SN74LXCH8T245: VOH VIH

0.55 V × 4 mA / 32 mA = 0.07 V

The voltage drop for VOH is 4.5 V − 3.8 V = 0.7 V; so 0.7 V × 4 mA / 32 mA = 0.09 V, and 5 V − 0.09 V = 4.91 V.

Hi Vidya,

Vidhya J said:

VOH =3.8V and VOL= 0.55V levels for VCC=4.5V and 32mA current as per datasheet.

Above indicates ROH of (VCC-VOH) / IOH i.e 22 ohms and ROL of VOL / IOL i.e 17 ohms.

VOH for VCC 5 V is VCC - I(OH*ROH) i.e 5 V - (32 mA * 22 ohms) = 4.3 V.

New ROH with 4 mA is (5 V - 4. 3 V) / 4 mA = 175 ohms. I.e. VOH with 4 mA is still 4.3 V with ROH of 175 ohms.

Vidhya J said:

I am looking for VOH and VOL levels for Vcc=5V and <4mA

Hence, new VOH and VOL < 4.3 V  and < 0.1 V respectively.

You may also see [FAQ] How do I determine the output voltage (VOH, VOL) or output current (IOH, IOL) of a CMOS logic device?, thanks.

Best Regards,

Michael.