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I am trying to load the output of a AT28C64B parallel EEPROM into a 74HC194 shift register. The datasheet for the AT28C64B states that the minimum high level output is 2.4V, which I believe is not enough to drive the input pins of the HC type shift register.

Therefor, I considered running the 74HC194 on 3.3V, and using a level shifter for the other input. Then 2.4V should be enough. But when doing measurement, I found that the actual output voltage from the AT28C64B was 4.8V, which is too much for the 74HC194 running on 3.3V.

The AT28C64B is powered with 5V.

So what should I expect the output voltage of the AT28C64B to be, and how can I ensure that the output voltage of the AT28C64B and the input voltage of the 74HC194 matches.

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1 Answer 1

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Few different solutions. In general TTL outputs are made compatible with CMOS inputs by using pull-up resistors. Put pull-up resistors on EEPROM output to make them go up to 5V, and switch the shift register to 5V. Another solution might be to use HCT type shift register that has TTL compatible inputs, but it would still require 5V supply for the shift register.

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  • \$\begingroup\$ Thanks. I would prefer running everything at 5V. The 3.3V for the shift register was only to make 2.4 valid as high input. So adding pull-up resistors to the eight output pins of the EEPROM would make it? Would 1kOhm be a good start? \$\endgroup\$
    – harelabb
    Dec 7, 2019 at 17:52
  • \$\begingroup\$ No, 1kohm is too low value, EEPROM would need to drive too much current. Try 10 k or 4.7 k. \$\endgroup\$
    – Justme
    Dec 7, 2019 at 18:01
  • \$\begingroup\$ Thanks. I will try that. \$\endgroup\$
    – harelabb
    Dec 7, 2019 at 18:05

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