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亲爱的男士,
我正在使用斯巴达6(100)。 我的目的是让RESET具有低于3的特性 PLL的锁定输出延迟1秒。 使用全局时钟路由 不要在CLB内部使用逆变器 我的重置=(未锁定)和(延迟) DelayPowerupValue:1,1秒后变为:0(大约需要1秒) LockedPowerup值:0 Pll锁定后变为1.(少于几毫秒) 问题1:如何确保MY RESET具有3个上限特征。 问题2:将MY RESET信号连接到BUFG输入并使用BUFG的输出工作吗? 额外信息:我真的需要使用延迟重置我正在使用需要保持重置1秒的反序列化器。 Xapp1064.pdftemplate代码甚至在代码中使用了异步重置。 我知道FPGA会自动重置,如果我们真的不需要,我们不想引入另一个重置 我想使用有源高电平复位但不幸的是我需要对信号进行门控,但我不确定bufg是否有效。 以上来自于谷歌翻译 以下为原文 Dear Gents, I am using spartan 6(100). My purpose is to have a RESET that has below 3 characteristics
MY RESET = (not Locked) AND (Delay) Delay Powerup Value: 1, After 1 second becomes : 0 (takes about 1 seconds) Locked Powerup Value : 0 After Pll locks becomes 1.(takes less then miliseconds) Question1 : How can I make sure MY RESET has the 3 upper characteristics. Question2 : Will connecting MY RESET signal to BUFG input and use the output of BUFG work ? Extra info :I really need to use delayed reset I am using a deserializer that needs to be kept resetted 1 second. Xapp1064.pdf template code even has async reset used in the code. I know that fpga is resetted automatically and we dont want to introduce another reset if we dont really dont need And I want to use active high reset but unfortunately I need to gate the signal but I am not sure if bufg works. |
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我已经实例化了pll并锁定了输出,我使用计数器创建了延迟信号。
我担心的是,我需要通过AND两个信号产生的信号是否会使用全局路由? GSR信号在配置期间始终置为有效,并且可以在配置后使用STARTUP_SPARTAN6原语进行控制。 为了最大限度地提高设计灵活性和利用率,请使用GSR并避免本地初始化信号.UG384.pdf 我会尝试这个 以上来自于谷歌翻译 以下为原文 I already instantiated pll and have locked output also I create the delayed signal using a counter. My concern is I need the signal that I generate by AND ing two signals will use the global routing or not? The GSR signal is always asserted during configuration, and can be controlled after configuration by using the STARTUP_SPARTAN6 primitive. To maximize design flexibility and utilization, use the GSR and avoid local initialization signals.UG384.pdf I will try this one |
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我不清楚你的延迟来自哪里。
您是说设计的输入持续约1秒钟,或者您是否需要在FPGA内产生1秒的延迟? 我认为合理的方法是使用一个计数器,当PLL未锁定时保持复位,并在PLL锁定后计数。 当计数达到某个等于时钟频率(Hz)的数字时,则结束上电复位信号。 如果它只需要大约1秒钟,您可以使用计数器的MSB作为~RESET信号(或者如果计数器复位值具有MSB高电平,则复位)。 最后,我不确定为什么要在全球网络上路由重置信号。 你有太多的负荷吗? 负载是否都是异步复位? - Gabor 以上来自于谷歌翻译 以下为原文 I'm not clear on where your delay comes from. Are you saying there is an input to the design that stays on for about 1 second, or do you mean you need to generate a 1 second delay inside the FPGA? I would think that the reasonable way to do this is with a counter that is held reset when the PLL is not locked and counts up after the PLL is locked. When the count reaches some number equal to the clock frequency in Hz, then you end your power-up reset signal. If it only needs to be approximately 1 second, you might be able to use the MSB of the counter as a ~RESET signal (or RESET if the counter reset value has the MSB high). Finally I'm not sure why you want to route the reset signal on a global net. Do you have too many loads on it? Are the loads all asynchronous resets? -- Gabor |
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亲爱的Gabor,
- 我正在使用正在使用有效高异步复位的Xilinx解串器模板。 - 是的,我有太多的负载来自现有的具有低有效异步重置的代码 - 我编写的新块根本不使用重置。 只是初始化值 但独立于负载我想学习实现具有不同极性的延迟复位的最佳方法(我需要实现有源高延迟复位和有源低延迟复位) 以上来自于谷歌翻译 以下为原文 Dear Gabor, -I am using the template of the Xilinx Deserializer that is using a active high async reset. -Yes I have too many loads in also from an existing code that has active low async resets -New blocks that I am codding does not use reset at all. just initialization value But independent from the load I want to learn the best way to implement a delayed reset with different polarities (I need to implement both active high delayed reset and active low delayed reset) |
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