Hello

Our customer will use ADM3053 and ask minimum combined impedance and drive capability.

I think below. ( in the case of  cable loss =0)

Is my understanding correct ?

Datasheet P.1 says "Connect 110 or more nodes on the bus".

If load of CAN bus is compesed of 110 nodes of ADM3054 in parallel ( 20Kohm minimum Differential Input Resistance of ADM2053),

Combined resistor is 180 ohm.

When this 180ohm  is in parallele with !20 ohm termination resistor*2  on the CAN bus, final combined resistor is 45 ohm.

So, minimum impedance will be 45 ohm and drive capability is driver output voltage(MAX)　/combined resistor  =   3.0V/45 ohm =67mA.

Regards,

Terumasa

• Hello Terumasa-san,

The minimum drive capability with the maximum impedance of 45 Ohm, is specified in the datasheet as the minimum differential output voltage (Vod), of 1.5V. So the current in that case is 1.5V/45 = 33.3 mA.

As per your post, the driver differential output voltage could be up to 3V for the datasheet test circuit (current of 67 mA).

Best regards,

Conal

• Hello Conal-san,