Interesting article in Elektor

mictester · Post by **mictester** » 27 Feb 2015, 22:57

There's a design in Elektor this month for a guitar overdrive. They claim that it's different because it uses "current drive" to the clipping diodes:

Elektor Overdrive.pdf: (716.71 KiB) Downloaded 494 times

It's a bit over-complicated for what it does, but it would allow remote voltage control of the "drive" control. If they'd been really smart, they could have made everything remote controllable....

Dirk_Hendrik · Post by **Dirk_Hendrik** » 28 Feb 2015, 07:39

mictester wrote: It's a bit over-complicated for what it does, .

Pfffft.... say that again. Not to mention the Elektor employee that was resposible for readable schematics apparently retired.

DrNomis · Post by **DrNomis** » 06 Mar 2015, 04:05

Looks interesting, I wonder if the type of OTA IC is critical to the circuit operation, ie: would the circuit work if LM3080 OTA ICs were used?....

I noticed that the circuit uses a lot of SMD components, I can't see why standard components could be used in a Vero layout....

An idea would be to make the overdrive voltage-controllable using a control voltage derived from the input signal's envelope characteristics, I'm thinking that doing so might produce a tone sounding a lot like what you get from overdriving a Valve amp....

coda-effects · Post by **coda-effects** » 02 Apr 2015, 07:43

I have read the article, it seemed really over complicated to me too...
I do not get why they use SMD components either, not the easiest stuff to work with

lolbou · Post by **lolbou** » 02 Apr 2015, 08:11

I had my hands on the french version (the schem was poorly printed

).

Any thoughts about the diode matching/pairing procedure?

A.J. Köhler wrote:For the optimal functioning of the circuit it is important that the characteristics of the two anti-parallel connected diodes are as identical as is possible. For this purpose the author has tested hundreds of OA91s at an identical ambient temperature. In a test setup the voltage drop across the diodes was measured at a current of about 1 mA and again at a current of about 50 μA (ratio 1:20). You can do that yourself with the aid of a regulated power supply (for example 15 V) and two resistors (for example 15 k and 270 k). The objective is to find two diodes that have an identical voltage drop (or as close as possible) at the first test current, and an identical voltage drop (which will be different from the first) at the other test current.

Is it really that crucial?