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Help With Jfet Transistor

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Caissiejp

New Member
Hello, may somebody explain me the JFET Transistor. How can we determine its resistance and how do we polarise it?
Please and thank you, and for an example of an JFET transistor. the 2N5457.
 

crutschow

Well-Known Member
Most Helpful Member
A JFET is a field-effect transistor that has a diode junction for the gate. JFETs are normally full on with zero gate bias and thus the gate must be reverse biased to turn them off. The 2N5457 is an N-Channel JFET so it's gate must be negatively biased with respect to the source to turn off.

The channel resistance of a JFET depends upon the gate bias.

What do you mean "polarise it"?
 

Caissiejp

New Member
What do you mean by negatively biased with respect to the source to turn off? I do not understand that part. And How may I find the value of the resistance for that 2N5457?
 

crutschow

Well-Known Member
Most Helpful Member
You have to supply a negative voltage to the gate with respect to the source. If the source terminal is connected to common, then you would supply a negative voltage to the gate with respect to common.

Resistance across what terminals and under what conditions?
 
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audioguru

Well-Known Member
Most Helpful Member
Jfet transistors have a wide range of spec's. The max current in the 2N5457 is from 1mA to 5mA and it is cutoff when its gate-source voltage is from -0.5V to -6V.

That circuit uses two Jfets in what might be an odd Cascode circuit but the Jfets must match or it won't work.
 

Roff

Well-Known Member
The circuits in the linked schematic are common-source amplifiers (Q1-Q2 and Q3-Q4) with active (current source) loads. They have very high unloaded gain. Their output impedances are high, so they need to be lightly loaded to maintain high gain. I simulated Q1-Q2, and the capacitor values are all wrong for an audio amplifier. I'm not sure what the original intent was, but if you make C3=100uF and C1=10uF, the simulation shows approximately 55dB unloaded gain, with -3dB points at ≈20Hz and 90kHz. Harmonic distortion for ≈1V p-p output was extremely low, <-60dB.
As Audioguru pointed out, the FETs need to be fairly well matched. If the bias current of Q1 exceeds the Idss of Q2, the gate-source of Q2 will be forward-biased, and the circuit won't work properly.
 

Caissiejp

New Member
My intention of the circuit is for building an disortion pedal for a electric guitar.
I understand that the FETs need to be fairly well matched, yes,but I think that the FET are like resistance in this case, but that i'm not quite sure. I am having trouble finding their impedance values for the circuit.


Thanks

---------'
Caissiejp'
 

audioguru

Well-Known Member
Most Helpful Member
The datasheet shows the resistance of many Jfets but only when the signal level is 100mV and less. Then the Jfet can be used as a low level attenuator. But since the Jfets have such a wide range of spec's then you don't know which one you have.

The Jfets on the bottom are amplifiers and the Jfets on the top are their constant current source loads. The resistance of the drain-source is not used in this circuit.

The Jfets cause a smoother sounding kind of distortion when they clip the signal compared to a transistor that sounds harsh when it clips.
 

Roff

Well-Known Member
My intention of the circuit is for building an disortion pedal for a electric guitar.
I understand that the FETs need to be fairly well matched, yes,but I think that the FET are like resistance in this case, but that i'm not quite sure. I am having trouble finding their impedance values for the circuit.


Thanks

---------'
Caissiejp'
The FETs are being used in the saturation region, not the linear (triode) region, so they will not look like resistors. Why are you hung up on their impedances?
9474-image1123.jpg

This is a representative graph, to show the two regions of operation. To see the IV curves of the 2N5457, see the datasheet.
 
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