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AM26LV31E: How to always enable AM26LV31E

Shyam Daram
Expert 1160 points
Part Number: AM26LV31E

i just want to enable the AM26LV31E driver always, but i am confused how to do it, i see the G pin to be enabled high, then what for /G is ?

if i wan to enable it always may i tie pin G  to 3.3V with a 10K pull up ??


how can i calculate pull up resistor in this case, as i see the max current being only 5uA how to calculate pull up resistor for these cases, kindly help

  • 0
    TI__Intellectual 2885 points

    /G is an active low input, it does the same function as G, but when a low input is applied, the driver is enabled. You can see this in Section 8.4 under Table 1.

    As for the pullup resistor, the VIH (High-level input voltage) is 2 V. This can be found on pg. 4 in section 6.3. Any voltage equal to or greater than 2 V will be read as a high on the G pin. because your VCC is 3.3 V we can calculate the max resistor value you need.

    (3.3 - 2)/ 5μA = 260kΩ

    Any resistor value smaller than that will pull the G pin to a higher voltage than 2 V. So yes, a 10 kΩ resistor will work just fine.

  • 0
    TI__Mastermind 19360 points
    Hi Shyam,

    The AM26LV31E device has has both an "active high" and an "active low" driver enable pin. Since some systems are designed to use active high enable signals, while others are designed to use active low enable signals, having only a single enable pin will require an extra inverting circuit to be used for half of the applications. To avoid this the AM26LV31E provides the flexibility to be used in either case. Only one of the pins needs to be used as the primary enable input, while the other one can be pulled either high or low, but should not be left floating.

    When the driver is enabled through either G or /G, the output pins Y and Z will be either High or Low depending on the voltage applied to the A pin. However, the output pins Y and Z can be placed in a High-Impedance state if G is pulled low and /G is pulled high.

    Most applications will use a micro controller or other device to actively enable/disable the outputs of the AM26LV31E. Table 1 of the datasheet provides the full truth table. If you don't care about the High-Z state and only want to use a single pin to enable the driver, then you can do the following:

    Active High application:
    Connect the G pin to the MCU and pull /G up to VCC. This will ensure the output state will follow the input on G.

    Active Low application:
    Connect the /G pin to the MCU and pull G down to GND. This will ensure the output state will follow the input on /G.

    However, since you want to enable the driver and never disable it, you can do any of the following:
    1.) Pull up G to VCC and pull down /GE to GND.
    2.) Pull up G and /GE to VCC
    3.) Pull down G and /GE to GND.

    Option 1 sets both G and /GE to their active enabled state and either one of them will cause the driver to be enabled on their own. However this requires two pull up resistors, one to VCC and one to GND.

    Option 2 sets G to the active state and enables the driver and the /GE input is a don't care. This solution can share a single pull up resistor to VCC.

    Option 3 is the opposite of Option 2 and sets /GE to the active state and enables the driver while the G input is a don't care. This solution can share a single pull down resistor to GND.

    The use of either a pull up or pull down resistor is a good idea and recommended but not necessarily required if BOM cost or layout space is of concern to you. For your static application, the G and /GE pins could be connected directly to either VCC or GND.

    I might be overlooking the 5uA spec in the datasheet you are referencing and instead I am seeing the Input Current spec at +/- 10uA. This implies that if you connected the pin directly to VCC, the pin would sink 10uA of current. If your VCC = 3.3V, then that would be the equivalent of a 330k ohm resistor to GND. An external resistor is simply in series with this equivalent resistance but adds a bit of a buffer between the power and the input pin should there be some form of short or other error. But using a 10k ohm resistor for your application is perfectly fine.

    This is a good general application note that I know of that goes into good detail about how to calculate and chose the proper pull up resistor value. www.ti.com/lit/an/slva485/slva485.pdf

    Regards,
    Jonathan