# why connect a jfet gate to source?

#### george2525

Jan 30, 2015
170
Hello

does anyone know why a jfet gate would be connected to its source as shown?

I have never seen this before!

Thanks

#### Alec_t

Jul 7, 2015
3,589
It biases the transistor so as to make it a fixed current source.

#### george2525

Jan 30, 2015
170
It biases the transistor so as to make it a fixed current source.
OK so a high gate voltage turns on the current source right?

How would I determine what the current is?

#### duke37

Jan 9, 2011
5,364
Look up the transistor data for a rough idea. Measure the current for the actual fet that you have.

#### george2525

Jan 30, 2015
170
Look up the transistor data for a rough idea. Measure the current for the actual fet that you have.

Ok I measured it as around 0.88uA which doesnt correspond to anything obvious in the datasheet

what is this current called?, the technical name please

shouldnt it be the current when Vgs = 0 based on the jfet Id formula?

#### Alec_t

Jul 7, 2015
3,589
Ok I measured it as around 0.88uA which doesnt correspond to anything obvious in the datasheet
What was the Vds when you measured it? The ON datasheet quotes the current for Vds=15V and a pulse time <= 630mS.

#### Audioguru

Sep 24, 2016
3,656
The datasheet for the 2N5457 shows its IDSS (current when gate is shorted to source) as from 1mA for some of them to 5mA for some of them without a load. The datasheet also shows higher currents from 2N5458 and 2N5459.
The power supply voltage does not affect the current much, down to a few volts.

#### TedA

Sep 26, 2011
156
Q2 is wired-up in a manner to produce a constant current source, but this may not be how it functions here.

It would be interesting to see more of the circuit. What we do see would appear to operate entirely at large negative voltages, if Q2 is serving as a constant current source.

I suspect something else is going on. Where did you find this circuit?

The drain load resistor for Q2 returns to -12V. For Q2 to work as a constant current source, R17 must go to a much more negative voltage than -12V. The currents mentioned above (1 to 5 mA) will drive the gate of Q3 hundreds of volts negative. 5mA through 200k will give you a neat 1kV. This does not seem likely.

So I suspect that Q2 might be functioning more as a very leaky diode, with the gate forward biased.

Ted

#### george2525

Jan 30, 2015
170
Q2 is wired-up in a manner to produce a constant current source, but this may not be how it functions here.

It would be interesting to see more of the circuit. What we do see would appear to operate entirely at large negative voltages, if Q2 is serving as a constant current source.

I suspect something else is going on. Where did you find this circuit?

The drain load resistor for Q2 returns to -12V. For Q2 to work as a constant current source, R17 must go to a much more negative voltage than -12V. The currents mentioned above (1 to 5 mA) will drive the gate of Q3 hundreds of volts negative. 5mA through 200k will give you a neat 1kV. This does not seem likely.

So I suspect that Q2 might be functioning more as a very leaky diode, with the gate forward biased.

Ted

ok heres a link to the original circuit
http://musicfromouterspace.com/inde...MAINTAB=SYNTHDIY&SONGID=NONE&VPW=1252&VPH=500

I actually changed the section and ran int another mystery which im currently asking for help on here

https://www.electronicspoint.com/th...te-make-the-circuit-work.281856/#post-1717398

both versions actually do work

Moderator
Jan 21, 2010
25,510

#### george2525

Jan 30, 2015
170
thats odd. it goes right there when I tested it

Anyway its called "single buss 1V/oct Kbd. Ctrl."

#### (*steve*)

##### ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,510
OK, it does on my PC, but not on my phone.

Dec 9, 2016
8
So now that I own JFETs , is using them as a current source preferred for LEDs and laser diodes, vs BJ trans ?

#### (*steve*)

##### ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,510
You can use them as constant current sources. However there is quite a spread of Idgss between devices.

There are commercially made 2 terminal devices commonly known as "constant current diodes" which are essentially jfets with the gate tied to the source. These devices are made in a series of current ratings and I suspect that there is some "binning" involved in the creation of devices with known "reverse" currents.

There are several circuits using a jfet and an additional resistor or even an additional jfet to make better current sources. These are typically low current and a resistor is frequently a better option for a LED (unless the voltage range is large)

#### TedA

Sep 26, 2011
156

Thanks for the link to the schematic.

To answer your question, I don't believe that Q2 in the linked schematic actually does much of anything useful. This device will conduct at all times. It will act as a relatively low value resistor. ( Small relative to the series 39k and 200k resistors.)

The author talks of Q2 functioning as a low leakage diode, but it is not connected correctly to do this.

This sort of error is not uncommon in published circuits.

Since the gate is strapped to the source, the gate cannot be driven negative relative to the source to get this depletion mode device turned off. So the channel resistance will remain small with both high and low control voltages. This will be true as long as the circuit current is much smaller than the transistor's Idss.

The circuit may work fairly well anyway. No diode is actually required. The low impedance of the U2-B op-amp output will absorb any current driven into the gate of Q3 when the control voltage is positive. When the gate of Q3 is driven negative, the gate is reverse biased and the leakage current is small.

The error introduced by the Q2 circuit might be reduced by eliminating Q2 and R14, while increasing the value of R17. A pull down resistor on 7 of U5-B might reduce leakage through Q1 when it is supposed to be off.

I think you have already mentioned good results after adding a large value resistor in series with the Q3 gate terminal. The main drawback to increasing the resistance there is that it makes the sample switch slower to turn on and turn off. For this application, pretty slow is still fast enough.

Ted

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