Circuits & Schematics: www.aaroncake.net - 12

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  1. .. C1’s connection between Q1 and Q2 collectors effectively bypasses Q2 and its stage of signal inversion, making the net polarity of loop feedback around Q1 and Q3 positive rather than negative as I originally intended. C1 thus ...
    08-09-2023
  1. .. Fortunately, Q3 has a matching but opposing 2 mV/ C tempco which, when cascaded with D2, results in good cancellation of net drift. But this scheme only works because the temperature setpoint reference is current, rather than voltage. This ...
    19-10-2023
  1. .. mode in which an input waveshape that’s already 50:50 symmetrical is doubled before input to the OG for net frequency quadrupling. When frequency doubling J2 is selected, the combination of RC delays (R1C4 in the IP and ...
    10-10-2023
  2. .. 50 ppm/ C (typical), but R W ’s tempco is orders of magnitude worse at ~3000 ppm/ C. Therefore, R W dominates net tempco for any N 230. Suffice to say, cancellation of R W would make worthwhile improvements to DPOT performance in ...
    16-11-2023
  3. Circuits Interfaces Texas Instruments CD74HC08 CD74HC390 CD74HC40105 CD74HC4052
    .. cable that uses a complementary, differential bipolar RZ (return-to-zero) waveform (Figure 2). The voltages are the net differentials that the biphase drive develops: For example, the differential is 10 V when you drive the Data A ...
    04-12-2023
  4. .. path, while a 45 phase lag is introduced into the negative-signal path by R C , R D , and CB. This results in a net 90 quadrature relationship between the two opposite polarity signal pathways. When the full-wave rectified ...
    22-12-2023
  5. .. shortcoming and provided an elegant computational numerical solution that virtually obliterates the problem and makes the net response all but perfectly linear, in his design idea ( Ref. 2 ). Nicely done, Mr. Dimitrov! However, a consequence ...
    29-12-2023
  6. .. the subsequent negative half-cycle, again: Depositing another charge onto C3 of: Thus, each full cycle of F PUMP deposits a net charge onto C3 of: Which, if I OUT = 0, forces Q = 0 and therefore: However, for the (much more interesting) case ...
    27-03-2024

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