Hi Team,
Could you help confirm if VOH is affected by the pull down resistor in output?
The datasheet says “During a low-to-high signal rising edge, the O.S.circuits turn on the PMOS transistors (T1, T2) to increase the current drive capability of the driver for approximately 30 ns or 95% of the input edge, whichever occurs first. This edge-rate acceleration provides high ac drive by bypassing the internal 10-kΩ pull-up resistors during the low-to-high transition to speed up the signal. The output resistance of the driver is decreased to approximately 50 Ω to 70 Ω during this acceleration phase.”
Per my understanding, when there is a rising edge on the input A, T2 will be turned on while N2 is turned off, R2 is bypassed, so the VOH equals to VCCB. If a pull down resistor is connected to ground on output B, the VOH still equals to VCCB. Could you help check?
Thanks and Best Regards!
Hao
