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  1. No, it does not have to be. Firstly I'll explain the prefetch architecture, then a read to precharge scenario with a single burst. Many DDR memory systems use prefetching technology to reduce the internal memory clock while still allowing for high transfer rates. The prefetch architecture uses an internal memory bus that is wider than the I/O bus by however many times the prefetch architecture used is. On DDR3 and DDR4 and 8n prefetch architecture is used, this means that internal memory bus is 8 times wider than the external I/O bus. The prefetch architecture works by having the data stored transferred from the internal core memory into prefetch buffers for reads and the data transferred from the prefetch buffers to the internal memory for reads. It takes a single internal memory clock cycle to transfer this data both ways, meaning that when the read command is addressed, after 4 I/O bus clock cycles the data will be in the prefetch buffers, and ready to transfer. Due to this having a CAS latency below 4 on DDR3 is not possible. This means that the DRAM is free to be precharged just 4 clock cycles after the read command is addressed, even though the burst has not occurred yet. Now I'll explain a single read burst to a precharge. This is a diagram of a read to precharge scenario I made my that applies to both DDR4 and DDR3 memory systems. This is a hypothetical situation where CL = 16, tRCD = 16, tRP = 16 and tRTP = 4, these timings are all legal and viable. So in this scenario the memory is firstly activated, which opens the row that is going to be read from, then tRCD clock cycles later, the read command is addressed which chooses the column to read from, and then starts the internal data transfer from the internal memory to the prefetch buffer. As DDR3 is 8n, the internal memory bus is 8 times wider, and as DDR3 is well DDR, the internal memory clock is 4 times slower then the physical I/O bus clock. This means that the data is transferred to the external prefetch buffer just 4 I/O clock cycles after. This means after this point, the memory can be precharged when ever. When the read command is addressed you can then see tRTP is the read to precharge delay, whilst CL is the delay to the start of the burst, both commands starting simultaneously but not caring about the other. tRP is then the recovery from the precharge to when the memory can be activated or refreshed again. So the min value for tRTP is just 4, not (CL+BL) - tRP, CL and BL don't even need to be accounted for since they go down a different path, and tRP happens when tRTP is expired. I hope this helps, I'm happy to answer any other questions you might have
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