Amp Input Impedance Matching

[ecisthebest]

To start it off, the amp I am working on is not a Dumble clone. It is a Dumble inspired single channel amp with three gain stages. I really like the tone that's in between the famous overdrive tone and the sparkle clean tone, with a strat's neck pickup. and I believe the three gain stage design can get there. but i know there's something missing.


I've stumbled across something called "Impedance Matching" for the input stage. now, i know what impedance matching is from electrical engineering standpoint.


engineering (math): the output impedance of the pickup with the volume full on is the pickup's DCR ohm + jwL, L being the inductance of the pickup. im guessing the load impedance refers to the 1M grid leak resistor? and to get the best power dissipation (voltage x current) the Z load has to be the complex conjugate of Zin, or DCR ohm - jwL...


so putting a cap&resistor in series, putting it in parallel with the grid leak resistor (grid leak for voltage reference) is what it means? or am i thinking too hard?


then there is Two Rock's pickup loading switch. i am speculating this as a switchable cap&resistor network to cut enough of the highs to give an even output from the pickups (basically). it can't be just a few resistors on a switch though right?


I would appreciate any input. i want to think this could be a pretty big deal when tuning a particular amp for certain pickups. either underwound/overwound single coils or various humbuckers. Thanks!

[Structo]

Not entirely sure what you are asking but my understanding is by using a large grid leak on the input, which in most amps is 1Meg, you maintain a pretty close match to the pickup which is usually around 5K-8K.

I know that the DC resistance of a pickup is not the same as it's impedance but for this instance lets assume it is.

R = 1/{(1/R1)+(1/R2)

Using a value of 8K for the pickup and 1Meg for the grid leak we get:

7936.5 ohms or 7.9K ohms

[alvarezh]

Hi ecisthebest, here is a recently published article that could be of help:

http://proguitarshop.com/andyscorner/un ... explained/

The guitar's pickup impedance makes a voltage divider with your tube's grid. Check out figure 2 and 3 of the article, they are using a pedal as an example, for your case substitute the pedal for an amp.

All the best.

[SoundPerf]

I'm actually in the middle of building a buffer based on the front end (buffer) of a certain Pete Cornish pedal. So far I pretty impressed with what I hear. Which is in essence, nothing. Just very nice unity gain restoration on my guitar as if I was plugged straight in to the amp. The only thing I'm undecided about is how bright I want it. Using the exact same values adds a bit of brightness, but I've messed with some different caps and not sure what I like better yet.