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- From: tom@cpac.washington.edu (Tom May)
- Subject: Re: Tube swapping/amp question
- In-Reply-To: kew@cbnewsg.cb.att.com's message of Mon, 21 Dec 1992 16:26:18 GMT
- Message-ID: <TOM.92Dec21114809@bailey.cpac.washington.edu>
- Lines: 19
- Sender: news@u.washington.edu (USENET News System)
- Organization: Center for Process Analytical Chemistry, UW
- References: <BzGywo.yE@acsu.buffalo.edu>
- <1992Dec21.162618.24450@cbfsb.cb.att.com>
- Date: Mon, 21 Dec 1992 19:48:09 GMT
-
- In article <1992Dec21.162618.24450@cbfsb.cb.att.com> kew@cbnewsg.cb.att.com (steve.w.askew) writes:
-
- In the Carvin Im not sure, but in Marshall and Fender amps you could
- always remove the 2 outside power tubes to reduce the power by 50%.
- But you had better change the Impedence selector to compensate.
-
- So, I'm running with four tubes into a 16-ohm load, with the impedance
- selector set to 16 ohms. If the plate current of each tube is I and
- the 16-ohm speaker load as seen through the output transformer is R,
- the voltage dropped across the transformer primary is 2IR (since there
- are two tubes in parallel providing a total current of 2I). If I remove
- the outside tubes without changing the impedance selector, the voltage
- across the primary would be just IR, which is probably a bad thing.
- In order to maintain the V=2IR relationship beteen plate current and
- voltage to keep the tubes happy, I have to twiddle the impedance
- selector to make R look twice as large (V=I*2R), which I can do by
- setting the selector to 8 ohms. Right?
- --
- Tom, tube hacker in the making
-