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puls reshaper - RF Cafe Forums

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Post subject: puls reshaper Posted: Sat Apr 28, 2007 5:08 pm


Joined: Sun Feb 04, 2007 2:58 pm

Posts: 9

Hi all,

Let me tell you what circuit i need.

I am working on a transmitter with very lage fets.

It is a push-pull design,where the fets are going te be driven by fet drivers.

So i need 2 ttl signals

1 in phase,and 1 180 degrees out of phase.(3.5-7MHz)

The push-pull design is class E,so i need a perfect 50% duty-cycle.

Just got a DDS with a on board schmitt-trigger for creating asquare wave.

But this has ofcourse no 50% duty-cycle.

My knowlidge of logic components is only very basic,so i can not design a circuit myself.

Anyone who can give give some ideas?

Thanks a lot!



Post subject: Posted: Mon Apr 30, 2007 10:40 am


Joined: Fri Feb 17, 2006 12:07 pm

Posts: 218

Location: London UK


The way this is usually done is to run the clock at twice the desired output frequency using the usual feedback gate arrangement in TTL (74LS). That generates a logic clock at 2*F

Then drive a D-Type toggle in 74LS, and the output will be at F, with a mark space ratio of 1:1, with very little width error.

BTW: the load capacitance of high power FETS at HF is high and so you will need a large load driving current capability in the drivers. (I = dQ/dT)



Post subject: Posted: Mon Apr 30, 2007 2:31 pm


Joined: Sun Feb 04, 2007 2:58 pm

Posts: 9

Thanks for your reply Nubbage!

I understand the concept now.

First double and the split.

The last part of splitting with a D flip-flop is no problem.

Just played with it last week.

But can you descibe the part of doubling a little more?

Or maybe a link with an example?

You are right about the drive current of the fets.

I use very large fets...FQA11n90 with high capacitance at the gate.

Just got some IXDD414(TO220) from the USA,they should do the job.

1 driver could drive 2 fets at 3.5MHz and 1 driver/fet for 7MHz.

The class E goeroe from classeradio.com did it so it has to work




Post subject: Posted: Tue May 01, 2007 4:36 am


Joined: Fri Feb 17, 2006 12:07 pm

Posts: 218

Location: London UK

Hi extreme

If your output bit rate is critical, and you have some sort of reference clock, then phase-lock a feed-back gate clock generator to that reference, but running at 2*F. If rate is not critical, then just free-run the 2*F clock. It could be any of a range of feed-back type NAND gate pulse generators in TTL 74LS gates. When the output is fed into a D-type flip-flop, on the D port (or maybe the clock port, I can't recall the exact arrangement) then you get 1*F output with an exact 1:1 M/S ratio.

I have a logic circuit somewhere here, and I will look it out, scan it, and upload it later today. I have seen this used for a 3.5MHz Weaver type SSB transceiver, in an RSGB (UK) publication "Radcom".



Post subject: Posted: Tue May 01, 2007 3:40 pm

Site Admin

Joined: Mon Jun 27, 2005 2:02 pm

Posts: 373

Location: Germany

JKFF (JK Flip Flop) in which both J and K are tied to '1' will divide the frequency by 2.

This arrangement is called T (Toggle) FF

Here is an example:

http://hyperphysics.phy-astr.gsu.edu/hb ... pflop.html



Post subject: Posted: Fri May 18, 2007 1:52 pm


Joined: Sun Feb 04, 2007 2:58 pm

Posts: 9

Thought i had it ....for very short time

I made a doubler from a 74hc86 XOR(quad)

Used 3 for a delay,and the last for doubler.

Followed by a 74HC74 D-flipflop,also worked fine as a divider.

But no 50% duty-cycle....got the same signal back

Now i see the problem,the pulses out of the XOR do not start at the same period.

So the flip-flop toggles me almost the same signal back.

Anyone other options to get me 50% duty-cycle?

Greetings Bert



Post subject: Posted: Wed May 30, 2007 9:45 am


Joined: Fri Feb 17, 2006 12:07 pm

Posts: 218

Location: London UK

Hi Extreme

I have found one of my circuits buried in the archive her at Nubbage Labs, one that uses a JK flipflop rather than a D type.

Tie the J and K together to the positive supply as an enable.

Feed the double frequency into the clock input, and look at the signal on Q or Not Q.

The other circuit uses a D type FF. The 2*F circuit is just as you have done, with 4xXOR, then the output of this feeds the clock line of the FF, the Not Q feeds the D input, and the output at F is taken from the Q output with a 1:1 MS ratio.

That's what the original designer said, anyway. This came from a Electronic Design Ideas magazine.

I will look for more.



Post subject: Posted: Wed May 30, 2007 9:49 am


Joined: Fri Feb 17, 2006 12:07 pm

Posts: 218

Location: London UK

After thought: maybe you need to follow the first D type with a second D type connected as before, Not Q to D and output of previous D FF to the clock line..

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