TPS79533 LDO Transient current

Hi Edison,

The TPS795xx is rated for 500mA steady state but has an internally set current limit of several amps.  It is certainly conceivable that during a large load transient that 2-4Amps could be drawn from the battery and thereby pull down the battery voltage.  However, assuming that the transient is very short lived - which is typical - the problem is solved by say increasing the capacitance at both the input and the output of the LDO - and then it is more the capacitance that supplies the energy for the transient than the battery.  

Regards

Bill

Hi Bill

Thanks for the reply. It's interesting that you mentioned 2 - 4 Amps. In my application the LDO supplies power to two components but when it is enabled the devices are not enabled so the load on the LDO is not very significant and is in the mA range (about 3mA). The transient current lasts about 50us. When my LDO is connected to lits loads, it peaks at 4A. But when it is completely isolated from its loads the transient current which I measured is about 2A. Is it normal for the LDO to give a 2A transient even when it is unloaded.

Can the transient current be supressed without affecting the normal functioning of the LDO.

Thanks

Edison

Hi Bill

Thanks for the reply. Sorry about my last reply, when I thought I had unloaded my TPS79533 LDO, I had a 10uF cap in parallel with a 0.1uF cap as output capacitance. When I finally removed the 10uF cap, the 2A-peaking transient disappeared. I would like to implement a soft-start-up external circuit to increase the start-up rise time from 50us to 10ms. I am using the following design notes:-

<http://www.ti.com/lit/an/slyt096/slyt096.pdf>

It describes two soft start-up architectures but I would like to use the first one which has an PFET switch which seperates regulator output from load and an NFET which controls the PFET. My question is, how did the design notes arrive at the formula for R_T {R_T = (V_out - V_th)/(V_out * (C_T/t_rise))}?

Thanks

Edison