101-1006 Rabbit Semiconductor, 101-1006 Datasheet - Page 214

• Manufacturer Part Number: 101-1006
• Description: MODULE POWERCORE FLEX 3800
• Manufacturer: Rabbit Semiconductor

Specifications of 101-1006

• Module/board Type: MPU Core Module
• For Use With/related Products: RCM3800
• Lead Free Status / RoHS Status: Contains lead / RoHS non-compliant

Option 5—Half-Wave Rectifier
Option 5 requires unregulated AC or DC. This option places a half-wave rectifier and fil-
ter before the regulators. The half-wave rectifier is necessary to rectify AC voltage into
DC voltage. This option has a friction lock connector installed on the PowerCore to accept
external power.
The half-wave rectifier causes a 0.7 V voltage drop before voltage is applied to the regula-
tors. This means that the 0.7 V drop will need to be taken into account.
With this option, unregulated DC voltage is input to pin 2 of friction lock connector J3.
The negative side of the input voltage is applied to pin 4 and/or 6 of friction lock connec-
tor J3. The selection of the +5 V regulator determines the voltage range for the unregu-
lated DC input power as follows.
• 8–43 V DC for 2 A regulator option
• 9–40 V DC for 1 A regulator option
AC voltage is applied to the same pins of J3. Because the peak AC voltages are 1.42 times
higher than the rated AC voltages, the AC voltage applied will have lower upper limits.
Also, because AC voltage inputs go down to zero voltage during each cycle, a higher min-
imum voltage than the DC voltage will be required. The selection of the +5 V regulator
determines the voltage range for the AC input power as follows.
• 17–30 V AC for 2 A regulator option
• 14–28 V AC for 1 A regulator option
The current required is calculated based on the power consumed by the PowerCore as
optionally configured: I = Power/Vin. The PowerCore will consume between 1 W and 2.7 W,
depending on the PowerCore options. Additional power may be needed to provide for
other user's circuitry external to the PowerCore. That additional power needed from the
input power supply is calculated as follows.
• For the user's 5 V circuitry, additional power from supply =
• For the user's 3.45 V circuitry, additional power from supply =
Additional power is taken by the input diodes.
Thus, as an example, if the user's 5 V circuitry consumes 2.5 W and the user's 3.45 V cir-
cuitry consumes 0.5 W, the total power needed from the power supply is (2.5 W × 1.33 +
0.5 W × 1.92 + (PowerCore consumption as optionally configured)) = 5.3 W to 7.0 W. If
the external power supply is a 24 V supply, then it will need to source I = Power/Voltage
current. Bear in mind that voltage presented to the regulators has a 0.7 V diode drop. Thus
the voltage in this formula must be reduced by 0.7 V, so current = Power/23.3 V = 5.3 W/
23.3 V = 227 mA; 7.0 W/23.3 V = 300 mA. An AC supply may not need to deliver that
much current because the filter capacitor may raise the DC voltage input into the regulator,
allowing less current to be consumed. For a 24 V DC input, this is what current the power
supply would need to deliver.
206
(5Vpower_to_user's_circuit) × 1.33.
(3.45Vpower_to_user's_circuit) × 1.92.
PowerCore FLEX