A question for anyone who knows about electricity.....again....

[newbie~wan]

I have a question about this circuit, but I am not sure if it will come out. It says this circuit is a voltage divider, but I dont quite understand how it is. All the examples in the section about voltage division involved circular circuits, which for some reason make more sense to me.

[Wizzard~Of~Ozz]

Ok, I'll try to make this legible,

Voltage divides on a series curcuit,

For example, if you have 2 3k resistors in series, and +12v applied to one side and neg to the other, you will get 2ma of current, the 2ma of current * one resistor is then 6v, so you get 6v dropped across the first resistor, 6v across the second, so from point input in your diagram to - will be 6v, by making the first resistor variable you can make the transistor saturated (on/Off style) or if you were to apply a signal to the input side, you get inputvoltage +/- the voltage across the second resistor, effectively you are boosting the low end of the signal so it doesn't fall below the base cutoff voltage.

To try to explain this a little better

Input = 60Hz sinewave at +3v to -3v

You have +12v across 2 equal resistors, the voltage in the center (at input) is +6V.

You sine wave now goes from +3v to +9v (It's effectively riding on top of the DC Bias, DC bias being +6v from the voltage divider.

Edit, Just to add, the purpose of using a voltage divider instead of applying straight voltage is current limitation if I remember correctly, and also it's the only way to bias a circuit.

[BFlurie]

That's odd. Newbie's image opens up as a .jpg in Mspaint instead of IExplore on my box.

[newbie~wan]

For example, if you have 2 3k resistors in series, and +12v applied to one side and neg to the other, you will get 2ma of current, the 2ma of current * one resistor is then 6v, so you get 6v dropped across the first resistor, 6v across the second

This I understand, because the simple examples in the book have full-circle circuits like the one you describe. After this point, you lose me.

so from point input in your diagram to - will be 6v

Where is the (-) on my diagram? There isnt one explicitly shown, where is it implied? When you say "first resistor", do you mean the one with the (+) near it? Do you suggest that the ground is the same as (-), at least in the case of this circuit?

You have +12v across 2 equal resistors, the voltage in the center (at input) is +6V.

There are no numbers in this schematic. I know that you are merely trying to offer a concrete example, so let me see if I understand. You are suggesting that the (+) terminal is at +12V and that the input is a +3V to -3V ac signal.
On a side note, what does it mean that a "signal" is +3V to -3V at input? I mean, no point has a voltage w/ respect to itself, so what point does this signal have a voltage w/ respect to......Ground?

I see what your saying now about biasing being like "piggybacking" the +3V to -3V signal on top of an existing +6V at that point to give a biased signal of +3V to +9V, though.

Sorry for the convoluted questions, but that's sort of the way electricity is. Thank you for helping me try to understand.

[Graham]

Not to steal Wizzard~Of~Ozz's thunder, but I have attached a modded diag,
Yes ground, unless otherwise stated, is usually considered to be the -ve side of the supply.
The base of the transistor will be at half supply voltage if the resistors (R1 and R2) are of the same value (any value), half the voltage being developed across each resistor.

If an A.C. signal of -3 to +3 applied (decoupled) to the base of the transistor will swing though (6-3)V to (6+3)V or +3 to +9V.

G

[newbie~wan]

Good to see you again Graham! Thanx for dropping in and chiming in some more support. I'm sure wiz of oz doesn't feel you're stealing his thunder. Everyone at techIMO knows that you can't have too many responses. Each offers a unique way of saying the same thing and you never know which one will "click" and make sense for whatever reason.

[Wizzard~Of~Ozz]

Thanks for clearing that up, I tried to do it with words, but sometimes a picture is worth a thousand of them, Graham is right about the ground, Sorry, neglected to mention it, opened the picture and saw it as a neg and closed it. Yes Almost every voltage is measured with respect to ground unless stated (for example, if I said "input to +'ve" then the reference (or negative lead is connected to +'ve" If I said what's the voltage at input then it is assumed "input to ground"

Surprising, I haven't had to touch this theory in about 5 years and I still remember some of it.

No probs Graham, I was out of town and couldn't reply anyways =]

[Wizzard~Of~Ozz]

I'm pretty sure this is how the whole circuit works, (all values being hypothetical) Maybe graham can confirm the signal hook up, though I'm pretty sure the other end of the signal is put to ground.

[Graham]

The circuit is not complete, it is just to illustrate the bias on the transistor base. An ac signal would need to be decoupled by a capacitor to the base, with, as you say, the other "leg" to ground.
There would also need to be a resistor in the collector to develop the OP signal across.

G

[Wizzard~Of~Ozz]

I forgot about the decoupling cap, (prevents DC from flowing back through the signal circuit).

Also the calcs are probably way off for the circuit, however the fundimentals of the circuit are achieved, I think I toasted all my notes from college, Haven't had to touch them since.

Graham, Is this a hobby or something you do for work?