedback such as RE is used to stable a transistor against temp. changes, and Volt divider bias resistors are used to design a stage independant of transistor parameters, like Beta.

The bias resistors are for the purpose of getting a transistor as independant as possible of any inherent pparameters, and temp. changes, so that the design is mostly programmed by the value of resistors chosen.

As yu have or will learn, the bias resistors RC/RE gives a stage gain, again like programing the circuit, for a specific gain, the values of RB1 and RB2, help programs the input impedance to the stage, so as you see chosing resistor values is all part of programming the circuit, to meet requirements you chose to meet.

That's why it is very important to understand why each value is chosen in a transistor stage, it's more than just getting it to work in it's linear region as an amp, but each resistor must work with input and output requirements, stage gains, than capacitors for frequencies, and positive feedback for ac signals, ect...

As you learn these things you see how it all fits together, like the RE is used to fight temp drift, but at the same time it will lower the gain of the amp, so a bypass capacitor, is put across RE to pick up the signal gain, and yet keep Dc drift at a minimum, but the capacitor chosen will also decrease neg. feedback, wich makes the stage unstable, so that capacitor is chosen to match the freqencies of the signal, as well as to keep the gain of the stage in its proper perspective, there is always tradeoffs in component values to make parameters work as close as possible, for a given design requirement.

 

Yes, Hobbyist, that's what we've been learning about in theory classes, and seeing in lab, how voltage divider bias is not dependent on β.

I've been wondering what those caps across Re were for, we haven't covered that capacitor yet. We've only discussed the capacitors and the effects of them on the input and outputs of the circuit.

Well here's what I've come up with:

I went to a proper voltage divider bias circuit. I did this based on the related reading, and that IS the title of the lab, even though it seems that the teacher may actually want an emitter biased circuit.

For reference (on the left): http://mouloudrahmani.com/images/untitled204.gif

All resistor values are E12 standard values, not actual calculated values, since that is part of the exercise.

Rc, as determined by (Vcc-Vc)/Ic = (18V-8.07V)/14.6mA = 680Ω

I chose a value of .5V for Re, to keep the Q-point as close to the middle of the load line as possible (my understanding anyway).

Assuming Ie=Ic=14.7mA:
Ve/Ie = .5V/14.7mA = 33Ω

Vb = Ve+Vce = .5V+.7V = 1.2V

I chose a value of 330Ω for Rb1, 10 times larger than Re as suggested.

Ib = Vb/Rb = 1.2V/330Ω = 3.6mA I thought this value was a bit large, since all previous exercises in class and lab have been in the μA range, but I checked the datasheet for the 2N4404 transistor, and if I am reading it correctly, this is within specified value for Ib.

So then using the formula given by Hobbyist, (Vcc - Vb)/(Vb/Rb1) = (18V-1.2V)/(1.2V/330Ω) = 4700Ω

If I did my Thevenin equivalents correctly, that means that I can use this as both a Voltage divider bias and Fixed Bias simply by removing or adding Rb2.

How's it look?

 

Hi,

I usualy like to actually build the circuit to test it's voltages and waveforms, but Im' not in my shop right now, so I'll simulate it on my compiuter to check for bias voltages and waveforms.

And I'll post the results.

 

Yeah, the assembly and testing will be in lab tomorrow.

If my power supply would go to 18V I would test it now, but sadly it only goes to 15V with a load of only a few mA. I need a store bought power supply that goes to 30V...

Thanks for your help.

 

Hi, give me a little time here, Im' back in my shop, so I will build athe circuit using a 2n3904 transistor, and check bias voltages and even put it througha basic signal test with my osciloscope and audio generator.

 

I just calculated β from the above values, and I'm getting a β of 4 Ic/Ib.

This concerns me, since my teacher has said that a β of less than 50 requires or is a property of a different transistor than what we are dealing with in this class, at the present time.

Most of the time we have seen or used β values of 100 to 200, in theory and calculated from lab experiments.

Should I use a larger value of Ve, to reduce Ib? This will push the Q-point farther away from the middle of the load line if I'm not mistaken, reducing headroom.

No worries on how long you take to test. That's WAY more in depth than I was looking for when I started this thread. lol

 

measured values,

VC ~ 8.8v
VE ~ 0.48v
VB ~ 1.19v

Av ~ 20 @ (2v pk. / 0.1v pk) = 20

used a input coupler cap of 100uf, and Freq, = 4khz NON distortion symetrical waveform.

Circuit tested under NO load.

Good bias design.

 

Your calculated value of beta is using the wrong variable, for IB.

Ib is actually not calculated in the above equations I gave, the current flow through the Rb1 and Rb2 is the divider current (ID)

That current ID the divider current this is the stable current that needs to be at least 10 times greater than the actual base current Ib.

However if you were given a set Beta value to work to, then you would have to rework the RB1 and Rb2 values to make this current at least 10 times larger than a given Ib value.

I posted the results in the above post.

 

Well, since you say it's good, I'll go with those values for now, and if the teacher wants something different, I'll re-calculate.

Thanks again for all your help!

 

Your calculated value of beta is using the wrong variable, for IB.

That current is ID the divider current this is the stable current that needs to be at least 10 times greater than the actual base current Ib.

I posted the results in the above post.

Oh?

Is there a different formula for β when using a voltage divider circuit, than β = Ic/Ib?

I'm all ears, if there is.

I just went through a few ideas of my own and don't seem to see how to get a proper β from this.

 

Your welcome,

What i gave you was basic practical design for a bias circuit, the teacher may be teaching you the ideal design using data sheets and learning from the ground up from there, to have a better understanding of how to design in a practical manner, later on in your course.

 

Beta is Ic/ ib

but if you were to be given a design constraint of design the circuit with a beta of 200 for example, then you would need to use the
(IC / Beta ) to give required Ib,
then you can recalculate the divider current to be at least, 10 times larger.
Which means recalculating RB1 and Rb2, as well.
So the divider current keeps the base current stable, and does not allow the base current to load the divider.

Always remember a good design is trying to avoid depending on Beta values.

But it's best to design the circuit to work equally well with a low value of Beta, as well as a high value, by choosing the right values for bias resistors.


That's the tradeoffs in designing a circuit, you design according to the constraints given.

 

I've just run through all the posts in this thread (somewhat quickly) and I'm seeing a lot of apples and oranges and ships crossing in the night, though I think Hobbyist has done a pretty good job of steering you along.

Six Shooter is posting schematics (that are noted as being the closest he can find) that don't have an emitter resistor (Re). Then he is tossing out equations involving two voltage divider resistors, Rb1 and Rb2, that involve relationships to Re. But elsewhere he specifically talks about there being just three resistors, Rb, Rc, and Re. So what's the story? You can't pick the values for a circuit unless you have a circuit to pick values for!

Here's an idea -- let's all get on the same page as far as what circuit is being discussed. Here is my best guess and I think it is pretty consistent with at least what Hobbyist has been thinking in terms of. It took all of three minutes to throw together in Paint, so it ain't pretty. But at least it gives us a starting point.

Now, you've calculated Rc to be 680Ω. You appear not to have been given enough info to really calculate anything else, so you get to make some "executive decisions" at the risk of perhaps having to redo things later. But, if you UNDERSTAND why you picked the values you did, then you can redo the design pretty quickly. But that means understanding what the equations mean, not just plugging and chugging along.

Now, assuming that the 14.6mA is your Q point, then let's pick a target parameter and work from there. There are several parameters you can pick, I'll pick one that hasn't been discussed, namely the output voltage swing. Let's target the ability to swing ±5V from the quiesent point without taking Vce below 0.5V. Our quiescent output voltage is basically 8V, so we want the output to go from 3V to 13V. At 3V we have an collector current of 15V/680Ω=22mA. If we have a Vce of 0.5V, then our emitter resistor can drop at most 2.5V with 22mA. Let's make it 2.2V with 22mA which gives us an Re of 100Ω. We can go smaller if we want. The 100Ω should give us a gain of about 7 (the voltage gain is roughly Rc/Re for this configuration). So let's go with Re=68Ω to give us a nominal gain of about 10.

Our quiescent emitter current is going to be about 15mA. and our emitter voltage is going to be about 15mA*68Ω=1.02V (call it 1V). That means the base voltage is going to be aout 1.7V. The base current is going to be less than 1% of the emitter current (if β=100, which it will probably be twice that or so). We want the voltage in the voltage divider circuit to be considerably larger than the base current (so that the base voltage is stable). If we want the divider current to be 10x the base current, then we want it to be 10*1%*Ie, or Ie/10. Do you see where one of the equations you were batting around had a mystry factor of 10 in it? Guess where it came from?

This means we want our divider to have a current of about 1.5mA. We also want Vb to be 1.7V, so our Rb2 is going to be 1.7V/1.5mA. Let's call it 1.7mA and get a nice round 1kΩ. Our top resistor, Rb1, has to drop 18V-1.7V=16.3V with a bit more than 1.7mA (since the ~0.1mA base current is also going through it). So that would be something around 9600Ω. If you're using the E12 series, your choices are 8.2kΩ or 10kΩ. Let's use the 10kΩ.

Now that you have component values, you can walk back and see if the expected results are acceptable. Neglecting the base current your base voltage would be about 1.63V. It will be a bit lower, say about 1.55V, if we allow for the base current. That will give an emitter voltage of between about 0.85V and 1.0V, which will give an emitter current of between about 14mA and 19mA. The spec'ed collector current of 14.6mA is in that range. And, remember, we are working with 10% tolerance on resistor values, so we can't be expected to hit is exact.

 

Oh?

Is there a different formula for β when using a voltage divider circuit, than β = Ic/Ib?

I'm all ears, if there is.

I just went through a few ideas of my own and don't seem to see how to get a proper β from this.

β, by definition, is Ic/Ib. But it is not fixed, it is not very well controlled, and it is not linear. So you design for a range of anticipated β values. Generally, though not always, you design for a β that is smaller than you expect to ever see in your circuit. If the actual β is higher, then life is good. Usually. There are times when having too high of a β can cause problems, but those are generally in circuits that are receiving a lot of design attention for a variety of reasons.

For most run-of-the-mill designs, small signal transistors at non-RF frequencies are assumed to have a β>100 and larger power transistors are typically assumed to have β>10. Typical values for each are usually two to three times that, giving you a good safety margin.

 

Believe me, I'm as confused about this lab assignment as you are.

If you read post #23, I made the decision there to use a voltage divider bias design, based on the assigned text book reading and the title of the lab.

It was a chart that was provided in the lab notes and naming of circuits in a past lab has me questioning what the teacher really wants.

I posted my values and calculations in post #23 and at this point I'm going to stick with them, unless I'm told that I need to do something different for the lab.

Thanks

 

Study post #34,

That teaches everything you need to know to bias a transisitor properly, as well as how to design it for specific output voltages, because in the end result the transistor will be used to amplify a input signal, so to have a good foundation of how to design for the output amplification is very important.

 

I meant to update this post before now, but have been busy trying to catch up on all of my school work.

Anyway...

It turns out that at some point we were told that if Vcc/2 does not produce a usable Vce, based on given Vc, that we are supposed to use 10% of Vcc. I haven't found this in any of my notes, but that doesn't really matter.

Also It seems we were supposed to just assume the value of β to be what we found in the previous week's lab. This was no where in the lab notes, and the way it read was that we were supposed to find the values without using β. *shrug*

Anyway, the calculated (and given) values I ended with are as follows:
Resistance values are given in E12 standard values, not actual calculated values. β value of 152.
Rc = 680Ω
Ve = 1.8V
Re = 122Ω
Rb1 = 18kΩ
Rb2 = 1.8KΩ
Vcc = 18V
Vc = 8.07V
Vce = 6.27V

When I actually tested in the lab, I had to reduce the Rb1 value closer to actual calculated value to get measured voltages closer to spec'd values. I ended up with a Rb1 value of 14.86KΩ actual calculated was 16.2kΩ.
I also ended up with an Rb2 value of 2.26kΩ.
I was able to get my Vc to 9.07V, which was close enough for the purposes of this lab, according to the instructor.

Final measured values:
Vb = 2.22V
Ve = 1.581V
Vc = 9.07V
Ic = 13.25mA
Vce = 7.52V

 

It turns out that at some point we were told that if Vcc/2 does not produce a usable Vce, based on given Vc, that we are supposed to use 10% of Vcc. I haven't found this in any of my notes, but that doesn't really matter.

Congrats on getting the lab done.

I suspect that both of those "rules of thumb" are seldom going to get invoked because they strike me as rules that would only apply when you don't have anything else to go on. But, in the "real world" you will be designing an amplifier to accomplish something that the goal you are trying to achieve will lead you down a design path that will render these kind of "fallback" rules of thumb immaterial.

 

Congrats on getting the lab done.

I suspect that both of those "rules of thumb" are seldom going to get invoked because they strike me as rules that would only apply when you don't have anything else to go on. But, in the "real world" you will be designing an amplifier to accomplish something that the goal you are trying to achieve will lead you down a design path that will render these kind of "fallback" rules of thumb immaterial.

I agree, it seemed like an almost arbitrary decision to use 10% of Vcc, though I suppose this could be an easy way for the instructor to have some way of being able to easily follow what a student is doing when the situation arises and provides a quick way to verify where any problems may be.