Re: High speed, high voltage amplifier for cap. load


"Winfield Hill" <Winfield_member@xxxxxxxxxxx> a écrit dans le message de
news:daoe860cv8@xxxxxxxxxxxxxxxxxx
> jwodin@xxxxxxxxx wrote...
> >
....
> I can advise you on the use of high-power wideband 50-ohm RF amps
> in your application: it might be reasonable. First, consider,
> 25 watts on a 50-ohm terminated line is 35.3Vrms = 100Vpp, which
> is exactly your required voltage. You'd put the power terminator
> as close as possible to the electrodes. The main problem is your
> huge 400pF capacitance, which with a 50-ohm source (we assume the
> RF amp isn't back terminated) places you down -3dB at 8.33MHz.
>
> But 8.3MHz is close to 10MHz and this gives us hope that a simple
> series peaking inductance can deliver the signal unabated to the
> electrodes up to 10MHz. For example, a 0.63uH series inductance
> would resonate with 400pF at 10MHz. Your 50-ohm termination can
> provide the needed Q-damping for the L-C resonance.
>
> Playing with spice to evaluate the best flat response curve in the
> 4 to 10MHz region, 0.7uH looks good, assuming a pure 50-ohm source.
> It peaks to +4% at 5MHz and then drops to -18% at 10MHz (compare
> to -36% at 10MHz with no series inductance). You can experiment
> with this stuff on your computer.
>

Going further you can start with a 4th order cheb matching filter and then
taylor it for just 400pF output cap and best output.
The best case would be to have an additional 50R load.

This gives you something like this:


.--------------. 680nH 1uH
| ___ | ___ ___
| .-|___|--------------UUU--+--UUU-----+--------.
| | 50R | | | |
| /+\ | | .-. | electrode cap
| ( ) | --- 50R | | ---
| \-/ | 520p --- Load| | --- 400p
| | | | '-' |
| | | | |
'--------------' | | |
HF amplifier === === ===
GND GND GND


There you have 0/-4% over a 10.5MHz BW.




> It'd probably be best to break up your capacitances into groups
> and feed each one with its own inductor. For example, with your
> 16 electrode sets, a simple calculation gives us 25pF each, which
> means 11uH in series at each electrode set. Or eight sets, 50pF
> + 22uH, etc. You can measure each one, at the 50-ohm terminator,
> including its coax cable (which just looks like more capacitance),
> and calculate the best inductor for that set. Perhaps you'll use
> small tunable inductors, adjusted to flatten the frequency sweep.
>

You can do that too and the components values become 11uH, 16uH, 33p, and
800R for each of the 16 cells.
The load resistor then have only 1.5W power dissipation.



--
Thanks,
Fred.


.

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