Electronic Parts [Explained] pls..

have you had a chance to look over the tutorial section in this site? the list of components used is extensive and in each circuit they can have different uses.
a capacitor is a great example, in alot of cases it stores energy, but in AC it is used to filter out DC signals allowing AC to pass.
get a circuit and ask us how it works, that would be the simple way of doing this

 

Firstly, forget November. You are not going to understand electronis circuitry in a couple of weeks.

Do you want to look at the first circuit or the second?

The first circuit is very dubious. It has wires crossing without a connection. The wires must be connected at this point so needs a blob adding.
DC analysis (Bias)


The two resistors in series are intended set the base voltage at 1V. You can work out the ratio with Ohms law. The upper resistor at 830k is not a preferred value so you should use a 820k.
With 0.7V base/emitter voltage, the emitter voltage will be about 0.3V so the current will be 0.3/360 = 0.83mA. Most of this current will go through the 10k resistor which will drop 8.3V BUT this voltage is not available. The emitter current must therefore be reduced, if you use 680 (a preferred value) the current will be 0.3/680 = 0.44mA and the drop across the 10k will be 4.4V, putting 9 - 4.4 = 4.6V, enough to give an AC output of +- 1V easily.

Do you understand this? Get the DC conditions right first, then consider the AC conditions.

 

Duke: You've explained that in such a way that I'm finally starting to understand that last wee bit about amplifiers (the respective roles of the collector and emitter resistors). Its one of those things that just wouldnt sink in when read from other sources. Cheers

Question: When you say 8.3V is not available, why not? If the emitter is at 0.3V then that leaves 8.7V. What else is dropping voltage on that end?

 

R.L.

You need about 1V across the transistor Also, the calculations are only aproximate.
If you want a reasonable output swing then the collector should sit at about 4V or 5V.

Gain will be 10k/680 = 15, a lot less than the original circuit

Time for tea!

 

The second one uses an IC, to understand how it works you have to dig into what composes the circuit in the LM386...

If you don't understand what duke said above trying to even start explaining this isn't going to work...

Time to learn and master baby steps before running in this hobby...

 

If you Google 'transistor as amplifier' you will be presented with a plethora of very well written articles on the subject...

 

You asked about biasing a transistor and I explained as best I could using you faulty circuit and how it should be corrected.
When you have the bias right, you can consider the AC signal.

I asked whether you understood. You did not deign to reply.

 

The design will depend on the needs of the circuit. If you are using a small battery, you may need to use very little current, if you have a car radio with a big alternator supplying the power, then current consumption is not important.

The amplifier will need to have a certain gain which you will have to decide. You need to know the input level of the signal and the requirements of the load.

Your amp was originally designed with 1V on the base, we can stick with that.
The emitter will be about 0.6V less than this (silicon bjt) i.e. 0.4V.
Suppose that we pass 0.5mA through the transistor, the emitter resistor will be R = V/I = 800. The nearest preferred value is 820 so we will use this.

The transistor is now passing 0.5mA and, with a 10k collecror resistance this will drop (V = I*R) 5V so leaving 4V on the collector, a nice value to let the collector swing up and down with a signal input.

Most of the emitter current will go through the collector so any variation of voltage on the emitter will vary the emitter current and will be reproduced in variation of the collector current. So the circuit voltage gain will be 10k/820 = 12.2

Assume the transistor has a current gain of 100, then the base current will be the emitter current (0.5mA) divided by the transistor gain 0.5mA/100 = 5uA.
The rule of thumb is that the base voltage divider should pass 10 times the base current i.e. 50uA. To pass this current, the resistors will total (R = V/I) = 9/50u = 180k. However we can be more generous than this say we put 1V on the base and pass 0.1mA, then the lower resistance will be 10k and the upper resistance will be 8/0.1m = 80k neglecting base current through the upper resistor. Chose 82k as the nearest preferred value.

Once you have the DC conditions set, you can look at the AC signal. The signal comes in via a capacitor which you have chosen to be 0.15uF. The reactance of a capacitor Xc = 1/(2*pi*f*C). To find the roll off of the lower frequency, the reactance should equat the input resistance which will be somewhat lower than 10k. Lets choose 10k for simplicity.
F=1/(2*pi*X*C) = 106Hz.
At medium frequencies the voltage gain will be 12 but as the frequency is lowered, the gain will drop and will be down to 6 at about 100Hz.

The output drives a 100k pot with a similar capacitor so the output roll off will be 10Hz.

Much of this design can be done by rule of thumb when you have done a few.

Time for tea!

 

I put 1V on the base because your original circuit had 1V on the base and because I can count that high! Chose any lowish value you like but it will affect all subsequent calculations.

A npn silicon bjt transistor needs the base to be about 0.6V above the emitter to turn it on. You can look up the data for your specific transistor.

1V - 0.6V = 0.4V across the emitter resistor.

Since the transistor is likely to have a high gain (over 100), the base current will be small and most of the emitter current will pass out of the collector. Thus the circuit voltage gain will be collector resistor / emitter resistor. V = I*R

 

www.technologystudent.com/elec1/transis1.htm

Depends on the transistor you use, the minimum amount will be on the datasheets, its usually something VERY small

when a voltage is applied to the base, it will allow electrons to flow from the collector (i remember it COLLECTS the power) through out the emitter (where it EMITS the electrons as such)

The link i sent is very good for the basics and teaches in animation and fairly good detail