fet switching questions

Started by dschwartz, July 14, 2009, 09:59:03 AM

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dschwartz

Hi all, i´m thinking over about electronic bypass, instead of a 4066, i want to experiment now with fet switches..

i have a couple of questions for the experts, though...
Which fet characteristics are needed for a good switch? which models suits this nicely?

also..
are mosfets like 2n7000 good for this job , too? they seem to have lower drain-source resistance when on, and a lower gate threshold...
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Mark Hammer

DOD preferred the J113 for many years, apparently because its "on" resistance was very low.  Check here: http://hammer.ampage.org/files/dodswitch.gif

dschwartz

i have access to 2N5458, J111, and 2N7000 at cheap prices....j111 looks like it has been used on switches.. but i still wonder is mosfets will do a nicer job as switch....
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R.G.

Quote from: dschwartz on July 14, 2009, 09:59:03 AM
Which fet characteristics are needed for a good switch?
There is a problem there in that "good switch" is highly application dependent. For switching audio, fast switching is not needed, and the slowest JFETs will be much faster than you need. Also low channel resistance (rdson) is not particularly needed, because most JFETs get down to a few hundred ohms, plenty for acting like a "short" to audio. Specialized switching JFETs have channel resistance as low as a few ohms. The biggest thing that suits a JFET to switching in pedals is having a Vgsoff of less than 9V, and preferably about 3V to 5V. That way you can bias them to a reference voltage, and still have enough of the power supply left to turn the channel off. Also, Idss is an absolute limit on the amount of current that can flow through the channel, so you have to ensure that when the JFET is on, channel current never gets to Idss. Some small signal JFETs have quite small Idss. The J201 can be as low as 1ma, which is the same as 1V across 1K.

Switching a JFET off means that you have to pull the gate more than Vgsoff below (for N-channel) the bias voltage and signal. If the signal gets as big as you have the gate reverse biased, it's going to break through by lowering the channel voltage down closer to the gate voltage.
Quotewhich models suits this nicely?
Bottom line, you want FETs much like you want phaser JFETs. Good devices are 2SK30/2SK30A, 2N5292, 2N5485. 2SK118 is good too.
Quoteare mosfets like 2n7000 good for this job , too? they seem to have lower drain-source resistance when on, and a lower gate threshold...
Unfortunately, no, not discrete MOSFETs. All MOSFETs have an unavoidable parasitic diode allowing reverse conduction around the channel. For DC switching, MOSFETs are a clear choice. But switching AC with them is limited to voltages much smaller than a diode drop to keep the body diode from even hinting at conduction. In a CMOS switch chip, the actual switches are an N-channel and a P-channel device connected in parallel. In this application, the body diodes are tied to the power supplies, not the drains, so they can be used anywhere within the power supply.

It is possible to make a back-to-back switch with opposing-direction body diodes out of a two MOSFETs, but this requires a two floating gate drive signals. Ugly.
R.G.

In response to the questions in the forum - PCB Layout for Musical Effects is available from The Book Patch. Search "PCB Layout" and it ought to appear.

dschwartz

oh..i´ll use the j111´s then
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earthtonesaudio

I like J113's for switching, but I've only compared them to J201's.

bioroids

I have ued BF245 with good results (they are easier to find than the "Jxxx" series in Argentina).

Regards!

Miguel
Eramos tan pobres!

R O Tiree

Don't dismiss P-channel JFETs - if you have a signal oscillating around GND, then raising Vgs above ground with a P-channel does the same trick as lowering Vgs for an N-channel for a signal with a positive DC bias.
...you fritter and waste the hours in an off-hand way...

dschwartz

well.. in Chile, getting a jfet is very, very hard. There used to be mpf102 and k30a´s around, but they sold out long ago and never see them again..
there´s one big supplier who sells 2N5458 (which are useless..i think) and PN4392 at very good prices (like 17 cents each)
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dschwartz

Quote from: R O Tiree on July 14, 2009, 06:19:48 PM
Don't dismiss P-channel JFETs - if you have a signal oscillating around GND, then raising Vgs above ground with a P-channel does the same trick as lowering Vgs for an N-channel for a signal with a positive DC bias.

well..P channel fets are very rare birds!!
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dschwartz

last question.. what´s that diode for? the one on the gate of the fet switch pointing to the flip flop? protection?
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earthtonesaudio

Two reasons:
1: to keep the switch/control voltage from forward biasing the PN junction formed by the JFET gate and it's channel (protection, yes).
2: to keep signals from being clipped/rectified/clamped or whatever else happen if you otherwise had a diode there (signal fidelity).

dschwartz

ok..
i simulated on livewire a circuit just like Mark H posted..but the fets didn´t switch when using reverse diodes..of course, the positive 9v voltage from the flip flop wasnt allowed to pass through the diode and the gate stayed allways at 0v
i don´t understand....
if i turned the diode forward (pointing to the gate), the fet worked OK..but ithink that way is just like having no diode at all...

help?
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bioroids

The diode must be backwards, that's OK.

When the flip flop is at high state, the diode is reverse biased and it doesn't conduct, so the Gate is kept floating at whatever voltage is at the source, so Vgs = 0 which means minimun resistance of the channel.

When the flip flop is at 0v, the diode becomes forward biased and the gate sits at one diode drop from the flip-flop low voltage. The source of the FET should be biased at Vref, so Vgs should be around -4v which turns the FET off. That's why you need FETs with Vgs(off) lower than 4v (and lower is even better, because you may be running the circuit with a low battery...)

I assume you are using the "Boss" configuration, where the Drain and Source of the FETs are biased at half the supply.

The diode is there because if you put 9v at the Gate with the source at 4.5v, Vgs=4.5v ! Vgs should always be negative for a proper behaviour of the FET and to avoid damaging it.

Regards!

Miguel
Eramos tan pobres!

dschwartz

oh, i understand.. thanks miguel..
seems that livewire simulations sucks then....
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