Step 5: Designing JFET amplifier using a constant voltage source... on the source!
The final method I am going to show you about JFET amplifier design is using a constant voltage on the source. This will maximise the gain but will not produce a perfectly symmetrical waveform depending on what JFET you use.
How this works is that the source resistor is being replaced for a diode. Different kinds of diode will work better with different JFETs. For example, a standard 1N4001 or 1N4148 diode will have a Vf (forward voltage drop) of about 0.6v-0.8v. Another type of diode is a BAT54 schottky diode which has a Vf as low as 0.3v. Each of these different types of diode will bias the JFET differently depending on the JFETs Vgs(on).
In the case of the 2N5457 JFET, using a 1N4148 biases the JFET colder than the resistor method (Cold biasing means the JFET is being starved of current and therefore, the voltage at the drain is going to be higher than Vcc/2. If I use the BAT54 diode, this still forces the JFET to bias cold, though less so than the 1N4148. With an LED, the circuit failed to work entirely.
How it works: This circuit works because a diode only starts to conduct until the Vf is exceeded. Therefore, if one is put between the source and ground, it will only conduct once its Vf is exceeded, that being around 0.6v so the gate is now at 0.6v with respect to the source, biasing the jfet. With the diode, the Vf hardly changes with the increasing and decreasing current, essentially keeping the source voltage constant.
I have included a picture of the Vf chart. The Vf increases by about 100mV at a change of 150mA which is going to be minute at the current used by the JFET.
How this works is that the source resistor is being replaced for a diode. Different kinds of diode will work better with different JFETs. For example, a standard 1N4001 or 1N4148 diode will have a Vf (forward voltage drop) of about 0.6v-0.8v. Another type of diode is a BAT54 schottky diode which has a Vf as low as 0.3v. Each of these different types of diode will bias the JFET differently depending on the JFETs Vgs(on).
In the case of the 2N5457 JFET, using a 1N4148 biases the JFET colder than the resistor method (Cold biasing means the JFET is being starved of current and therefore, the voltage at the drain is going to be higher than Vcc/2. If I use the BAT54 diode, this still forces the JFET to bias cold, though less so than the 1N4148. With an LED, the circuit failed to work entirely.
How it works: This circuit works because a diode only starts to conduct until the Vf is exceeded. Therefore, if one is put between the source and ground, it will only conduct once its Vf is exceeded, that being around 0.6v so the gate is now at 0.6v with respect to the source, biasing the jfet. With the diode, the Vf hardly changes with the increasing and decreasing current, essentially keeping the source voltage constant.
I have included a picture of the Vf chart. The Vf increases by about 100mV at a change of 150mA which is going to be minute at the current used by the JFET.
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