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I am currently trying to impedance match a capacitor as a load. In detail, it is an insulating thin film grown on top of a conductive layer (grounded). We now sputtered small Pt electrodes on top and want to apply a voltage to these. We expect the capacitance in the area of 0.5 - 1 pF and our source has a 50 Ohm impedance. We want to apply square shaped voltage pulses with < 50 ps rise time (20 GHz) and up to 20 ns (50 MHz) length. Thus, the impedance matching circuit has to be very broadband.

Since I am not an electrical engineer, I would be very grateful if you explain any attempts to solve this problem as easy and detailed as possible.

EDIT:

I did the calculation to check the phaseshift of the back reflection. I obtain this graph:

enter image description here

it seems like I have the jump right in my bandwidth (approx 50 MHz to 50 GHz).

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Unfortunately, you cannot match a purely reactive load. You could cancel out the reactance at a particular frequency using an inductor where $$Z_L=j*\omega*L$$ and $$ Z_C=\frac{1}{j*\omega*C}=\frac{-j}{\omega*C}$$

However, that would just give you a short circuit. In any case using the formula for reflection coefficient of a load on a 50 ohm line $$ \Gamma=\frac{Z_{Load}-50}{Z_{Load}+50}$$

The magnitude of Gamma for any purely imaginary Z_Load will be 1 (complete voltage wave reflection). For example, if Z_Load is j50, you'll get a Gamma of j. If Z_load is j25 you'll get a Gamma of -0.6 + j0.8. If it's -j33 you'll get -0.39-j0.92, etc.

I would just try to keep your transmission line very short and hope your source has sufficiently low output impedance to deal with this very non-ideal load.

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  • \$\begingroup\$ Hi MikeP, thank you very much for your answer! A complete voltage wave reflection would not be the worst case as long as the amplitudes of both waves simply add up. But as far as I understand it, there will be a phase shift depending on the capacitance. Is there any way to somehow fix the phase shift such that I will always have twice the voltage of my source pulse amplitude at the load? Thank you very much! Cheers, Marvin. \$\endgroup\$ – Marv_92 Oct 10 '18 at 13:25
  • \$\begingroup\$ You're very welcome Marv! I think that as long as the capacitance is low, and thus you have a large negative reactance all the way to very high frequencies, then the -50 and +50 ohm terms in the expression for Gamma will be negligible, and you will have a relatively consistent phase. For example, at 1 pF, you'd have over 1000 ohms of reactance out to almost 160 MHz. For 50 MHz, this is good, but square pulses need a number of harmonics, and by the 7th harmonic (350 MHz), the reactance is down to just about 450 ohms (about 12.5 degree shift, which might be OK). \$\endgroup\$ – MikeP Oct 10 '18 at 15:35
  • \$\begingroup\$ Hi MikeP, thanks for the answer! Can you please show me the equation to calculate the phase shift? And my pulses need to have very sharp rise times (less than 50 ps). Thus, I will have frequencies in the GHz regime that experience a large phase shift. I am afraid now, that my rise time gets significantly extended due to phase shifts in the reflection. \$\endgroup\$ – Marv_92 Oct 16 '18 at 6:09
  • \$\begingroup\$ No problem Marv, just use Zload = 1/(j * w * C) where w is 2 * pi * frequency, and C is the capacitance, then plug that into the equation for Gamma above and find the angle of that complex vector. For example, at 100 MHz and 1 pF: Zload = 1/(j*2*pi*100e6*1e-12)=-j1592, Gamma is then 0.998-j0.0628, atan(-0.0628/0.998) = -3.6 degrees \$\endgroup\$ – MikeP Oct 16 '18 at 11:23
  • \$\begingroup\$ Hi MikeP, thanks again for the help! I did the calculation and plotted a graph for my bandwidth (20 ps rise time = 50 GHz; 20 ns width = 50 MHz). It appears that the large jump (in phaseshift) is right in the middle of my bandwidth and thus, will most probably screw my pulse at the sample? I assume I will get a longer rise time as well as ringing and other instabilities in the pulse amplitude (at least at the beginning of the pulse). Is there anything I can do about that? \$\endgroup\$ – Marv_92 Oct 16 '18 at 12:30

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