According to a Junction Field Effect Transistor (JFET) article in ElectronicsTutorial, the voltage across the source resistance is equal to 1/4 the supply voltage applied to the drain/gate for a class "A" amplifier. Quoting from the article:
This common source (CS) amplifier circuit is biased in class “A” mode by the voltage divider network formed by resistors R1 and R2. The voltage across the Source resistor RS is generally set to be about one quarter of VDD, ( VDD /4 ) but can be any reasonable value.
I set out to verify this equation by using it in a question that I was attempting to solve, a simple DC biasing equivalent voltage-divider JFET circuit with 2N3819 as the JFET component for my circuit.
- R2 = 150k Ohms
- ID (Drain Current) = 3mA
- IDSS (Saturated Drain Current) = 10mA
- VDD (Supply voltage for DC bias) = 20V
- VDS (Drain-source voltage) = 7V
- VGS (Gate-source voltage) = -3V
I set out to find the unsolved resistance values and the required voltage values. Based on using VS = VDD/4 and other equations like Shockley's Equation and the KVL equations concerning the gate/gate-source/drain-source voltage:
- Vgs = Vg - Vs
- Vds = Vdd - IdRd - IdRs
- Shockley's Equation
- Voltage divider rule for finding the gate voltage, Vg
I was able to find the calculated voltage values (which will be used to compare with the simulated results):
- Vg = 3.643V
- Vgs = -1.357V
- Vs = 5V
- Vd (drain voltage) = 8V
The calculated resistance values were:
- R1 = 673459.2173 Ohms
- RS = 1666.67 Ohms
- RD = 2666.67 Ohms
Plugging the calculated resistance values into the simulation gives me:
It seems that VS = VDD/4 is correct, albeit with slight differences in the value between the calculated and simulated results:
- Vg (simulated) = 3.643V [matches calculated value]
- Vs (simulated) = 5.117V [differs from calculated value with difference of 0.117V]
- Vgs = -1.474V [differs from calculated value with difference of 0.117V]
- Vd = 8.19V [differs from calculated value with difference of 0.19V]
- Id = 3.07 mA [differs from original value with difference of 0.7mA]
Now that I have verified that the equation is actually correct, I want to ask:
- How is the equation VS = VDD/4 derived?
- What causes the difference between the simulated and calculated values? Is it because of additional factors (like wire resistance) that the simulation software considers when showing the simulated results?
- Is there another way to find the calculated resistance/voltage values without having to use the aforementioned VS equation or the graphical method?