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我的项目consistsinprogramming PSoC作为信号adquisitor利用ADCs不同类型才能看到它们之间的差异。 现在我尝试用sar6 ADC但我不了解它是如何工作的时候得到的样品。 当我阅读数据表和API函数我知道样品的当我叫sar6_cgetsample(),它用一些微秒转换示例。那么,如何影响模拟列时钟采样率?看来,采样率将是我的sar6_cgetsample()函数代替模拟柱时钟的相关代码。 另一个疑问是为什么数据表表说,转换时间为20微秒,与CPU 12 MHz和250千赫fclock条件,若数据表说,这个时间是周期的采样时钟的六倍。它是不是可变的参数? 最后,了解fclockmentioned的datasheet inthetables作为采样时钟相同(Φ和Φ),为什么是限制为333千赫如果它取决于模拟列钟吗?它会是几兆赫,不是吗? 我误解了什么? 谢谢您, 以上来自于百度翻译 以下为原文 Hello, My project consists in programming the PSoC as a signal adquisitor using the different types of ADCs in order to see the differences between them. Now I am trying to use the SAR6 ADC but I don't fully understand how it works when it get the samples. When I read the datasheet and the API functions I understand that the sample is taken when I call the SAR6_cGetSample(), and it takes some microseconds to convert the sample. Then, how does the Analog Column Clock influence in the Sample Rate? It seems that the Sample Rate would be related to the code where I have the SAR6_cGetSample() function instead of the Analog Column Clock. Another doubt is why some tables of the datasheet say that the Conversion Time is 20 microseconds with the condition of a CPU of 12 MHz and fclock of 250 kHz, if then the datasheet says that this time is six times the period of the sample clock. Is it a variable parameter or not? And finally, understanding that fclock mentioned in the tables of the datasheet is the same as sample clock (phi1 and phi2), why is it limited to 333 kHz if it depends on the Analog Column Clock? It would be some MHz, couldn't it? What I am misunderstanding? Thank you, |
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6个回答
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SAR由内置硬软件组成。写在数据表的时钟频率是模拟柱时钟除以四。在数据表的名字是fCLOCK,并列时钟/ 4。CPU的时钟的依赖已经跟SAR的工作方式,它在每个摊位的CPU转换步骤以得到准确的结果。所以(我阅读数据表)和模拟时钟速率为1.3 MHz,你将获得约20µS. @ 5V采样时间,24兆赫。鲍勃
以上来自于百度翻译 以下为原文 The SAR consists of built-in hard- and software. The clock frequency written in the datasheet is the analog column clock divided by four. The name in the datasheet is fclock, which is given as column clock / 4. The dependency of the cpu-clock has to do with the way the SAR works, it stalls the CPU while each conversion step is made to get accurate results. So (as I read the datasheet) with an analog clock rate of 1.3 MHz you'll get a sample time of about 20µs. @5V, 24 MHz. Bob |
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嗨,鲍伯,
谢谢你的回答。但我仍然怀疑萨尔。 虽然我调用Sal6xCGETSAMPLE()函数的连续时间(例如在BULE中),但是当CPU被释放时,样本将被获得,并且当转换时间(APROX)通过时将得到。因此,改变采样率的方法是改变转换时间(采样率=1 /转换时间?)通过模拟列时钟(对流时间=6×周期(模拟列时钟/ 4)),是正确的吗? 谢谢, 以上来自于百度翻译 以下为原文 Hi Bob, Thank you for answering. But I still have some doubts abut SAR6. Although I call the SAR6_cGetSample() function serveral times (for example in a bucle), the sample only will be obtained when the CPU is released, and it will be when the Conversion Time (aprox.) has passed. So, the way to change the Sample Rate is changing the Conversion Time (Sample Rate = 1/Conversion Time ¿?) through the Analog Column Clock (Conv. Time = 6 * Period of(Analog Column Clock / 4)), is it correct? Thanks, |
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你的假设是正确的。采样率=1 /转换时间。GETSAMPLE()函数等待直到进行转换(阻塞)。与ADCINCthere不同的是,没有办法启动转换,并寻找结果稍后准备好。
为什么你想增加转换时间?是的,当然你可以根据需要减少/调整列时钟,在大约125千赫的情况下模拟列时钟的下限。 鲍勃 以上来自于百度翻译 以下为原文 You are right with your assumptions. Sample rate = 1 / Conversion time. The GetSample() function waits until the conversion is made (blocking). Unlike ADCINC there is no way to initiate a conversion and look for a result ready a bit later. Why do you want to increase the conversion time? Yes, of course you may reduce/adjust the column clock at your needs, there is a lower limit for the analog column clock at about 125 kHz. Bob |
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好的,我明白了。
我不想增加转换的时间,我的问题只是为了了解它是如何工作的,因为在我的项目,我将不得不改变采样率,得到不同的结果和比较。 谢谢你的耐心。 以上来自于百度翻译 以下为原文 Ok, I get it. I don't want to increase the Conversion Time, my question was only in order to understand how it works, because in my project I will have to change the Sample Rate to get different results and to compare them. Thanks for your patience. |
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请问你为什么要使用SAR6?
它前面没有样本/HODD。当你触发它,它拥有24栏时钟周期CPU时钟。没错,CPU是停滞的那段时间,a1mhz列时钟,将24usec。这会严重混乱的其他用户模块快速响应中断的储藏。 如果是样品的速度,我们可以做一个delsig adcto有百分之五十cpuloading做60ksps @八位。(我们还没有发布新的Delsig,但是坚持下去!)我可以在125ksps做6位但CPU负荷接近100% 戴夫范斯 以上来自于百度翻译 以下为原文 May I ask why you are using the SAR6? It doesn;t have a sample/hodd in front of it. and when you trigger it, it holds the CPU clock for 24 Column clock cycles. That's right, the CPU is stalled during that time. For a 1MHz column clock that would be 24usec. This could seriously mess up other User Modules that rely on fast response for interrupts. If it is sample speed, we can make a DelSig ADC to do 60Ksps @ 8 bits with a 50% cpu loading. (We haven't released the new DelSig yet but hold on!) I could do 6 bits at 125ksps but the CPU loading is close to 100% Dave Van Ess |
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是啊,
我知道一个&;H在sar6前是必要的,这是强制性的快速信号但不慢,我想。 对于CPU的举行,我想用thesar6与UART和DAC模块共。UART是响应很好但对DAC的我没有yetimplemented它。如果我有问题,CPU响应我会尽量增加为sar6采样率减少转换时间,因此CPU响应时间。 为什么我在前面的其他转换器的原因是我的项目指南说,因为目的是比较不同转换器获得的结果,例如50/60 Hz干扰或噪声。 我真的很感激你对该变换器的特殊特性的评论。 费尔南多 以上来自于百度翻译 以下为原文 Yeah, I know that a S&H in front of the SAR6 is needed, which is mandatory for fast signals but not for slow, I think. Regarding to the CPU held, I want to use the SAR6 with an UART and a DAC modules altogether. The UART is responding quite well but about the DAC I haven't yet implemented it. If I have problems with the CPU response I will try to increment the Sample Rate of the SAR6 in order to decrease the Conversion Time and consequently the CPU response time. The reason why I use this converter in front of the other is that my project guidelines says it because the objective is to compare results obtained from different converters, for example the 50/60 Hz interference rejection or noise. I really appreciate your comments about the special characteristics of this converter. Fernando, |
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