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DIY Stompboxes => Building your own stompbox => Topic started by: Captainobvious99 on June 29, 2009, 05:54:26 PM

Title: Preamp booster?
Post by: Captainobvious99 on June 29, 2009, 05:54:26 PM
Hey guys, I was wondering if there is a loayout/schematic floating around here already that would be best suitable for what Im trying to do. I'd like to build a compact preamp gain circuit so that I can take the low voltage output from an IPOD and end up with at least 4v of output to go to an amplifier. I'd like it to be as transparent as possible.

I was wondering if Gus' NPN boost could be tweaked for this...

Recommendations?

THANKS!
Title: Re: Preamp booster?
Post by: JasonG on June 29, 2009, 06:09:38 PM
Why 4V output?
Title: Re: Preamp booster?
Post by: Captainobvious99 on June 29, 2009, 06:39:33 PM
It doesnt have to be exact, Id just like it to be adjustable via a pot to at least around that range.
It will be used in a car audio application. The current source unit puts out up to 4v and I'd like the preamp for the Ipod to be able to get in the same range.
Title: Re: Preamp booster?
Post by: JasonG on June 29, 2009, 07:23:08 PM
There are boosters that would be ok with an Ipod but I think you would do better doing a search on google. There are forums dedicated to diy pre-amps. Have you tried plugging  it straight into the amp? What is the peak output on the Ipod?
Title: Re: Preamp booster?
Post by: BAARON on June 30, 2009, 02:52:47 PM
Why, look what I have hiding in my bookmarks... perhaps something here would be of use to you?

http://www.zen22142.zen.co.uk/Circuits/Audio/audio.html
Title: Re: Preamp booster?
Post by: valdiorn on July 01, 2009, 09:59:53 PM
very easy, just use an op-amp and a 9V battery (you can get about 7 volt peak-to-peak with that).

Use inverting design. maybe 100k resistor at the input, then 500k audio pot in the feedback loop (remember to create a voltage reference at 4.5v and couple the input signal to that).
If the gain is not enough, use a smaller resistor (remember, gain in an inveting op-amp is ResistorFeedback/resistorInput, so 500k/100k = 5)