JFETs as switch in 2-channel tube amp

Status
This old topic is closed. If you want to reopen this topic, contact a moderator using the "Report Post" button.
Hi Guys

Some here seem to be confused about the jfet gate control.

The most important detail to remember about jfets is that they are "normally-on". The function of the switch control circuit is to _actively_ turn the jfets off. The control circuit allows _passive_ turn-on.

For the series switching above, the jfet must be in a low-resistance mode, or 'on', for the signal to pass through. This means the gate has to be at the same potential as the source, and will be so more or less if the gate is allowed to float. The gate diode assures this. However, when the signal goes above ground, the diode will conduct and hold the gate at zero volts, turning the jfet 'off'. To keep this from happening, the cathode of the diode must be sitting at a voltage higher than the signal will ever reach. This is the passive turn on.

To open the signal path, the gate is pulled negative by a BJT - active turn off - and the jfet channel reverts to a high resistance and blocks the signal.

If you read the above carefully, you see how the signal fits within the control circuit voltage window. The control voltages must be higher than the peak signal of any point controlled BUT control voltage must be below the Vgs rating of the jfet.

Why did one poster use +/-7V? Did he use CMOS chips for control or for switching? If so, then he is stuck with that voltage level if he uses CMOS to directly control the jfets. BJTs are cheap and simple level shifters, so whether the control circuit runs from positive, negative or very low voltages, there is no problem to provide the correct gate control range

Have fun
Kevin O'Connor
londonpower.com
 
To keep this from happening, the cathode of the diode must be sitting at a voltage higher than the signal will ever reach.

I see. Right now, and as U can see in the latest schematic with the voltages I attached, I have +/-7.15 VDC. So, since I use the 2N5638 and since it turns off at Vgs=-4VDC (correct me if I'm wrong, please) the signal must not exceed ~6 Vpp.

Am I getting it right? :confused:

So, how could I increase the control voltage of the control circuit at the gates??
 
Hi Guys

The control voltage range is only limited by the gate-to-source voltage rating NOT the cutoff voltage NOT the pinch-off voltage. You need the gate voltage to well exceed the rated voltage for cutoff if you want the highest channel resistance possible.

Please leave your linear preconceptions of how to use a jfet at the door - this is a switching circuit and the limits and requirements are actually far less stringent. The jfet should be all the way on or all the way off. The RC gate control provides a very quick linear ramp between the two states.

If the jfet has a 25V rating - usually it will be the same for Vds and Vgs - then that is the only spec other than Ron that you need to know or to concern yourself with. This sets the maximum control circuit supply to +/-12V. If you had a 50V jfet, then +/-25V.

Have fun
Kevin O'Connor
londonpower.com
 
Last edited:
Sorry for the late reply, I somehow missed the message...

The Jfets I use (2N5638) have a rating of 30V. So, if I understand correctly it's the JFET itself that goes to -7.15Vdc during the OFF state and will not go lower, right?

So, if I wish for a larger control window, should I replace them with devices with larger voltage rating? (Which means that I also have to replace the 15V zeners as well, correct?)


Best Regards!
/Kle
 
Hi Guys

If you wish for a larger control window, please read post 134 again, which explains how to do this. The jfet does not limit this beyond what its actual voltage _capability_ is - that would be 30V for the device you specified. The control circuit limits the control range and signal window size.

You need +/-15V control circuit rails.

Have fun
Kevin O'Connor
 
Hi Guys

If you wish for a larger control window, please read post 134 again, which explains how to do this. The jfet does not limit this beyond what its actual voltage _capability_ is - that would be 30V for the device you specified. The control circuit limits the control range and signal window size.

You need +/-15V control circuit rails.

Have fun
Kevin O'Connor
But we haven't reached #134, we are at #106. Perhaps you are referring to #104?

The thing is I HAVE +/-15V control circuit rails. It's after the 1MΩ gate resistor that it becomes +/-7.15. By what you're saying, changing the 15V Zeners with say 18 or 20V Zeners will suffice for raising the final gate voltage a bit...

Cheers,
Kle
 
Hi Guys

Yes, 104 - the one directly before yours asking the Q. Must have been a time-zone difference...

Where are your zeners? If they are attached to the jfet or the jfet-end of the control-line series-R, then remove them completely as they are the cause of voltage loss here. They are not required anyway as the control circuit provides protection to the gate.

You may be thinking of mosfets, which require close voltage limiting of their gates. A mosfet cannot be used in a plug-n-play wiring for the jfet in this circuit, but has many other uses in other switching circuits.

Have fun
Kevin O'Connor
 
It's Z1 and Z20 on the schematic I am attaching, near the bottom. The one we had plotted in a previous conversation. :)
 

Attachments

  • switchingRails.png
    switchingRails.png
    103.1 KB · Views: 73
Hi Guys

Measure the voltage at the BJT collectors to make sure they are going from rail to rail when switched. If they are not, then the series base resistance is too high and needs to be reduced.

If the BJTs are switching rail-to-rail as they should and the gate end of the 1Ms reads much lower, there is a leakage path somewhere that needs to be addressed. The 10M input of the meter will cause a misreading itself, but by no more than 10%.

Have fun
Kevin O'Connor
 
Hi guys, still struggling with this?
*If* the OP wants to, I may suggest a simple and working configuration, but previously need him to re-post here the schematic of the preamp where he wants to add switching, in fact showing conventional switches drawn in place, to see what the *actual* task the Fet will have to fulfill.
There is no "one solution fits all", so suggestions will differ depending on circuit point to be switched.
Many good ones have been suggested above, and in some cases even CD40xx chips might work, but not in others.
The only truly "universal" switch is a metal-to-metal one (hand/foot actuated switches or relays), followed by LDR if high "ON" resistance is acceptable; all others *can* work very well but with distinct limitations which must be considered.
Please post the schematic already converted to black-on-white so it can be easily edited; those colorful black background .sch type are good within the PCB design world, but not so much to exchange ideas in Forums.
Thanks.
 
Hi Guys

The ideal switching element for a given circuit varies both with the circuit and the aesthetics of the design. Devices that break DC paths, such as mechanical switches, often require the addition of DC paths which might alter the circuit gain or frequency response. So, the decisions must be made with care.

Table 9-1 of the Switching Methods chapter of TUT shows the characteristics of the various switch elements commonly used. The OP's circuit is just a recto clone, which itself is a copy of an SLO with some of the extraneous bits removed. LDRs have been replaced with jfets, if you read the thread here and the secondary one at PowerScaling.com forum. The double switching of the EQ - both input and output _ is not needed except when using LDRs. The output switching using a series-shunt jfet configuration works much better, providing better channel isolation.

CMOS switches would definitely _not_ work in this app.

Have fun
Kevin O'Connor
 
Struth said:
Measure the voltage at the BJT collectors to make sure they are going from rail to rail when switched. If they are not, then the series base resistance is too high and needs to be reduced.
Yes, the collectors are going from rail to rail. Checked!
Struth said:
If the BJTs are switching rail-to-rail as they should and the gate end of the 1Ms reads much lower, there is a leakage path somewhere that needs to be addressed. The 10M input of the meter will cause a misreading itself, but by no more than 10%.

I'll check. The only thing that stands between the 1MΩ resistors and the diodes are the 10nF caps to GND. Just to be clear, I am posting once more the Switching circuit in B&W using red only at the voltage reading points.

Generally, the switching circuit Struth proposed performs correctly, even at high gain situations! :)
 

Attachments

  • SwitchingcircuitVoltages.png
    SwitchingcircuitVoltages.png
    353.5 KB · Views: 65
Status
This old topic is closed. If you want to reopen this topic, contact a moderator using the "Report Post" button.