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Re: PP 2A3/6B4G anybody? [message #31473 is a reply to message #31469]

Thu, 24 November 2005 11:01

Damir
Messages: 1005
Registered: May 2009

Illuminati (2nd Degree)

Quick "analise" gives Raa ~ 2*(Ugk*µ/Ia - rp) ~ 8kOhms, for OP 300V/50mA/-60V.
Theoretical Pout ~ Paa = Ia^2*Raa = 0,035^2*8000 = 10W, or
Ua=Ugk*µ/(1+2rp/Raa) = 42,4*4/(1+1600/8000)=141,4Vrms, and then
Pout ~ Paa = Uaa^2/Raa = 282,8^2/8000 = 10W

But, realistically, (losses, distortion) - Pout ~ 8W

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[Message index]

		PP 2A3/6B4G anybody? By: PakProtector on Wed, 23 November 2005 18:22
		Re: PP 2A3/6B4G anybody? By: 2wo on Wed, 23 November 2005 23:07
		Re: PP 2A3/6B4G anybody? By: PakProtector on Thu, 24 November 2005 06:13
		Re: PP 2A3/6B4G anybody? By: Manualblock on Thu, 24 November 2005 07:21
		Re: PP 2A3/6B4G anybody? By: PakProtector on Thu, 24 November 2005 08:13
		Re: PP 2A3/6B4G anybody? By: Manualblock on Thu, 24 November 2005 08:36
		absolutely! By: PakProtector on Thu, 24 November 2005 09:17
		Re: absolutely! By: Manualblock on Thu, 24 November 2005 09:39
		winter projects By: PakProtector on Thu, 24 November 2005 13:50
		Re: PP 2A3/6B4G anybody? By: Damir on Thu, 24 November 2005 11:01
		Re: PP 2A3/6B4G anybody? By: PakProtector on Thu, 24 November 2005 11:47

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