如果供电为10V/100mA,那要怎样支持大约300mA的GPRS发射？

Hi All,
Currently I am designing a power management system for GPRS modem.
The input power is 10V/100mA. The VBATT programmed from DC-DC Buck Regulator (LM3845) is set to 4V. DC-DC Buck Regulator has current limit protection function. So i programmed it to 100mA also.

Pout = Pin x efficiency.
Assume efficiency = 90%.
4V x IBATT = 10V x 100mA x 90%
So the expected IBATT should be 225mA.

However, GPRS current drawn average is around 300mA in Class 8.
Therefore, the extra current should be supported by Lithium Battery and supercapacitor.
Any one has the idea of how to choose super capacitor or the lithium Battery?
How about the hardware connection for Battery? Parallel or Series from DC-DC Buck Regulator? I am using IWOW TR-900 GPRS module.
Thank you.
Regards,
Elfrec
大家好
最近正在设计供电系统于GPRS 数据机。
供电设定为10V/100mA,假设我们用DC-DC Buck Regulator， 把VBATT 调成4V 和 电流保护为100mA.
Pout = Pin x 效率
如果效率是90%。
那算法应该是：
4V x IBATT = 10V x 100mA x 90%
那 IBATT 应该是于 225mA 之间.
但是，GPRS CLASS 8 的电流应该是在300mA 之间，那么多余的电流应该由SUPER CAPACITOR 或 LITHIUM BATTERY 承受巴。
问：谁知道如何选择SUPER CAPACITOR 或 LITHIUM BATTERY?
它的电路结法又如何? 是从DC-DC BUCK REGULATOR 接平行还是连接行?
在下用的是IWOW的TR900 GRPS MODULE。
谢了