Message-ID: <002701c06928$217ca8c0$3d1d893e@lvm>
From: "LAWRENCE MAYHEAD" <LAURIE@g3aqc.freeserve.co.uk>
To: rsgb_lf_group@blacksheep.org
Subject: LF: Re: Guard Circuit
Date: Mon, 18 Dec 2000 19:23:26 -0000
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<DIV><FONT size=2>Yes please Jim, I am considering an 800W amp so this guard 
circuit could be very useful. 73s Laurie </FONT></DIV>
<BLOCKQUOTE 
style="BORDER-LEFT: #000000 2px solid; MARGIN-LEFT: 5px; PADDING-LEFT: 5px">
  <DIV><FONT face=Arial size=2><B>-----Original Message-----</B><BR><B>From: 
  </B>James Moritz &lt;<A 
  href="mailto:j.r.moritz@herts.ac.uk">j.r.moritz@herts.ac.uk</A>&gt;<BR><B>To: 
  </B><A 
  href="mailto:rsgb_lf_group@blacksheep.org">rsgb_lf_group@blacksheep.org</A> 
  &lt;<A 
  href="mailto:rsgb_lf_group@blacksheep.org">rsgb_lf_group@blacksheep.org</A>&gt;<BR><B>Date: 
  </B>18 December 2000 14:53<BR><B>Subject: </B>LF: Guard 
  Circuit<BR><BR></DIV></FONT>Dear Andy, LF Group,<BR><BR>Aha... wondered why 
  you hadn't mentioned it before - unfortunately, the guard circuit is not shown 
  in the simplified diagram in the LF handbook.<BR><BR>What it is - the tank 
  coil has a link winding on it. This feeds the 'AC' side of a bridge rectifier 
  made up of 4 fast diodes, via a series leakage inductance tuning capacitor. 
  the 'DC' side is connected to some big plastic film reservoir capacitors, and 
  then across the PA supply rail. In fig 3.20 in the handbook, it is assembled 
  on the 3 heatsinks largely concealed by the tank capacitor boards. I think it 
  works like this:<BR><BR>W<?/color>hen the tank circuit current (determined by 
  the load <BR>impedance) exceeds a certain ratio with the supply voltage, the 
  <BR>voltage across the link winding forward biases the guard circuit 
  <BR>diodes, and returns power to the DC supply. This effectively <BR>clamps 
  the voltage across the tank circuit, and due to the <BR>impedance transforming 
  properties of the LC circuit, appears to the <BR>PA as a non-linear resistance 
  in series with the load which increases rapidly when the output current 
  exceeds a certain point. If you monitor the 'guard', PA input, and supply 
  currents as the load is decreased, the guard and PA currents increase, but the 
  difference between them (ie. the supply current) decreases. With a dead short 
  on the output, the PA current is increased by about 50%, eg, after the unit 
  was re-tuned:<BR><BR>With 50ohm load, 60V supply -<BR>Total PA supply current 
  for 3 PA's = 21.8A = PSU supply current<BR>Total guard current = 
  0.1A<BR><BR>With short circuit load -<BR>Total PA current = 34.9A<BR>Total 
  guard current = 28.3A<BR>Supply current from PSU = 6.5A<BR><BR>With the full 
  supply voltage and a short circuit load, the PA modules run quite a bit warmer 
  than normal, but not dangerously so, so this is an impressively rugged 
  design!<BR><BR>I can send you a photocopy of the full diagrams if you like - 
  also the details of how I tuned it up for a different frequency and load 
  impedance, if anyone's interested.<BR><BR>Cheers, Jim Moritz<BR>73 de M0BMU<?color><?param 0100,0100,0100><BR><BR></BLOCKQUOTE></BODY></HTML>