• State Resolved
  • Locked Locked
  • Replies 3 replies
  • Subscribers 21 subscribers
  • Views 1277 views
  • Users 0 members are here
Support feedback
Options
Options
Related

This thread has been locked.

If you have a related question, please click the "Ask a related question" button in the top right corner. The newly created question will be automatically linked to this question.

Original question:

TINA/Spice/CD4093B: Classic 4093 Schmitt trigger "gated oscillator" trouble

CD4093 incoming pulse shaping

Jerry Dobson
Prodigy 10 points
Other Parts Discussed in Thread: CD14538B

I have a simple question. Positive impulses with the duration 10 microseconds and the amplitude about 5 volts comes from out "Q" of the CD14538B (monostable multivibrator) to the pin number 5 "C" (one NAND-Schmitt) of the CD4093. Between pin number 5 "C" and pin number 6 "D" included the resistor 10 kOhm and  fast impulse (Schottky) diode - its cathode connected to the pin 5 "C"of the NAND-Schmitt and its anode respectively connected to the pin 6 "D" of the NAND-Schmitt. Also, pin number 6 "D" connects with the ground wire via the capacitor 150 pF. So the inverted negative impulses gathering from the pin 4 (out "K") of the CD4093. What was the purpose of connecting the resistor and the diode between pins number 5 and number 6, and for what the capacitor 150 pF was connected with the pin number 6 of the NAND-Schmitt and the ground wire? Wouldn't it have been easier tied together pins 5 and 6 and connected them with the out "Q" of the CD14538B?  In another (old) version of the circuit it was done as I described. How does this affect the waveform at the inverter output ( pin 4 out "K") of the CD4093? Maybe it reduces the noise level? Or is the impulse-shaping depends on it? Could someone explain it for an old student?) Above described part of the circuit was used in the  analog survey meter. Thanks!

  • 0
    TI__Genius 10651 points

    Hi Jerry,

    Can you please provide a schematic? Its hard for me to visualize your issue with just text.

    Best,

    Ian

  • +1
    TI__Guru* 97046 points

    Hey Jerry,

    I'm not entirely sure, but I believe you are talking about a circuit like this:

    If I am correct, the circuit provides a one-way delay. As I have drawn it, this would be a falling-edge only delay. The diode provides a bypass path for current on the rising edge, but on the falling edge, the capacitor must charge from the 1k resistor.

    Here's a simulation of the above circuit showing the input pulse and capacitor voltage:

    Given that this is a NAND gate and the inputs are connected with this circuit, it's likely that the circuit is being used to generate a pulse of a particular width only on the rising or falling edge, depending on the diode orientation.

  • 0 in reply to Emrys Maier
    Prodigy 10 points

                                                                                                                                                                                                                  Hi Ian and Emrys! Thank you both for your responses. The circuit I've described is the small part of the Bicron Micro-R Survey meter scheme it`s a micro-Rentgen counter.

    Best regards,   

    Jerry