Dear All,

This is a Technical question regarding the APD3367 regulator IC. We have incorporated your IC in our circuit & have tested many times. It is running as per the Datasheet. Recently we have encountered one issue & We want your clarification regarding it.

We have one circuit where the IC will convert 15 V input to 3.3 V output & We need 30 mA output current. Initially we were using TRACO POWER TSR 0.5-2433 IC for this purpose. The current consumption was 24 mA with Traco Power IC.

Now We soldered ADP3367 smd IC & the current consumption increased to 66 mA. The IC started to heat up & output of 3.3 V is also not achievable. This happened several times & but every time the same result. Kindly guide us regarding this issue. Why current is increasing everytime when we are replacing Traco IC with your ADP3367 IC. Help & Support is needed.

Parents
• Hi, Nitesh.

Here are some thoughts/suggestions you might consider in your circuit.

• Load Current: I think your load current is 100mA instead of 30mA. Using your previous circuit wherein you are using a Traco IC switching regulator, the load current resulted to nearly 100mA as calculated above. We cannot always assume the load current to be nearly equal to the input current. This only holds true to linear regulators. This load current value must be determined exactly as this is needed in building your circuit using the ADIsimPower tool.
• Heating issues: Selecting a 1W series resistor dropper for ADP3367 is not yet optimal as power dissipation reached 0.97W (at minimum VIN) as shown below. You still need to increase your resistor power rating.
• PD = ((VINVINmin) × ILOAD)
• PD = ((15V – 5.3V) × 100mA)
• PD = 0.97 W

Regards,
Errgy

• Hi, Nitesh.

Here are some thoughts/suggestions you might consider in your circuit.

• Load Current: I think your load current is 100mA instead of 30mA. Using your previous circuit wherein you are using a Traco IC switching regulator, the load current resulted to nearly 100mA as calculated above. We cannot always assume the load current to be nearly equal to the input current. This only holds true to linear regulators. This load current value must be determined exactly as this is needed in building your circuit using the ADIsimPower tool.
• Heating issues: Selecting a 1W series resistor dropper for ADP3367 is not yet optimal as power dissipation reached 0.97W (at minimum VIN) as shown below. You still need to increase your resistor power rating.
• PD = ((VINVINmin) × ILOAD)
• PD = ((15V – 5.3V) × 100mA)
• PD = 0.97 W