I'm looking at this Intel 320GB SSD:
I see six 470uF, 6V Kemet Organic (KO) capacitors in parallel.
The total capacitance is about 0.003F.
Assuming that they are charged to the supply voltage (+5V), and assuming that the SSD can still function down to 4V, then the amount of time before shutdown would be given by ...
dt = C x dV / I
Assuming that the current draw is 0.5A, then ...
dt = 0.003F x 1V / 0.5A = 6 milliseconds
The SDRAM on the PCB is a 64MB hynix H55S5162EFR.
If we assume that the write transfer rate is 200MB/s, then in 6ms the SSD could write 1.2MB of data to the array. This is far short of the buffer size.
I notice that there is a PWM controller (marking CAW) nearby.
TPS61041DRV, Texas Instruments, marking CAW, 1MHz, 1.8V - 6V, low power DC/DC boost converter:
It appears that it might be charging the capacitor array. If so, and if it works down to 1.8V as specified, then the voltage margin would be 3.2V, in which case the available housekeeping time would be increased by however long the main PSU was able to maintain the supply voltage above 1.8V.
ISTM that protecting the user's data from corruption is not a primary design consideration. If it were, then the capacitor bank would need to be a great deal larger so that the entire SDRAM cache could be reliably flushed. Instead I believe that the capacitors merely attempt to ensure that the drive's Flash Translation Layer (FTL) is not damaged. The FTL is updated as a necessary consequence of wear levelling, so it would be at risk of corruption during a power failure. The FTL maintains a table which maps LBAs to physical NAND block addresses. If the table were to be corrupted, then the SSD would disappear from BIOS and the user's data would become inaccessible.
Edited by fzabkar, 11 June 2013 - 11:03 PM.