To: | [email protected] |
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Subject: | LF: G0MRF 300 w Tx / DDS input |
From: | [email protected] |
Date: | Sun, 1 Feb 2004 03:32:22 EST |
Reply-to: | [email protected] |
Sender: | <[email protected]> |
In a message dated 1/31/04 1:47:14 PM GMT Standard Time, [email protected] writes:Hi all at LF, Hello Gary. Have just seen your question and Alan's reply. The external input is designed for CMOS levels. However if you have a DDS with less than 12V P-P then Alan's suggestion of two resistors is a good one. My only thought would be that these should not be of equal value as it would cause false triggering of the 4013. So, how about 47k and 33k. This would give 4.95 Volts on the input which would allow a 1V margin and would prevent false triggering from noise etc. However a 1V P-P DDS will still be too low. You need around 3V P-P to be sure. Assuming you wish to use the DDS and not the internal VXO (permanently) then there's a better choice. You can use IC2 on the circuit to amplify the DDS. IC2 has a gain of 10 and would be ideal for a 1V P-P output DDS To do this you need to break the connection between L1 C11 and C12. Feed your DDS into C12, having isolated L1 / C11 and you will have enough signal to drive the CMOS logic. I also suggest that to avoid the false triggering problem you add a resistor in parallel with either R8 or R9 so that the output of IC2 is not sitting at 6.0V. If you select a value to give 5 Volts or 7 Volts that would work well. ( 27k in parallel should be about right) Hope that helps. 73 David G0MRF |
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