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SN75188: Converting +-7V Square Wave to 1Vpp and Vice Versa Using TI SN75188 and TI SN75189

user3855443
Prodigy 60 points
Part Number: SN75188
Other Parts Discussed in Thread: SN75189

I am trying to design a circuit which has an input like an +-7V square wave and changing it to a 1V peak-to-peak square wave (+-0.5V or 0-1V not important) with saving its shape of signal. Circuit will work double-sided (1V peak-to-peak signal can be converted to +-7V also). I have to design this cirucit TI SN75188 Driver (for +-7V to 1V peak-to-peak) and TI SN75189 Receiver (for 1V peak-to-peak to +-7V). Usage of these ICs is a restriction, so i can not use any opmap to make this circuit. Do you have an idea related to how i can obtain this levels using these ICs? Do i need anything else to get my specifications? The frequency of +-7V signal is 16kbit/s. This is a similar signal to RS-232 but not it actually.

Driver has +-9V and receiver has +5V supply voltages. I am considering that taken this voltages from a transformer and full-wave bridge rectifier and proper regulators.

If you share your ideas i will be happy. Thank you so much.

  • 0
    TI__Guru 62920 points
    Hello,

    The SN75188 is able to give output voltages of at least +/- 7.5 V, but it would need an input voltage that is large enough to exceed the specified input high-level and low-level thresholds. You may be able to use a logic level translator in order to increase the voltage level, or you could potentially use your 0-1 V signal to control the gate of a FET that then translates the logic level (this would invert the signal as well).

    The SN75189 can receive the larger signals, but would give a full logic-level output that would scale with the supply voltage. Have you considered attenuating this output voltage using a resistive voltage divider? (I'm not sure how close you need to be to 0 V and 1 V specifically, or if some inaccuracy is OK.)

    What did you mean that the shape of the waveform needs to be retained? The outputs of these devices will always be driven fully high or fully low (e.g., in order to represent binary high/low levels), so if any intermediate values are needed then we would need to take a different approach.

    Max