What would be a suitable enable circuit for the LMZ14201 that is NOT powered from Vin? I need to drive this input from an external 5V and got totally confused by this statement in the datasheet:
"The EN pin is internally pulled up to VIN and can be left floating for always-on operation"
Does it mean I cannot drive this from a 5V input as there will be current flowing back into this input since VIN is usually higher than 5V? I need the switcher to be OFF when 5V input is not active, and turn it ON when the 5V input becomes, well... 5V
The EN pin is internally tied to Vin but you can tie a 5V source to it to control turn and tun off. The pull down current is 6uA, in significant compare to the switching current.
EN needs to be high (>1.2V) for the part to work. If you decide to tie to vin, make sure not to exceed 6.5V on EN by using a zener diode.
You can also have a voltage divider from Vin to En in reference to 1.18V to change the UVLO option.
Thanks for the prompt reply! The 5V input is actually an optoisolator transistor output, and I am not sure the current can flow backwards. What is the VIN pullup resistor value, and is VIN the source for the 6uA current?
depending on the optoisolator transistor, if iti s open drain or open collector it will take current. What is the part number for the opto?
Vin is the source of the 6uA current.
The optoisolator PN is ACPL-217-500E. It has a NPN output photo-transistor, and (unlike a MOS transistor in saturation) I don't think I can swap the collector and emitter as far as the current flow is concerned. Is there another circuit you could recommend?
In order to pull EN high using the opto mentioned above, just tie the collector to a rail lower than 6.5V and the emiiter to EN. when the light turns ON, the BJT is ON and the EN is pulled High from the rail. In this configuration, you will need a resistor to pull it low. Make sure you size the resistor to En is above 1.2V
If you have the opto as pull down, connect the collector to En and Emitter to ground, when the light turns ON, the BJT is ON and the En is pulled to GND.
Thanks for your answer! I am not sure I understand what you mean by "you will need a resistor to pull it low"? Could you please be a little more specific?
When you tie a node low (to GND), you need a resistor in the K Ohm range to limit the amount of current drawn from that pin. Ohm's law principle.
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