accapacitordctransistors
I came across circuits probably using an AC at the base to probably amplify it. I also came across the voltage divider to add a DC bias.
simulate this circuit – Schematic created using CircuitLab
Neglecting the values of the components and supplies. Is the reason we add this DC bias to not have the voltage go to a negative value because that would damage a transistor?
And just to be sure, the voltage at the base would be a sine wave with the DC bias as the baseline, correct?
The diode is in the best position, and is of an appropriate type.
It conducts when the input is negative, the same as the transistor base conducting when the input is positive. The 47K resistor is about 1/10 of a normal RS-232 load. One could also block the voltage, but then a -100V spike (ESD say) could break down the 1N4148 and break down the E-B junction, causing irreversible damage.
Also, a 1N4148 is an appropriate diode for this application. It's a "switching diode", low capacitance and fast reverse recovery. A 1N4001 would also likely work okay, at least at slow baud rates. The 200mA rating means that even if a very high voltage were to appear at the input the transistor is fully protected, at least up until the resistor arcs over,.
My guess is that you have the Base and Emitter swapped over. Here's an example that is similar to yours. Note how inductor L3 is connected (through R5) to the Emitter, and the erase head is connected from ground (through R3) to the Base.
The circuit is based on a Common Collector Colpitts oscillator. The erase head in combination with C6 and C7 form a tuned circuit. The tap between the two capacitors lowers the impedance on the Emitter side, providing the power gain and positive feedback required to make it oscillate.
Best Answer
The reason for the DC bias is to ensure the transistor stays in the linear region and therefore works as an amplifier.
The idea with biasing is that you set the DC operating point of an amplifier somewhere in the linear region, preferably in the middle, and you superimpose a "small" varying ac signal so that it gets amplified. If the signal isn't small enough, you may end up steering your transistor out of the linear region, it will no longer work as an amplifier.
Take a look at the following graph. If you bias the transistor so that it works in active mode, you can utilize it to amplify small varying signals. In the graph \$v_{be}\$ is the ac signal to amplify and \$V_{BE}\$ sets the DC bias point and the amplified signal output is \$v_{ce}\$.
Yes. You essentially have \$V_{BIAS}+v_{ac}\$ at the input. That's why the input signal has to be small enough so that you don't disrupt the DC bias point.