Dave wrote:
??? How do you get to that 45 ohm figure. The input has a pair of 100R in parallel, so I can't see how it can be 45R input impedance.
Also in parallel with the 2x 100 ohm resistors (R2 and R3) is 10k (R5) in series with 100nF, plus 475 ohms (R1) to a pair of (parallel) back-to-back diodes in series with a 100nF capacitor (C5). Both 100nF capacitors essentially sit at the half-supply bias voltage (there is a very small AC component on C3, ~1mV, which can be ignored). So, for peak input voltages of less than a diode drop, the input impedance is ~49.75 ohms (100||100|10k). But for peak voltages greater than a diode drop, the input impedance is 100||100||475||10k, or about 45 ohms. (There is a slushy transition zone of ~100mV as the diodes turn on.)
The TBolt puts out nominally +13dBm, or 1Vrms (2.8v peak-to-peak, 1.4v peak) into 50 ohms.
So, the TBolt sees a nonlinear load of ~50 ohms for the first ~600mV (plus and minus) of voltage excursion, then ~45 ohms from 600mV to 1.4v (plus and minus).
I'd be inclined to change R2 and R3 to 113 ohms for use with sources that put out +4dBm or more, although the practical effect in most cases is probably minor. Note, however, that if one feeds the divider and another instrument using a simple BNC "T", the nonlinearity of the divider's input impedance will raise the distortion floor of the sine wave seen by the other instrument to ~ -40dBc even if the source is perfectly pure.
Best regards, Charles _______________________________________________ time-nuts mailing list -- [email protected] To unsubscribe, go to https://www.febo.com/cgi-bin/mailman/listinfo/time-nuts and follow the instructions there.
