您好，
1、下面TI 高精度实验室系列课程的噪声 6讲了如何创建运放的噪声模型：
https://edu.21ic.com/video/2546
2、该宏模型开环输出阻抗是多少？
您是怎样反算的，可以提供具体计算方法吗？请注意其闭环输出阻抗与频率、增益有关
谢谢！我是根据Tim Green的运放稳定性分析里提的输出阻抗换算的。
这个是我反算出来的值，特别是低频部分差异巨大
对延迟回复对您表示非常抱歉。
我将此问题咨询了美国的资深工程师，他的回复如下，总的意思是：很可能是数据手册给出的开环增益曲线不准确造成的，他的回复翻译成汉语为
当试图将dB值线性化时，仅基于估计值可能会有相当大的差异。数据表上的图形似乎是从物理打印输出中扫描出来的，这并不符合最佳近似值，但存在可以以相当合理的精度进行估计的软件。
按照Tim Green列出的相同步骤，计算10kHz时的开环输出阻抗，我根据每条曲线得出以下结果，并查看一个频率，以便于获得相同的Aol值。
Av=1: 0.3*（1+398/1）=Ro=119.4欧姆
Av=10: 1.2*（1+398/10）=Ro=48.96欧姆
Av=100: 10.2*（1+398/100）Ro=50.7欧姆
由于这些都是提供的数据点，您可以看到Av=1曲线与其他两个曲线不对齐，并且Ro很可能在50欧姆左右。所以我对开环增益的猜测可能不是很准确。
“There can be a pretty significant difference just based on estimation when trying to linearize a dB value. The graphs on the datasheet appear to be scanned from physical printouts, which doesn't lend itself to the best approximations, but there exist software that you can estimate with fairly reasonable accuracy.
Following the same procedure listed by Tim Green, and calculating the open loop output impedance at 10kHz, I got the following results based on each curve, and looked at one frequency for convenience of having the same Aol value.
Av = 1 : 0.3*(1+398/1) = Ro = 119.4 Ohms
Av = 10 : 1.2*(1+398/10) = Ro = 48.96 Ohms
Av = 100 : Ro = 10.2*(1+398/100) 50.7 Ohms
Since these are all the data points provided, you can see that the Av = 1 curve doesn't align with the other two, and most likely the Ro is around 50 Ohms. so it's likely my guess for the open loop gain was not very accurate.”
