To: | [email protected] |
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Subject: | Re: LF: RE: 500 kHz JT4A |
From: | "Johan H. Bodin" <[email protected]> |
Date: | Thu, 28 Jan 2010 09:25:53 +0100 |
In-reply-to: | <[email protected]> |
References: | <[email protected]> <BF4A524700075746A6467658DFC7102C1284D4B328@ICTS-S-EXC2-CA.luna.kuleuven.be> <[email protected]> <[email protected]> <BF4A524700075746A6467658DFC7102C1284D4B32B@ICTS-S-EXC2-CA.luna.kuleuven.be> <[email protected]> |
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Andy, such a doubler will work but the flip-flop is not a guarantee for a perfect 50/50 square wave in this case. The flip-flop will be clocked by both edges of *one* of the XOR input signals, which one depends on the clock polarity (rising or falling edge) at the flip-flop input so the flip-flop output will have the same duty cycle as one of the XOR input signals. The traditional C.T. transformer will work equally well (or bad). 73 Johan SM6LKM ---- Andy Talbot wrote: > I was looking at driving push-pull transmitters from fundamental drive > signals quite recently. > > How about a frequency doubler? The output doesn't have to have 1:1 > mark-space ratio as it only clocks a flip-flop. And simple doubler can > consists of a delay line (CR network with around 0.5us time constant) > followed by a squarer to give a resulting waveform shifted in phase from > the 500kHz drive signal. Then apply this one, along with the original > to a two input XOR gate whose output will then consists of a 1MHz train > suitable for dividing by 2 for a perfect 1:1 squarewave drive. > > > Andy > www.g4jnt.com <http://www.g4jnt.com> |
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