TI高精度实验室-带宽 2.pdf.pdf

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he open-loop gain, or Aol, of an op amp represents the gain applied by the amplifier to the voltage difference between the inputs of the device. Aol for an ideal amplifier is infinite. Modern real-world op amps, however, have open loop gains in excess of 1 million volts per volt, or 120dB e In order for an amplifier to be useful, negative feedback is required. This is accomplished with Rf and R1. Sometimes this is referred to as 'closing the loop Rf and r1 represent the beta network, or feedback factor, of the op amp circuit Beta is a measure of how much of the output voltage, Vout, is fed back to the inverting terminal of the op amp. In this circuit we see that rf and R1 create a voltage divider. Therefore beta is equal to R1 divided by r1+Rf o In addition to the open loop gain of the op amp, we now have what is called the closed-loop gain, or Acl, of the op amp circuit. The equation for Acl is Aol divided by 1+ Aol"Beta, where Aol" Beta is known as the Loop Gain. This equation can be re-arranged as shown o The close loop gain equation can be simplified for very large values of loop gain or Aol x B. Looking at the loop gain equation you can see that as aol x B increases towards infinity, you can ignore the 1"term in the denominator. This leaves Aol over Aol x B, and the Aol terms will cancel, leaving only 1 over beta. Substituting the equation for B from above yields a closed loop gain of 1+ Rf /R1. This is the common closed loop gain formula for a non-inverting amplifier. The main point to understand is that this equation is only true for very high values of loop gain. Later we will see what happens for low values of loop gain Op Amp Bandwidth R1 1k R 1K Aol de Acl 1M t20 30 yak ccK 101 Freq Nancy (H7 HU TEXAS INSTRUMENTS 3 然而在现实世界中,运算放大器的开环增益具有低频的 dominate pole(主极点)。如图 所小,这可以被看作是一个 RC filter(滤波器)。 ●此仿真描绘了頁实世界的运算放大器开环回路增益。在直流或低频,Aol是非常大的。 在这种情况下它是120分贝,或1,000,000V/V,随着频率的增加,Ao以 20dB/ decade的的速度降低。我们看到,在10MHz时,开环回路增益为0dB,或 In the real-world, however, the open-loop gain of an op amp has a low-frequency,or dominate pole. This can be thought of as an RC filter as shown here This simulation depicts the open loop gain of a real-world op amp. At dc, or low frequencies, Aol is very large. In this case it is 120dB, or 1,000,000 V/. As frequency increases, Aol decreases at a rate of -20dB/decade. We see that at 10MHZ, the open loop gain is OdB, or 1V/ Loop Gain, Closed Loop Gain, Aol 回踣增益,闭环回暗增益&开环回踣増益 A lol Loop Gain=AdB)-l/β(dB 40 1/B 0 f0M Frequency (Hz) og(A B)= log(A ) log(3) Closed Loop Gain og(A01阝)=log(A)-log 电 TEXAS INSTRUMENTS ●现在我们明白,开环冋路増益会随着频率降低,此现象如何影响我们的闭环回路增益 呢? ●回想一下,Ao邝β被称为回路增益。如果我们绘制了开环回路增益和1邝β在对数轴,回 路增益为两糸曲线之间的差值。数学证明回路增益是Aol-1/β,如底卜的方程式所式 在这个例子中,1邝β为水平虚线在20分贝。注意,闭环回路环增益在低频率时为1β, 但在高频率吋为Aol由线。还要注意,在这1/β曲线与Ao曲线交错的点是闭环回路 带宽。让我们更深入的看看为什么闭环回路增益在低频率为1/β,而在较高频率为Aol o Now that we understand that the open loop gain decreases with frequency, how does that affect our closed-loop gain? Recall that the term Aol Beta is known as Loop Gain. If we plot the open loop and 17B on a logarithmic axis, loop gain is the difference between the two curves. The mathematical proof that loop gain is Aol -1/B on a logarithmic curve is given at the bottom of the slide. In this example, 1/B is shown as a horizontal dashed line at 20dB. Notice that closed loop gain follows 1/B for low frequencies, but follows the Aol curve for high frequencies. Also notice that the point at which the 1/3 curve intersects the aol curve is the closed loop bandwidth Lets take a deeper look at the reasons why closed loop gain follows 1/B for low frequencies and follows Aol at higher frequencies Loop Gain, Closed Loop Gain, Aol 回暗增益,闭环回暗增益&开环回暗增益 A lol Large 60- AclB 40 1/B Small 0 foM Frequency(Hz) Acl=1/B for Large Aop Acl= Aai for Small AolB Aa-Aap∞1+ABB-7、B 1 A Aol p-o1+AoB 电 TEXAS INSTRUMENTS 在低频时,回路增益,或Aoˆβ很大。请记住,闭环回路增益为Aol除以1+Ao|*β,所 以对于较大的Ao*β,则可以忽略“1”这一项。公式可简化为Ao除以Ao*β,将分子 分母的Ao相消,可得邝β。在这种情况下,1邝3是一个熟悉的同相放大器增益方程,值 为1+Rf/R1。 ●在高频率下,Aoβ很小。请记住,闭环增益的Aol除以1+Ao*β,所以对于小的AoB 值,则可以忽略Aorβ”。这样一来,就只剩下Aol除以1,或者说就是Aol。因此, 当Aoβ值变小,闭环回路增爷跟随Ao曲线 ●我们定义了电路的带宽为1β和Ao曲线相交的频率。因此,从运算放大器数据表中 的Aol曲线可以近似到电路所需的闭回增益带宽。 ●然而,请注意,ⅹ轴是对数的。因此,以图形方式选定的带宽可能不够准确。 At low frequencies, loop gain, or Aol"'B is large. Remember that closed loop gain is Aol divided by1+Ao|β, so for large values of Ao"β, you can ignore the“1”term This leaves Aol over Aol" B, and the Aol terms will cancel. Thus, the closed loop gain for large values of Aol 'B is approximately equal to 1/B. In this case, 1/B is the familiar gain equation of a non-inverting amplifier, 1+Rf/R1 At high frequencies, Aol"'B is small. Remember that closed loop gain is Aol divided by1+Ao|"阝, so for small values of ao|β, you can ignore the“Ao|*β"term.This leaves Aol over 1, or just Aol. Thus, the closed loop gain follows the Aol curve for small values of ao|阝. e We define the bandwidth of the circuit as the frequency at which the 1/B and Aol curves intersect. So, given the Aol curve from an op amp data sheet you can approximate the bandwidth of your circuit for the desired closed loop gain e However, notice that the x-axis is logarithmic. Therefore, graphically determining the bandwidth can be inaccurate Gain Bandwidth Product 增益带积 STANDARD GRADE HIGH GRADE PARAMETER COND TIONS IN MAX MIN MAX UNIT FREQUENCY RESPONSE GBW G-4 GBW〓 GailBW In this ex ample, for any gain irom CdB to Avol OPEN-LOOP GAIN AND PHASE VS FREqUENCY where GBW- Gain bandwidth in hz Gain-cl sed loop volt age gain BW-Bandwiath in h Gain= 100= 40dB For example 40 ain〓100 135 Closed Loop Bandwidth is calcuated BW= GBW 22MHZ 1101001k10k100k1M10M100M H Gain Frequency (Hz)[BW=220kH 电 TEXAS INSTRUMENTS 6 ●另一种方法来确定带宽是使用运算放大尜数据表中的増益带宽积产品规格 ●增益带宽积是线性增益和带宽的乘积。因比,在给定两个变量中的个,可得另·个 解 ●例如,让我们计算OPA827在100V/∨增益电路的带宽 ●从数据衣我们可以得知,增益带宽积为22MHz ●为求解带宽,从増益带宽公式告诉我们,带宽为增益带宽积除以线性增益将O尸A827 增益带宽积22MHz除以100/V的增益,可得220kHz的带宽。 此计算可由观察数据表中OPA827开环回路增益曲线米验证。如果我们画一条水平线 在100V/V,或40dB的闭环回路增益,直到它相交Aol,让我们找到了相应的带宽大 约为200kHz的。可以注意到,通过计算,我们发现带寬为220kHz,它解决了图解可 能锆误地解释带宽为200kHz。 ●请注意,计算出米的带宽,需当Ao|曲线以-20dB/ decade的速率下降才是有效的。虽 然大部分的运算放大器都符合,但也有一些只特定的增益带宽积是有限的范围。此外, 考虑到数据表有増益带宽积和Aol曲线的典型值。通常,可以预期该值在室温下的变 异数高达±30%,在规定的温度范围外,可能有附加±30%误差的变化。因此,当考虑 到放人器的带宽时,建议于设计时保留您的设计裕度。 Another approach to determining bandwidth is to use the gain-bandwidth product specification from an op amp data sheet o The gain bandwidth product is literally the product of the linear gain and the bandwidth. Therefore, you can solve for one of the variables given the other two For example, let's calculate the bandwidth of a circuit that uses the OPA827 in a gain of 100V/V From the data sheet we see that the gain bandwidth product is 22MHZ e Solving the gain bandwidth equation for bandwidth tells us that the bandwidth is the gain bandwidth product divided by the linear gain. Dividing the OPA827 gain bandwidth product of 22MHz by the circuit gain of 100V/ yields a bandwidth of 220kHz This calculation is verified by looking at the OPA827 open loop gain curve from the data sheet. If we draw a horizontal line at the closed loop gain of 100V/, or 40dB until it intersects Aol we find the corresponding bandwidth is approximately 200kHz Notice that solved graphically you may incorrectly interpret the bandwidth to be 200kHZ, though by calculation we found it to be 220kHz o It should be noted that the calculation approach to solving for bandwidth is only valid if the Aol curve decreases at a rate of -20dB/decade. While this is true for most op amps, there are some that have a limited range where the gain bandwidth product is specified. Also, consider that the data sheet gives only the typical value for both the gain bandwidth product and the Aol curves. Generally, you can expect a variation of as much as t30% from this value at room temperature and an additional error of +30% over the specified temperature range. So, it is always advisable to include margin in your design when considering amplifier bandwidth Simulation: Non-inverting Gain of 100V/V ②日、岛国四台NPm R2 8 R18 5000- x11459k y 2.98 22.4 l1P140 Simulated bw=118kHZ 180.DJ Fraquency (z) AC Buue1 Actor B环≈GBW1MHz =110kHz Guin 100 TEXAS INSTRUMENTS 7 现在让我们与TNA-T仿真作比较。 ●这里,我们将一个OPA140改计为同相放大器组态,100V/∨或40分贝的闭环回路增 益 ●该OPA140具有11MHz的増益带宽积。代入我们的闭回增益100V/V,可计算出带 宽为110kHz。 ●在这里,我们仿真了电路的闭环回跤带宽。在-3dB点,或37分贝,仿真带宽为118kHz 虽然不完仝一样,但我们计算和仿真结果是正相关的。 Now let's compare a calculation to a tina-Ti simulation Here we have an OPA 140 in a non-inverting amplifier configuration with a closed-loop gain of 100V/, or 40dB The OPA140 has a gain bandwidth product of 11MHz. Given that our closed loop gain is 100V/ we find that the calculated bandwidth is 110kHz Here we simulate the closed-loop bandwidth of the given circuit. At the -3dB point or 37dB the simulated bandwidth is 11 8kHz e Though not exact, we have correlation between our calculated and simulated results

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