Hello,
I was reading a data sheet on some IC that has a Vcc pin, and an Enable pin (among others). The data sheet said that the input supply voltage should be applied prior to the application of the enable signal high. And then the enable signal must be set to low before removing the supply voltage.

Is there a simple way to do this?

Thanks!

 

Hello,
I was reading a data sheet on some IC that has a Vcc pin, and an Enable pin (among others). The data sheet said that the input supply voltage should be applied prior to the application of the enable signal high. And then the enable signal must be set to low before removing the supply voltage.

Is there a simple way to do this?

Thanks!

A "supervisor" IC from TI or Maxim should be able to do this function. There are various versions with different Vcc and logic styles. Take a look at TPS3836xx from TI and MAX6861xx from Maxim.

(* jcl *)

 

Okay, so it looks like I need the supervisor IC to enable/disable the primary IC, and a capacitor so that both of them have enough power to shut down after power down. I'm starting to think I'm in over my head.

Thanks for your help!

 

What is the specific IC that you are referring to ?

hgmjr

 

If you have a microcontroller (uC) in your system, it can take a role of the supervisor IC. Connect the enable pins to the digital output pin(s) of the uC and add a pul-down (or pull-up depending on your enable signal level) resistor. Many modern uCs have built-in supervisors (and brownout detectors) that wake up the uC, when the power is applied. The uC is waked up by it’s internal supervisor some time after the power is applied. Then the firmware waits for some and enables the rest of the system.

 

not to show any doubt or disrespect here, mr. wombat, but, are you sure they are saying supply voltage? because many ic's with enable say that the data inputs must be present (and stable) before the enable signal occurs. just curious.

 

Okay, so it looks like I need the supervisor IC to enable/disable the primary IC, and a capacitor so that both of them have enough power to shut down after power down. I'm starting to think I'm in over my head.

Thanks for your help!

Both ICs run off of the same Vcc which should be decoupled with a capacitor (or two)
anyway. The supervisor IC operates over a wider range of Vcc than the primary IC.
As the voltage starts to drop close to the Vcc(min) of the primary IC the primary IC is disabled.

The supervisor IC is very easy to use. Take a look at the application circuits in
the datasheet.

(* jcl *)

 

Hello,
Thanks for all the help you have given me so far! I've had to put this project on hold for a bit, but am trying to pick it back up now. The whole point of this project (I probably should have started with this) is to use a small electroluminescent lamp in a scale model. The EL lamp requires AC voltage, so I need an inverter to drive it from a battery. I found a D381B IC that does just this, in cell phones and PDAs, so I though I'd give it a try. I found a ZM331643GTA supervisor that I'm hoping will work for me (looking at the data sheet did make this a lot easier to understand). So I've got all the components I need except for the capacitors - how do I figure out how much capacitance I need?

Thanks again!

 

I second techroomt's question. One of my projects consisted of an A/D converter (ADC0804) controlled by a computer's parallel port. The converter had this Chip Select pin (~CS), which, when set HIGH, it would effectively turn off the IC even if it was powered. I think this allows for easier (less complex) and maybe faster turn-on/off.

So maybe you just need to delay the activation of your Enable pin after powering up the device (and completely ignore the power-off procedure). Check the datasheet more thoroughly, maybe it tells something about it's exact purpose.

 

Straight from the data sheet:
"The DC input supply voltage (V+) should be applied to pin 10 of D381B IC module prior to the application of the enable signal high to pin 1. Conversely, when powering off the device, the enable signal must be low prior to the removal of V+ signal."