### TI高精度实验室-带宽 3.pdf.pdf

TI高精度实验室-带宽 3.pdfpdf,TI高精度实验室-带宽 3.pdf
●主极点是在Ao图中,当Ao开始随频率下降的点。于发展 macro模型时,此 参数非常重要。极点频率可以从Ao曲线来仙计,但更准确的方法是使用GBW 及Aol来计算 ●以∂PA827为例,有22MHz的增益带宽积和126分贝的开回路增益。我们可 以运用这个公式转换126分贝以线性W)来表小,代入22MHz增益带宽及 1995乘以10的6次方开回路增益,得到11.03Hz的主极点频率。此计算与 数据表中的图形致。 STANDARD GRADE HIGH GRADE OPA827AI PA827)2 PARAMETER CONDITIONS N TYP MAX TYP MAX UNIT 卜KEQU〓 NCY RESPONS上 Gain bandwidth Product MHZ OPEN-LOOP GAIN AND PHASE S FREQUENCY 140 Slope = 20dB/decade 100 Gain Bandwidth is constant for closed loop gain from o to 120dB 60 -90 180 110100 Frequency (z TEXAS INSTRUMENTS o In this slide we further examine the oPA827 open loop gain curve. We see that the dc open loop gain is 120db and remains constant until we reach the dominant pole. At frequencies greater than the dominant pole, the open loop gain decreases at a rate of -20db per decade. Notice that for the opA827 the slope of aol is constant until we cross unity gain. therefore the gain bandwidth product is constant for closed loop gains from o to 120 decibels o While it is common to have open loop gain curves decrease at a constant rate of 20dB per decade, it is not always the case. For example, let' s take a look at the high-speed OPA847. ● Origina| notes: This slide emphasizes the fact that gain bandwidth is only defined over the range where Aol rolls off at a rate of 20d B/ decade. In this example Gain Bandwidth is defined over the entire range of Aol (odb to 120dB). Later we will show an example where gain bandwidth is not defined ●于此投影片,我们进一步研究OPA827的开回路增益曲线。我们看到,直流 开回路增益一直保持120分贝,直到我们到达主极点。当频率高于主极点, 开回路增益以-20dB/ decade的速率减小。请注意,OPA827的Ao斜率是定值: 直到我们越过 unity gain(单位增益)。因此,增益带宽积从0到120分贝的闭 回路增益为定值 ●虽然开回路益曲线以-20dB/十倍频的恒定速率下降很常见,但情况并非总是 如此。例如,让我们来看看高速的OPA847。 OPAB47ID, IDBV YP AIN/MAX OVER TEMPERATURE 0 C to-40@C to PARAMETER CONDITIONS +25C+25 C(2)I+85CI2 MAX Ac PERFORMANCE (see Figure 1) Closed-Loop Bandwidth G=+12.R6=3929,Vo=200mvp 600 MHZ G=+20,R。=3920,Vo=z00mvPF MHZ min G=+50.R=3929.Vo=200mV Gain Bandwidth Product(GBP) G2+50 3900 3100 3000 MHZ OPEN-LOOP GAIN AND PHASE Gain Bandwidth=12 X 600MHZ=7200MHZ 120 in x Bandwidth =20x350MHZ 下m Gain Bandwidth=[50 x78MHz 900MH log (A is-2Od B/decade Gain× Bandwidth=100×39MHz:390MHz Gain x Bandwidth=1000 X3.9MHz-3900MHzG Gain =50 8 lope below Gain= Not stable 20dB/decade for Gain< 12 F TEXAS INSTRUMENTS o This slide shows the Aol curve for the oPA847, whose gain bandwidth product is defined for only a portion of the aol curve. In this case it's defined only for closed loop gains greater than 50V/V. o Looking at the open loop gain curve, we see that for gains greater than 50V/V, or 34dB, the slope of the aol curve is -20db/decade. therefore the gain bandwidth product is equal to 3900MHz for all closed loop gains greater than 50V/V. o However, as the gain decreases below 50v/, the slope of Aol changes Therefore there is no gain bandwidth product specified. Instead, the closed loop bandwidth for particular gains is specified Also notice that for gains less than 12V/V the phase margin indicates that the device is not stable The table illustrates how the product of the gain and bandwidth is not constant for gains less than 50v/V, but is constant for gains of 50 or greater. ●投影片显示OPA847的Aol曲线,其增益带宽乘积只定义一部分的Ao出线 在这种情况下,只定义了闭回路增益大于50V/的增益带宽乘积 ●我们看到开回路增益曲线,当增益大于50V/V或34分贝,Aol曲线的斜率为 20dB/ decade的。因此,所有死循环増益大于5oV∧∨下的增益带宽积等于 3900MHz。 ●然而,随着增益小于50V/,Aol的斜率发生变化。因此,没有特定的增益带 宽积。相反地,特定增益的闭回路带宽是被指定的。还要注意的是,对于增 益小于12V/V,其 phase margin(相位裕度)表示系统并不稳定。下表说明了增 益低于50/时,増益和带宽的乘积不是定值,但当人于等于50ν∨为定值 For Inverting Amp For Buffer Non Inv Gain= 2 Non Irmy Gain= 1 BW GBW 22MHz E 11MHz GBW 22MHZ n Iry Gain 2 BW= =22MH BW Simulated 10 9MHz BW Simulated= 21.94MHz 中 TEXAS INSTRUMENTS e In part 2 we calculated closed loop bandwidth for a non-inverting configuration using the gain bandwidth product. You might be surprised to learn that the bandwidth calculation for the inverting configuration is calculated using the non-inverting gain. Note that the non-inverting gain is typically referred to as noise gaIn. e This example shows the same amplifier connected in both an inverting and non-inverting configuration. The inverting configuration has a gain of -1 and the non-inverting configuration has a gain of +1. Let's start by calculating the bandwidth for the non-inverting configuration. The bandwidth for the non-inverting amplifier U1 is calculated by taking the gain bandwidth product and dividing by the non-inverting gain So, for this example the bandwidth is 22 MHz divided by 1 which is equal to 22MHz o On the other hand the bandwidth of the inverting amplifier, U2, is calculated using the non-inverting gain. The gain with respect to the non-inverting input is calculated as rf/R1 +l, which is 2 in this example. So, the bandwidth of the inverting amplifier is 22MHz divided by 2 which is 11MHz. A common mistake is to consider the gain seen by the signal source rather than the noise gain for bandwidth calculations involving inverting amplifiers o This example is simulated to prove that the hand calculations are correct Notice that the simulation and hand calculation results are very close to each other. This simulation uses a simple single pole amplifier model to illustrate the relationship between bandwidth and circuit configuration. Later, we will see that more comprehensive models include secondary effects that have additional impact on the bandwidth ●在带宽系列视频第2部分,我们于同相组态使用増益带宽积计算岀闭回路带 宽。你可能会惊讶地得知,在计算反相组态的带宽也是使用同相增益的计算 方法。请注意,同相增益通常被称为 noise gaIn(噪声增益)。 ●这例子显示相同的放大器设置为反相和同相组态。反相组态具有-1的增益和 同相组态为1的增益。让我们先来讣算同相组态的带宽。同相放大器U1是 透过增益带宽积及同相增益来计算。所以,在这个例子中,带宽是22MHz 除以1,等于22MHz ●另一方面,在反相放大器U2的带宽是使用同相增益来计算。相对于所述同 相输入端的增益的计算公式为Rf/R1+-1,在这个例子是2。因此,反相放大器 的带宽是22MH除以2,等于11MH。一个常见的借误是以反相放大器的信 弓源增益来计算带宽,而不是噪声增益。 ●这个例子是用仿真证明手算的结果是正确的。请注意,在仿真和手算结果非 常接近。此仿真使用了一个简单的单极放大器模型来说明带宽和电路结构之 间的关系。稍后,我们将看到更全面的模型,包括具有对带宽额外影响的二 次效应。 Vin 阝 Aol 1M × Vin Allf) yout ==C Feedforward F= Feedback Factor R Vin ⊥M丈 Q vout TEXAS INSTRUMENTS o On the previous slide we demonstrated that you always need to use the non-inverting gain, or noise gain, when doing bandwidth calculations. This leads one to wonder what the theoretical basis is for always using noise gain regardless of the circuit configuration. First lets consider the inverting configuration The transfer function can be derived using either the controls system representation shown on the left or the circuit representation on the right. The controls system representation is intuitive so let's focus on that. The transfer function of inverting configuration is aol divided by 1 AOl x Beta multiplied by the feed forward factor alpha. aoL is the only term in this equation that changes with frequency. Now let's compare the inverting and non-inverting configuration ●在上一张投影片,我们证明,你需要使用同相增益或噪声增益来计算带宽。 这导致我们好奇是什么样的理论基础,得使用噪声增益来计算带宽而不管电 路组态。首先考虑反相组态。转移函数可以使用左边的控制系统或是右边的 电路表示。控制系统的分析表示较为直觉。反相组态的转移函数是AOL除以 1+AOL*B乘以前馈因子α。AoL是方程式中唯一随着频率变化的变量。现 在让我们比较反相和同相组态。

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