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LMZ14203H Negative Voltage application

Other Parts Discussed in Thread: LMZ14203H, LMZ14203

Hi,

I would like to use the LMZ14203H to provide a -12 V Vout on a 24 V Vin. On the "Inverting Application for the LMZ14203 SIMPLE SWITCHER Power Module" application report, it shows an example circuit where the EN pin has a voltage divider that supplies 10.5Vin on turn on?

I am not quite understanding what value of Rent I should be choosing for my circuit. The LMZ14203H data sheet says the max voltage into the EN chip should be 6.5V.

Thanks.

This is the Inverting Application sheet: http://www.ti.com/lit/an/snva425a/snva425a.pdf

  • The value of 10.5V is the input voltage at which the device will turn-on.  The maximum voltage on the EN pin, with the values shown in the app note, is about 5V.  If you need a different value of turn-on voltage, we can help you calculate the values for the divider.

  • Thanks Frank.
    You are saying that the 10.5 V at the VIN turns on the device. I was confused because, assuming a 12V Vin, the voltage divider shows 10.5 on the EN pin which was over its threshold. I should then choose a Rent value that would supply approximately 5V for the EN pin?
  • Take a look at the attached simplified schematic.  With the device off, the vout=0V.  With 10.5V on the Vin, the voltage VEN=1.18V=10.5*(11.8/(11.8+93.1)).  With the device on and the input voltage all the way up to 37V, the total voltage across the device is 37-(-5)=42V; the max for the device.  The voltage from EN to GND, of the device, is then VEN=4.7V=42*(11.8/(11.8+93.1) wich is safe.  The voltage limits are defined from the EN pin to the GND pin of the device and not to the application ground.   0880.Scanned from device SCPR310.pdf