Jfet Troubleshooting

JFET TROUBLESHOOTING• Faults in self-biased JFET• Faults in D-MOSFET and E-MOSFET circuits

FAULTS IN SELF-BIASED JFET

SYMPTOM 1: VD=VDD

For this condition the drain current must be zero because the voltage drop across RD, as illustrated in this figure.

As in any circuit, it is good troubleshooting practice to first check for obvious problems such as open or poor connection as well as charred resistor. Next, disconnect power and measure suspected resistors for opens. If these are okay, the JFET is probably bad.

Any of the following faults can produce this symptom:

1. No ground connection at RS

2. RS open

3. Open drain lead connection4. Open source lead connection5. FET internally open between drain and

source

If the JFET is okay, the other possible fault is

SYMPTOM 2: VD SIGNIFICANTLY LESS THAN NORMAL

For this condition, unless the supply voltage is lower than it should be, the drain current must be larger than normal because the drop across RD is too much. This figure indicates this situation.

This symptom is cause by any of the following:

1. Open RG2. Open gate lead3. FET internally open at gate

Any of these three faults will cause the depletion region in the JFET to disappear and the channel to widen so that the drain current is limited only by RD, RS, and the small channel resistance.

FAULTS IN D-MOSFET AND E-MOSFET CIRCUITS

One faults that is difficult to detect is when the gate opens in a zero-biased D-MOSFET. In a zero-biased D-MOSFET, the gate-to-source voltage remains zero when an open occurs in the gate circuit; thus, the drain current doesn’t change, and the bias appears normal.

AN OPEN FAULT IN THE GATE CIRCUIT OF AS D-MOSFET CAUSES NO CHANGE IN ID.

WHY DOESN’T THE DRAIN CURRENT CHANGE WHEN AN OPEN OCCURS IN THE GATE CIRCUIT OF A ZERO-BIASED D-MOSFET CIRCUIT?

It is because VGS remains at approximately zero.

In an E-MOSFET circuit with voltage-divider bias, an open R1 makes the gate voltage zero. This causes the transistor to be off and act like an open switch because a gate-to-source threshold voltage greater than zero is required to turn on the device on.

If R2 opens, the gate is at +VDD and the channel resistance is very low so the device approximates a closed switch. The drain current is limited only by RD. this condition is illustrated in this figure.

If the gate of an E-MOSFET becomes shorted to ground in a circuit with voltage-divider bias, the device is off and VD=VDD