0
\$\begingroup\$

I'm considering switching options for an HF (3-15 MHz) tuner network. Pin diodes are interesting but their surrounding control circuitry is somewhat inconvenient, and they require relatively high power to run.

I'm reading on RF JFETs. Both generic JFET guides and specsheets for RF JFETs such as the J211 suggest that Vgs must be negative, which would require the addition of a negative power supply.

Can I get around this by connecting the N-channel JFET as I would a P-channel common-source MOSFET? Source to Vdd, gate to a TTL-level control line and drain down to a load. This would not require an additional negative power supply.

From dialogue it sounds like I need to move from this (please pretend the MOSFET is a JFET) :

single bias

To this:

dual bias

|improve this question
\$\endgroup\$
  • \$\begingroup\$ Vgs must be negative, which would require the addition of a negative power supply. Not if the Source and Drain are biased at say 5 V and the Gate is at 1 V. Voltages are relative. Look up some JFET bases circuits and note how few of them actually use a negative supply rail. I still need to see the actual circuit proposal to be able to judge if that would work or not. I also doubt if you can get the same isolation (when off) that you would get when using PIN diodes. \$\endgroup\$ – Bimpelrekkie May 22 '18 at 8:22
  • \$\begingroup\$ @Bimpelrekkie ok - I've hacked together a couple of diagrams. \$\endgroup\$ – Reinderien May 22 '18 at 11:39
1
\$\begingroup\$

To a certain extent you can use an N channel JFET with the source at V+ and control the conductivity of the channel with a gate voltage between 0 volts and V+. The problem is that you can't really have any significant DC level (or signal with peak levels) much below the positive rail on the drain. Take the 2N5951 characteristic: -

enter image description here

As you can see for various levels of VGS, you can get a reasonable 4 quadrant characteristic of drain current against drain voltage but this will start to become problematic in the third quadrant as VDS becomes more negative.

Reason: The JFET is a fairly symmetrical device and you can operate the device with drain and source swapped: -

enter image description here

And the problem is that when the drain gets several hundred mV below the gate you will begin to get strong gate current just as you would if you take the gate several hundred mV above the source.

This report explains how you can use some JFETs as voltage controlled resistors but they are pretty uncommon these days.

Personally speaking, if you are trying to switch in capacitors to tune a coil (as per your previous question) I think it will work because of the lack of DC and the shear low-level of the signal involved but, you can also find regular N channel MOSFETs that can do this and are good for a few hundred mV negative on the drain (until the body diode starts to kick-in).

|improve this answer
\$\endgroup\$
  • \$\begingroup\$ What do you mean by four quadrant characteristic? Aren't the curves present in only two of the quadrants? \$\endgroup\$ – Reinderien May 22 '18 at 11:02
  • \$\begingroup\$ Don't hang me on that! I used the term because all four quadrants are in the picture yet only two of them are useful! \$\endgroup\$ – Andy aka May 22 '18 at 11:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.