To: | [email protected] |
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Subject: | Aw: Re: LF: Re: My transmit converter description |
From: | "Eike Katzera" <[email protected]> |
Date: | Fri, 8 Feb 2013 17:08:15 +0100 (CET) |
Importance: | normal |
In-reply-to: | <CAKvcm2eWJJSWQ9gk0j4=SxpN_DKyjSOhn=c-1PDcXmnewjJnpQ@mail.gmail.com> |
References: | <CAKvcm2edSxcESuKJR6yux7M-pBXE63rUVPbE98M-L6Mb8--cLg@mail.gmail.com> <001301ce03c5$495bd9c0$0401a8c0@xphd97xgq27nyf> <CAKvcm2f4NNusmobnuMuHj03RSoKRY0ghY7L6fEZDku9x13s2_w@mail.gmail.com>, <CAKvcm2eWJJSWQ9gk0j4=SxpN_DKyjSOhn=c-1PDcXmnewjJnpQ@mail.gmail.com> |
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Hi Dimitris, have you ever thought about using an Class-D-audio-Amp as a powersupply for the Class-E-RF-Amp? Let the audi-amp follow the rf-envelope an the rf-amp follow the frequency. Is that accurate enough to work as an linear-rf-amp for digi-modes? 73s de Eike DL3IKE PS: what are the correct settings in wspr-software for using split mode in the transceifer in combination with an transverter to prevent data mismatch in the wspr-database? Transceifer RX: 474kHz, TX:10.474kHz
Gesendet: Dienstag, 05. Februar 2013 um 22:32 Uhr
Von: "Dimitrios Tsifakis" <[email protected]> An: [email protected] Betreff: Re: LF: Re: My transmit converter description
> The whole point of this design is that the output impedance is 50 ohm,
> therefore there is no need for impedance transformation! Try > calculating the values for a 100 V, 100 W Class-E transmitter and note > the output impedance. For those not familiar with Class-E, I can control the power output of a transmitter by the Vcc I use. The power output is proportional to the square of the voltage, so by reducing the voltage by SQRT(2), I get half power: 50 W from 70 volts or so. The calculated output impedance and C1/C2/L2 do not change with scaling the voltage up and down. That makes class-e an excellent client for amplitude modulation :-) 73, Dimitris |
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