JFET Current Source: Understanding & Calculations

[marc.orr]

Hello, I'm trying to brush up on my analog electronics a bit. I've been studying "The Art of Electronics".

In chapter 4 they show a jfet current source. It consists of two nfets. The source of the top one is connected to the drain of the bottom one. Both have their gates connected to ground. The bottom fet has its source connected to ground through a resistor. The drain of the top fet is where the load is connected.

It seems pretty straightforward. I just don't understand the calculations of it.

1) First of all, the book says that IDSS has to be larger for the top transistor than for the bottom one. Why is this?

2) General FET calculation question: In a bjt you know that there is a .7 volt drop between base and emitter and this makes calculations very straight forward. For the FET, i understand that the drain and source current depend on the voltage between the gate and the source. How do you know the voltage at the source?

Thanks for everyone's time.

-Marc Orr

 

[SGT]

For the bottom FET the source is connected to ground, so the gate-source voltage is zero and that is why the FET current is IDSS (Drain current with gate-source short circuited). This is the current source.
The upper FET is operating normally. Its source has a positive voltage and its gate is grounded, so Vgs < 0. It seems that it operates as a buffer between the current source and the load.

 

[ravenprp]

Hello Marc,

I'm glad to hear that you're brushing up on your analog electronics and studying "The Art of Electronics"! The JFET current source you described is a common configuration and understanding its calculations can definitely help improve your understanding of analog circuits.

To answer your questions:

1) The reason for having a larger IDSS for the top transistor is to ensure that it operates in the saturation region. When a JFET is in saturation, the drain current is relatively independent of the drain-source voltage. This allows the current source to have a relatively constant output current regardless of changes in the load resistance. If the IDSS for the bottom transistor was larger, it would not operate in saturation and the output current would vary significantly with changes in the load resistance.

2) Calculating the voltage at the source of a JFET can be a bit more complex compared to a BJT. In a BJT, the base-emitter voltage is typically constant at around 0.7V, making calculations simpler. However, in a JFET, the gate-source voltage can vary depending on the biasing conditions and the characteristics of the JFET. To determine the voltage at the source, you would need to use the voltage divider equation (Vs = VDD * (R1/(R1+R2)) where R1 is the drain resistor and R2 is the source resistor) or use Ohm's law to calculate the voltage drop across the source resistor.

I hope this helps clarify the calculations for the JFET current source. Keep studying and practicing, and don't hesitate to ask for help if you have any further questions! Best of luck.