常见IC设计/FPGA面试问题之：FIFO深度计算

原文可以看FIFO Depth Calculation 这篇文章。

FIFO深度计算的关键在于：
在规定时间内传输的数据等于接收的数据，写快读慢的情况下，突发burst写入的数据减去该burst时间内读出的数据，多余的数据需要能缓冲下来，让接收端在剩下空闲的时间能从容地把多余的数据读出来。

下面看几道例题。

case1：

• fA > fB with no idle cycles in both write and read.
• Writing frequency = fA = 80MHz.
• Reading Frequency = fB = 50MHz.
• Burst Length = No. of data items to be transferred = 120.
• There are no idle cycles in both reading and writing which means that, all the items in the burst will be written and read in consecutive clock cycles.

计算方法（本题假设写数据和读数据位宽相同）：

• 总传输数据为120
• 把所有数据全部写入FIFO需要时间为 120/80MHz=1500ns 120 / 80 M H z = 1500 n s
• 该时间段（1500ns）内可以从FIFO中读出的数据为 1500ns∗50MHz=75 1500 n s ∗ 50 M H z = 75
• 多余的数据为 120−75=45 120 − 75 = 45
• 所以FIFO的最小深度为45

case2

• A > fB with one clk cycle delay between two successive reads and
  writes.

这种情况和case1相同

case3

• fA > fB with idle cycles in both write and read.
• Writing frequency = fA = 80MHz.
• Reading Frequency = fB = 50MHz.
• Burst Length = No. of data items to be transferred = 120.
• No. of idle cycles between two successive writes is = 1.
• No. of idle cycles between two successive reads is = 3.

计算方法

• 写时钟的等效频率为 fW=fA/2=40MHz f W = f A / 2 = 40 M H z
• 读时钟的等效频率为 fR=fB/4=12.5MHz f R = f B / 4 = 12.5 M H z
• 最小深度为 120−120/fW∗fR=83 120 − 120 / f W ∗ f R = 83

case4

• fA > fB with duty cycles given for wr_enb and rd_enb.
• Writing frequency = fA = 80MHz.
• Reading Frequency = fB = 50MHz.
• Burst Length = No. of data items to be transferred = 120.
• Duty cycle of wr_enb (write enable) = 50 % = ½.
• Duty cycle of wr_enb (write enable) = 25 % = ¼.

和case3相同，只是问题描述方式不同

case5

• fA < fB with no idle cycles in both write and read (i.e., the delay between two consecutive writes and reads is one clock cycle).
• Writing frequency = fA = 30MHz.
• Reading Frequency = fB = 50MHz.
• Burst Length = No. of data items to be transferred = 120.
• There are no idle cycles in both reading and writing which means that, all the items in the burst will be written and read in consecutive clock cycles.

这种写慢读快的情况fifo的深度为1就够了。

case6

• fA < fB with idle cycles in both write and read (duty cycles of wr_enb and rd_enb can also be given in these type of questions).
• Writing frequency = fA = 30MHz.
• Reading Frequency = fB = 50MHz.
• Burst Length = No. of data items to be transferred = 120.
• No. of idle cycles between two successive writes is = 1.
• No. of idle cycles between two successive reads is = 3.

这个case看似是写慢读快，但是因为有idle周期的存在导致

• 写时钟的等效频率为 fW=fA/2=15MHz f W = f A / 2 = 15 M H z
• 读时钟的等效频率为 fR=fB/4=12.5MHz f R = f B / 4 = 12.5 M H z
• 最小深度为 120−120/fW∗fR=20 120 − 120 / f W ∗ f R = 20

case7

• fA = fB with no idle cycles in both write and read (i.e., the delay between two consecutive writes and reads is one clock cycle).
• Writing frequency = fA = fB = 30MHz.
• Burst Length = No. of data items to be transferred = 120.
• There are no idle cycles in both reading and writing which means that, all the items in the burst will be written and read in consecutive clock cycles.

这种情况不需要fifo，如果clka和clkb有相位差，可以采用两级dff或者深度为1的fifo即可。

case8

• fA = fB with idle cycles in both write and read (duty cycles of wr_enb and rd_enb can also be given in these type of questions).
• Writing frequency = fA = 50MHz.
• Reading Frequency = fB = 50MHz.
• Burst Length = No. of data items to be transferred = 120.
• No. of idle cycles between two successive writes is = 1.
• No. of idle cycles between two successive reads is = 3.

同样算等效时钟频率就行了。

case9

• Writing Data = 80 DATA/100 Clock (Randomization of 20 Data’s)
• Outgoing Data= 8 DATA/10 Clock.
• Burst size = 160

可能有下面几种情况，以及每种情况下写完160个burst所需要的周期数

考虑最坏的情况case4，160个cycle就要写完160个数据的burst，在这160个时钟周期内能从fifo中读走 160/10∗8=128 160 / 10 ∗ 8 = 128 个数据，所以fifo的深度为160-128=32

case10

• Frequency (clk A) = frequency (clk B)/4
• Period (en_B) = period (clk_A)*100
• Duty cycle (en_B) = 25%

假设clkb=100MHz, 则clka=25MHz, 看起来是一个写慢读快的情况，不需要fifo。

但是因为第三个条件en_B的duty cycle只有1/4个周期，也就是在这个burst期间只有1/4的时间能从fifo中读出数据，这1/4周期的时间内是读快写慢，fifo不需要缓存数据。其余3/4个周期只有写，没有读，所以fifo的深度要能缓存下这3/4个周期中写入的数据。

从第二个条件可以看出burst的数据为100，总burst的时间为 100/25MHz=4000ns 100 / 25 M H z = 4000 n s , 3/4个周期就是3000ns，所以fifo的深度为 3000ns∗25MHz=75 3000 n s ∗ 25 M H z = 75