0

I'm thinking to buy new RAM (for Lenovo 320 15AST, motherboard is LNVNB161216). I had 4GB, I need 8GB. My current RAM is m471a5143sb1 2133Mhz, CAS (CL) 17, PC19200.

I was looking at two with such specs:

Crucial [CT8G4SFD8213] 2133 Mhz, PC17000, 15-15-15-36

VS

Patriot Signature [PSD48G240081S] 2400 Mhz, PC19200, 17.

First is a bit cheaper, has frequency that no more that baseboard provides, lower latency, but lower PC (also I'm not sure which PC my baseboard provides).

(I think I'm gonna go with this one: SODIMM Crucial [CT8G4SFS824A] 8 GB)

(I have this question on superuser, I just had to move it here, there also answer already, but if you have something to add you're welcome)

1 Answer 1

4

The "PC" number is a DDR module name, it indicates module's data rate.

  • PC4-17000 modules have 2133 MHz data rate (2133 × 8 ≈ 17000)
  • PC4-19200 modules have 2400 MHz data rate (2400 × 8 = 19200).

The 4 indicates that it's a DDR4 module. 17000 and 19200 are peak transfer rates. DDR4 data bus is 8 bytes wide, so 8 bytes are transmitted each clock cycle, hence multiplication by 8.

Now, which is better: PC4-17000 with CL 15 or PC4-19200 with CL 17?

CAS latency (or CL for short) is measured in clock cycles, so we can calculate how much time fetching data takes for each of these modules.

  • 2133 MHz = 2,133,000,000 cycles ÷ 1 s

    Each cycle takes 1 ÷ 2133 µs

    Latency is 15 ÷ 2133 µs ≈ 7.0323 ns

  • 2400 MHz = 2,400,000,000 cycles ÷ 1 s

    Each cycle takes 1 ÷ 2400 µs

    Latency is 17 ÷ 2400 µs ≈ 7.0833 ns

So PC4-17000 has slightly better CAS latency.

But! All RAM has to work with the same clock speed, so once you've installed 2133 MHz module, your current 2400 MHz module will underclock to 2133 MHz too.

I'd probably go for 2400 MHz one. Latency difference is very small, but clock is 12.5% faster.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.