can't find good complementary transistors

[sawreyrw]

keantoken,

I'm a little reluctant to post this, because it is somewhat oversimplified. But here it is. This should at least help you get the biasing right.

Note that you have to do the math if you plan to be an engineer.

Rick

[keantoken]

Mostly I have horrible transistors from radioshack *shudders*. Like 2N3904/3906, 2N4401, but most aren't even those! They are substituted with MPSA42 and 2N5088/5087 and other... *trembles*... yeah, you get the idea. I have a few 2N3055's and some MPF102 FET's and a PN2907 and as far as diodes I've got 1N4001's and some ancient germanium ones, though I don't remember the part numbers. And that's all that I can find in the semiconductor department.

[poobah]

Well most of thise are the types I use... not for audio... and they're not from Shidio Rake.

I'll have a look...

[keantoken]

Ok, sawreyrw, I am starting to get this. "If currents through R1 and R2 are much larger than Ib", meaning that the amount of current drained off by the base affecting bias is insignificant, "then Vb~Vcc*R1/(R1+R2)". And I suppose that Vbe is nominally ~650mV? I keep forgetting that the base voltage is measured from the base to the emitter. *sigh* I am sorry to waste so much of everybody's time.

[poobah]

Keep on it dude...

I can't even see the schem and I know you're doin' it!

[keantoken]

Today something in my brain clicked (it may sound painful, but it is actually quite interesting). On the way home on the bus my brother was learning how to find the distance from two points on a coordinate plane. I looked and pointed out that he was actually using the pythagorean theorem. Then I looked at the work he was doing and remembered from previous years how area and the number of dimensions on a graph were related. He was also doing some 3-dimensional work and it made me think of how you would use the pythagorean theorem in 3-dimensional calculations: instead of A^2+B^2=C^2, for 3-dimensional it would be A^3+B^3+C^3=D^3. Then I we thought about how this could be used to find the area of the bottom of a rectangular right-pyramid. Though, it turned out that he was thinking about a triangular right-pyramid, so I pointed out that you would just have to find the area as if it was a rectangular pyramid and then just chop it in half because the area of a 2D right-triangle is (length*width)/2. Until today I hadn't realized how much potential I had! It's like my cerebral black hole got larger and sucked up everything that was near it. I know that most of you will say that it won't be that way when I get older, and I realize this. I am waiting in despair for that time.

About biasing, i think something clicked in that category too.
I knew that Vb~Vs*R1/(R1+R2), but since you said that this method was oversimplified, I decided to analyze it while I had time, and i came up with some stuff:

Ib=Ie-Ic

I remembered something about effective resistance in the amplifier section of my engineering book, and I wondered about Re, because there is a voltage drop between the emitter and base. So I formulated that since the voltage was lower, the current through Re would look lower seen through the base. So I formulated that the effective resistance of Re would be

Vbe/Ib

because of ohms law. But I had a problem: Vbe needs to be calculated acurately, so I made an equation:

Vbe=Vs*((R1+Re)/2)/(R2+((R1+Re)/2))

The (R1+Re)/2 part is to calculate the resistance of Re in parallel with R1. So once I got that finished, I moved on to the effective resistance equation:

effective resistanceof Re=Vbe/((Ve-Vc)/Re)

Where Ve-Vc is equal to the voltage drop between the base and emitter. /Re finds the current produced by Re after the voltage drop.

The above equations won't work on their own because they are all networked, for instance, Vbe can't be calculated without having the optimal value of Re already. To find the optimum values, the equations would have to be run over and over again storing results in variables to be used by the next equation. Therefore floating around until the optimum values are found, indicated by the variables staying the same each time the loop is run.

I'm holding my breath hoping that I am right this time. At leat I am thinking!

[keantoken]

Have I done it again? Am I really that bad at math? Maybe all my friends just work on this day... I thought that everybody would be on during the weekend but I guess I was wrong. I am not sure about the lifestyle of an actual engineer, but I am wondering what the normal workdays are and such so that I can know when to expect replies. Then again, I could just have made another one of my sadly disapointing posts, and noone wants to upset me... Well, whatever. Happy 'gineering
-Anthony

[sawreyrw]

keantoken,

You are getting closer, but you're still not quiet there. Generally, for bias calculations, the base-emitter drop of a forward biased transistor (not saturated) is considered a constant value of .65 to .75 volts. While this is not entirely correct, it is reasonable, especially when you consider variations of transistor parameters. It is left as an exercizefor the reader (keantoken) to find the actual relationship between Vbe and Ib.

Under this assumption, you may write a couple of simultaneous equation that relate Ib, Ic, Beta, Ie and all of the resitors involved. So pick an Ic and compute the resistor values. Note that if you let Re get very small, the circuit is very sensitive to Beta. This is why Re is almost always used in a CE stage without DC feedback.

The whole point of this post is to say "You will not become proficient in electronics by asking questions in a group like this, unless you are a genius, or you find someone who is willing to spend a lot of time answering your questions."

I will not respond to further comments from you, but I may answer a specific (meaningful) question.

Rick

[poobah]

Kean,

Here is a good link.

Study this... if you already have some 2N3904's; that will work fine. Learn the common emitter cicruit first. Then learn the emitter follower (sometimes rather stupidly called the common collector circuit).

If you get stuck... post. But BUILD these circuits and measure at the same time you calculate.

That's the best way to learn.

Learning engineering is not like most things... it's more like building with bricks. Every piece piles on top of the last piece... so you have to learn everything correctly before you move on to the next step.