TXB0104: TXB0104 drive strength question

Hi Dane,

Because the TXB0104 is an auto-bidirectional device, it always has active outputs on both sides of the device. To allow the user to force one side to a new state, these drivers have an added series 4kohm resistor.  This is shown in the datasheet functional block diagram:

Let's say the current state is "LOW," forcing B4 in the above image to 0V. When your driver turns on and attempts to force the same pin HIGH (for this example, let's make that 5V), then the device must output enough current to make this happen.

One side of the 4k resistor will be 0V, and the other side will ramp from 0V to the threshold that causes the device to switch (typically Vcc/2 or 2.5V in this case).  If we calculate the current required to drop 2.5V across a 4kohm resistor, we get 2.5/4000 = 625 uA. Of course, we must _exceed_ 2.5V, before the output will switch to the 'HIGH' state, but it won't take much more than the calculated 625 uA.

The next question is - how does the output actually do this?  It's not a magic box - there's probably a MOSFET in there that's driving the output to 5V, and it has a resistance associated with it. A strong driver will be rated to high currents (typically 64mA or 32mA), while a weak driver will be rated to lower currents (typically 1 mA or 100uA).

Let's say that the output is rated to output exactly 1 mA at VOH = 3.8V, which means it has an equivalent output resistance of (5-3.8)/1m = 1.2 kohms.

At what resistance will a device no longer be able to trigger the TXB device's threshold of 2.5V?  There's a voltage divider here -- so we look at when VDD/2 = VDD*4k/(4k + R)

R = 4k

But what if the threshold isn't exactly 2.5V (it isn't)?  What if the internal resistor isn't exactly 4k (it's probably not)? Does all this change over temperature (yes)?

A 2mA output drive recommendation provides ample room for these types of variations and guarantees that the device will function correctly.