As a source of information for this tool I have been using electronics-tutorials.ws, specifically the 'Common Emitter Amplifier Example No1' in the Common Emitter Amplifier article.
| Ω | Load resistance. | ||
| V | Supply voltage. | ||
| V | Emitter resistor voltage drop (why was a value of 1 V chosen in the example?). | ||
| V | Collector-emitter voltage at transistor saturation. |
\(I_{C_{MAX}} = {V_{CC}-V_{RE} \over R_{L}} \)
| A | Maximum Collector current. |
\(I_{C_{Q}} = {(V_{CC}-V_{RE}) / 2 \over R_{L}} \)
| A | Collector current at Q-point of the amplifier. |
\(I_{B} = {I_{C} \over \beta} \)
| Transistor gain. | |||
| A | Base current (at Q-point). |
\( R_{2} = { V_{RE} + V_{BE} \over (multiplier) * I_{B} } \)
| Potential divider current multiplier. | |||
| V | PN junction voltage drop. | ||
| Ω | Value of R2. |
\( R_{1} = { V_{cc} * (V_{RE} + V_{BE}) \over (multiplier + 1) * I_{B} } \)
| Ω | Value of R1. |
\( I_{E} = I_{C} + I_{B} \)
\( R_{E} = V_{RE} + I_{E} \)
| A | Emitter current. | ||
| Ω | Value of RE. |
(this component was not calculated in the example but was mentioned in the text)
A reactance value of maximally a tenth of the value of RE at the lowest frequency.
\( X_C = \frac{1}{\omega C} = \frac{1}{2\pi fC} \)
\( C = { 1 \over \omega X_C } \)
| V | Lowest frequency. | ||
| F | Maximum value of CE. |