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^!1=!!X!C%!9j %!jjjj"b-~jjjjClass-E 1KW PushPull Example 17 May 2004 19 Mar 2005 RevA
Prelims:
About page 1 vs page 2: Page 1 (Startup)- describes the BASIC Class_E values with Elmatch to desired RL.
It also references the PINET Ro so you can judge how the PiNet Ro changes parametricly.
Page 2(Alternate) - Adds the transformers for Basic or Pi. Note: The transformers are never reflected back to
page 1 since it would get very confusing ( especially from a programmers point of view).
Page 2 however is reflected forward to be used in the broadband analysis.
If you use page 2 and come up with a new LS2, you have to do LS2 manually (almost, see below).
Example:
Lets try a 3.8 mhz 48 volt , 11.67amp, 560 watt (each side) Q=8 design. Set Ron=0.1 ohms or a little less.
Assume push-pull to smush those pesky harmonics.
This will require two modules in series, so each one is designed with RL=25 ohms (dont forget).
Therefore: ( Im putting LS2 after CS2 since we will combine it with LS3 later)
CP1= 3787pf (N=1) CS2=2774pf LS2=0.75uh (Ro=2.23) LS3=.298u Cp3=5357p RL=25
Go to ALT (Page-2) to do the xfmrs: Set (BASIC) N=2 to multiply CS2, LS2 and Ro by a factor of 4
and also rematch Ro to RL (LS3 and CP3 are automatically recalculated )
CP1= 3787pf (N=2) CS2=694pf LS2=2.98uh (Ro=8.91) LS3=.501u Cp3=2251p RL=25
Combine the two 25 ohm designs: Note that each CP1= 3787pf remains unchanged
2X output impedances when summed (e.g. CS2/2 2*LS2 2*LS3 Cp3/2 and 2*RL):
CP1a= 3787pf CS2=347pf LS2=5.96uh (Ro=17.8) LS3=1.00u Cp3=1125p RL=50
CP1a= 3787pf
Sum coils for final: CS2=347pf Lmatch=LS2+LS3=6.96uh Cp3=1125p RL=50
Go back to Page 1 to calc the coil losses: (This is the fun part)
We need a calculation with proper series current in the new coils to properly estimate coil losses
so plug in Full Power 1120 watts, 50 ohms (dont forget) into a (Fake) design:
Change to P=1120 RL=50 (Leave Q and F untouched)
Everything is all wrong! (first impressions, you know)
Think!
If we design full power at the present voltage, the network impedances are low.
If we design full power at a higher voltage, the network impedances will go up.
If we design full power at a higher magic voltage, the network impedances will match LS2+LS and Cp3.
Increase V to 192 volts. Inspect only the output elements to the right of CP1(CS2, LS2+LS3, Cp3 and RL).
Bingo! The program did it again. Is this havin fun, or what?
Now you can Change the Choke (and LS2+LS3) diameter and length for more reasonable values, e.g. D=3 L=2.5
Ignore all the efficiencies, this is not the design youre building (you can probably get to 90% though).
Coil watts will be correct ( about 20 watts) (but add some more, ~10 W, for the shielding/coupling degradation)
Note that Capacitor dissipation is also substantial!
Try the design again with lower Q to lower the coil losses and also experiment with the PiNet alternate which
will allow you to ground all the caps. Remember, the main thing is to have fun and zorch things,
Regards,
BobbyT- NU2B L.I.NY
/= (Qcaps are fixed at 500) ]eGgpv
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dںںRLdjj@,~jjjjClass-E 1KW PushPull Example 17 May 2004 19 Mar 2005 RevA
Prelims:
About page 1 vs page 2: Page 1 (Startup)- describes the BASIC Class_E values with Elmatch to desired RL.
It also references the PINET Ro so you can judge how the PiNet Ro changes parametricly.
Page 2(Alternate) - Adds the transformers for Basic or Pi. Note: The transformers are never reflected back to
page 1 since it would get very confusing ( especially from a programmers point of view).
Page 2 however is reflected forward to be used in the broadband analysis.
If you use page 2 and come up with a new LS2, you have to do LS2 manually (almost, see below).
Example:
Lets try a 3.8 mhz 48 volt , 11.67amp, 560 watt (each side) Q=8 design. Set Ron=0.1 ohms or a little less.
Assume push-pull to smush those pesky harmonics.
This will require two modules in series, so each one is designed with RL=25 ohms (dont forget).
Therefore: ( Im putting LS2 after CS2 since we will combine it with LS3 later)
CP1= 3787pf (N=1) CS2=2774pf LS2=0.75uh (Ro=2.23) LS3=.298u Cp3=5357p RL=25
Go to ALT (Page-2) to do the xfmrs: Set (BASIC) N=2 to multiply CS2, LS2 and Ro by a factor of 4
and also rematch Ro to RL (LS3 and CP3 are automatically recalculated )
CP1= 3787pf (N=2) CS2=694pf LS2=2.98uh (Ro=8.91) LS3=.501u Cp3=2251p RL=25
Combine the two 25 ohm designs: Note that each CP1= 3787pf remains unchanged
2X output impedances when summed (e.g. CS2/2 2*LS2 2*LS3 Cp3/2 and 2*RL):
CP1a= 3787pf CS2=347pf LS2=5.96uh (Ro=17.8) LS3=1.00u Cp3=1125p RL=50
CP1a= 3787pf
Sum coils for final: CS2=347pf Lmatch=LS2+LS3=6.96uh Cp3=1125p RL=50
Go back to Page 1 to calc the coil losses: (This is the fun part)
We need a calculation with proper series current in the new coils to properly estimate coil losses
so plug in Full Power 1120 watts, 50 ohms (dont forget) into a (Fake) design:
Change to P=1120 RL=50 (Leave Q and F untouched)
Everything is all wrong! (first impressions, you know)
Think!
If we design full power at the present voltage, the network impedances are low.
If we design full power at a higher voltage, the network impedances will go up.
If we design full power at a higher magic voltage, the network impedances will match LS2+LS and Cp3.
Increase V to 192 volts. Inspect only the output elements to the right of CP1(CS2, LS2+LS3, Cp3 and RL).
Bingo! The program did it again. Is this havin fun, or what?
Now you can Change the Choke (and LS2+LS3) diameter and length for more reasonable values, e.g. D=3 L=2.5
Ignore all the efficiencies, this is not the design youre building (you can probably get to 90% though).
Coil watts will be correct ( about 20 watts) (but add some more, ~10 W, for the shielding/coupling degradation)
Note that Capacitor dissipation is also substantial!
Try the design again with lower Q to lower the coil losses and also experiment with the PiNet alternate which
will allow you to ground all the caps. Remember, the main thing is to have fun and zorch things,
Regards,
BobbyT- NU2B L.I.NY
/= ]eGgpv
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