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I'm finding tons of information and equations on calculating this for solid state devices, but nothing for good old fashion particle accelerator based amplifiers :P

Currently I'm revamping the K270 guitar preamp to use components that are readily available. See the original design here: http://s13.postimg.org/patw5krxj/k270.png

This is my current progress: https://123d.circuits.io/circuits/852190-jan-6418-vaccum-tube-valve-preamp

Here is a snapshot at the time of writing this: JAN 6418 tube Preamp

The problem I think I might run into is that my guitar's pizeo pickup requires a ridiculously high input impedance on the other side of the cable. I've been reading about jfet based piezo preamps and it's not uncommon to see people have 5 - 10 mohm input impedances on their piezo preamps.

1) How do I go about figuring out what the input impedance of my current circuit is?

2) How do I raise it if necessary?

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The input impedance is almost entirely set by R2/R1, so it's about 47K || 1M or about 47K.

You can increase R2, for example to 5M and leave out R1.

A tube behaves much like a JFET (which, in turn, is like a depletion-mode MOSFET with a diode from gate to source).

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    \$\begingroup\$ I'd suggest having a JFET preamp 'guitar side' of the cable. Having a very high input impedance on the amplifier side means there is a lot of opportunity for the cable to pick up noise. The added advantage is that the cable capacitance has much less effect on the sound - long cables will attenuate high frequencies which impact on the sound of your (assumed) acoustic guitar. A very simple JFET voltage-follower powered by a 9V battery will suffice. Example here: guitarnuts2.proboards.com/thread/3150. I've got a DIY one built into my (electric) guitar. Battery lasts for years. \$\endgroup\$ – RJR Jun 9 '15 at 6:10
  • \$\begingroup\$ @RJR The circuit is a preamp for a stomp box. \$\endgroup\$ – Spehro Pefhany Jun 9 '15 at 15:23
  • \$\begingroup\$ Hey one more dumb question, what did you mean by your "||" operator? In computer science that means boolean OR \$\endgroup\$ – Jonathan S. Fisher Jun 9 '15 at 15:52
  • \$\begingroup\$ In this context it's short for parallel (parallel lines). For two resistors R1, R2 Rp = R1*R2/(R1+R2). For many resistors Rp = 1/(1/R1 + 1/R2 + ... + 1/Rn) \$\endgroup\$ – Spehro Pefhany Jun 9 '15 at 19:39
  • \$\begingroup\$ @SpehroPefhany I know his circuit is. He mentions however he uses a piezo pickup - to get the best results out of that, a buffer at the 'guitar end' is best. This also means that the input impedance of the stomp box isn't really an issue anymore. \$\endgroup\$ – RJR Jun 10 '15 at 0:40

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