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SSD backup capacitors - how much time before shutdown?


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#1 fzabkar

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Posted 11 June 2013 - 11:00 PM

How much time does an SSD need to perform housekeeping before shutdown after an unexpected power loss, and do backup capacitors provide sufficient time?

I'm looking at this Intel 320GB SSD:
http://images.anandt...20/DSC_2617.jpg

I see six 470uF, 6V Kemet Organic (KO) capacitors in parallel.

http://www.kemet.com...ben/T520 KO-CAP
http://www.kemet.com/kemet/web/homepage/kechome.nsf/file/T520%20Series/$file/KEM_T2015_T520.pdf

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.

http://www.skhynix.c...lNo=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:
http://www.ti.com/li...3e/slvs413e.pdf

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.

#2 bards

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Posted 11 June 2013 - 11:43 PM

How much time does an SSD need to perform housekeeping before shutdown after an unexpected power loss, and do backup capacitors provide sufficient time?

I'm looking at this Intel 320GB SSD:
http://images.anandt...20/DSC_2617.jpg

I see six 470uF, 6V Kemet Organic (KO) capacitors in parallel.

http://www.kemet.com...ben/T520 KO-CAP
http://www.kemet.com/kemet/web/homepage/kechome.nsf/file/T520%20Series/$file/KEM_T2015_T520.pdf

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.

http://www.skhynix.c...lNo=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:
http://www.ti.com/li...3e/slvs413e.pdf

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.


You sound like you know exactly what you are talking about dude....but here goes;

I read somewhere that the majority of data in the DRAM is, typically, the indirection table and not user data. This appears to be backed up by this article;

http://www.anandtech...ller-analyzed/3

Edited by bards, 11 June 2013 - 11:45 PM.

#3 fzabkar

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Posted 12 June 2013 - 06:02 PM

Thanks for that. However, your information now introduces more conceptual problems for me. It now appears that the FTL (or "indirection table") is too large to be flushed to the array in its entirety, so this would suggest that, at any one time, there must be only a small portion that is not backed up.

There is a similar review for the 200GB Intel S3700 here:
http://www.storagere...rise_ssd_review

There is a single 4Gbit (= 512MByte) SDRAM, so this is 8 times larger than the SDRAM in my example. Yet the photo of the S3700 PCB shows two 47uF 35V electrolytic capacitors which amounts to about 100uF (2 x 47 = 94uF) compared with the 3000uF of the "lesser" SSD. I can't make out the part numbers of the ICs adjacent to the capacitors, so I can't envisage how these capacitors are being used.

I notice that the SSD's label shows a power rating of 1.44A @ 5V (= 7.2W) and 0.6A @ 12V (= 7.2W), yet the power consumption specifications are 6W (typical) when "active", and 650mW when idle. This suggests to me that the +12V rail may be used for backup, and that the 0.6A current draw only occurs momentarily during a power failure, or when the capacitors are being charged. Does this make any sense???

As for the voltage rating of the 47uF capacitors, if they were charged directly from the +12V supply rail, then 16V or 25V capacitors would have sufficed. This leads me to wonder whether the capacitors are charged to some higher voltage, say 25V, via a boost converter. The energy stored by a capacitor is given by ...

E = 1/2 x C x V x V

Therefore, in order for two 47uF capacitors to store the same amount of energy as six 470uF capacitors ...

E = 1/2 x (2 x 47) x V12 x V12 = 1/2 x (6 x 470) x 5 x 5

V12 = sqrt(30) x 5 = 27V

It would be interesting to measure the actual voltage on the capacitors to see if such an arrangement is indeed being used.

#4 fzabkar

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Posted 14 June 2013 - 04:56 PM

Just to put the economics into perspective, 470uF, 6.3V KO capacitors cost about US$8 - $9 each in single quantities, depending on whether one opts for the 10 milliohm or 4 milliohm ESR devices. Single quantities of the KZH 47uF, 35V, low ESR aluminium electrolytic capacitors cost 35 cents each.

This means that the backup capacitor bank for the enterprise SSD would cost me $0.70, but the backup capacitors for the consumer grade SSD would set me back around $50.

http://www.digikey.c...T530X477M006AHE

http://au.mouser.com...1z0wrk9Z1z0x3xj

Edited by fzabkar, 24 June 2013 - 04:56 PM.




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