TI高精度实验室-压摆率 3.pdf.pdfTI高精度实验室-压摆率 3.pdfpdf,TI高精度实

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详细说明：TI高精度实验室-压摆率 3.pdfpdf,TI高精度实验室-压摆率 3.pdfWG1 Input step= 10mv, 100mv, 250mv, 750mV, IV, 2V, 4V vou川们 Small signa t,=20ns, dv/dt=0, 4V/us sponse Niall t=20ns, dv/dt =5V/ 25m Wout 3 r=20ns,dv/dt=10v/μs 0.0 50m Betwe Youth Small and =254ms,dv/dt=15.7/μs Large signal 750m Vout(G) t,=34ns, dv/dt= 17. 6V/ Voute r=43ns,dv/dt-185v/μ 2.0 Vout[7 n5 dvf V/us slew limited 0.0 slew= 20V/us 4.0 Vout e t,= 162ns, dv/dt 198v/Hs 100u10.1u 7C.3 10.4u TEXAS INSTRUMENTS o This figure illustrates the output of the non-inverting buffer circuit from the previous slide versus different input step sizes o Notice that the output rise time is a constant 20ns for input steps of 10mvto 250mv. Because the rise time is constant we know that the response is small-signal. The device in this example the opa192 has a slew rate of 20v/us You can see that the rate of change for the small step response is lower than the slew rate. Also, notice that the output signal increases exponentially for the small signal response, as opposed to the linear increase for amplifiers that are slew rate- imited o For input steps between 500mV and 1v, the amplifier is transitioning between small-signal and large-signal response. In this region the rise time is no longer constant; however, the amplifier is not yet at the full slew rate of 20v/ us o For input steps greater than 1V, the amplifier is slew rate-limited. You can see that the rate of change of the output signal is at the slew rate limit of about 20V/us. Also, notice that the output signal increases linearly in response to the put step o In this example input signals less than 250mv caused a small signal response and signals greater than 1V caused a large signal response. However, this transition depends on the amplifier s design and technology. CMOS amplifiers tend to reach slew limit for signals greater than 100mv, and bipolar amplifiers can slew limit at even lower input voltages. Note that the industry standard for small signal response is a 100mv step but in practice the actual limit may be lower o Because of the way the output signal is scaled in this figure, it is possible to be tricked into thinking that the slew rate limited signal at the bottom of the figure is moving more slowly then the small signal response at the top of the figure This is not the case, and the next slide will provide further clarification 这一页幻灯片显示了前面电路的仿真结果。图中分别给出了不同阶跃量输入时,电路的上升时间和压摆率。 ●注意当输入电压从10mV变化刭250mV时,图中显示输出上升时间都是20ns, 是一个常数,而且电压变化率都小于PA192的压摆率20V/us,还可以看到这时输出电压随时间呈指数上升。 ●当输入电压从500mV变化到ⅣⅤ时,输入实现了从小信号到大信号的变化, 可以看到这时上升时间不再是常数,但电压上升速率还没有达到20V/us的压摆率 ●当输入阶跃幅值人于1时,输出电压的上升速率接近20V/us,受到压摆率限制,输出电压不再指数上升,而是以固定斜率线性增长。 ●这个例子中,包含了阶跃小于250mV时的小信号输入响应,也包含了阶跃大于1V吋的大信号输入响应。这之间的过度是由运放的设计结构和设计方法所决定的。CMOS型的放人器在输入电压人于100m时就开始受压摆率限制了,而BJT型的运放会在更低输入电压处受限。工业上,小信号响应的标准输入电压为100mV阶跃,而实际上往往更低 ●图中各个响应曲线的纵轴尺度不同,可能让人有个误区,觉得图中靠下的曲线斜率小于上面曲线的斜率,其实不是这样的,下一页幻灯片将进一步阐述这个问题 500m 4OUm- /dt=198v/p Vout 81 Vout 2 dv/dt=5v/μs dv/dt=5V/us 4.011 Vout[21 100 Wout(81 dv/dt= 19.8V/Hs 0 U 995u 0.00u 10.05 9.90 10.15u me(s) TEXAS INSTRUMENTS o The left-hand side of this slide shows the small-signal response to a 100mV input signal in green, and the large-signal response to a 4V input signal in blue The plots are scaled such that the two output signals have equal height. This makes the small-signal response look like it is moving faster then the large signal response. On the right, we combine the two responses to emphasize that the large signal response is slew rate-limited and is changing much faster then the small-signal response. In fact the large signal response is changing at 19.8 V/us, and the small signal response is only changing at 5v/us ●左图从上到下三条曲线依次是输入阶跃,如红色所示;100mV输入时的输出晌应,如绿色所示;4Ⅴ输入时的输岀响应,如蓝色所示。 ●为了显示方便,图中两条响应曲线高度相同,这让我们看起来觉得小信号的响应速度更快,但把这两条曲线放在一起,如右图所示,可以看到按压摆率线性上升的大信号响应速度远高于小信号响应速度。实际上,大信号的输出电压变化率为198Vus,而小信号为5V/us 50 5000 0 5000 Wos-50 100 st', 990 1.5u 18.8V/ 9.80u 10 1070u 1045u 4055 I me(s) TEXAS INSTRUMENTS o Let' s take a closer look at the output of the non- inverting buffer circuit with a 10V step applied to the input. o In this case the device is in slew-rate limit of 20V/us. However when the signal approaches the final value of 10v, the op amp will transition to a small signa response. Zooming in on the last 300mv of the signal swing, you can see that the rate of change of the output decreases from approximately the slew rate, 20V/us, to a lower rate of change such as approximately 5v/us. Also, you can see that the shape of the output changes from the linear rise associated with slew rate limit, to the exponential behavior associated with small signal response ●这里,我们对前面的同相电压跟随器施加一个10V的阶跃输入。 ●理论上,输岀信号应该按20v/us的斜率变化,然而,当输岀信弓即将增大到 10V时,运放将会以小信号输入时的电压变化率进行响应。我们把9.7V到 10V这段输出曲线放大,可以看到输出电压的变化率从20Vus逐步下降到约 5V/us。同时可以看到输出曲线从压摆率限制下的线性上升变成了近似于小信号输入时的指数方式上升。 20mvpp TEXAS INSTRUMENTS o Most of the examples discussed thus far have been basic buffer circuits also called unity-gain followers. Now let s look at amplifiers with different closed loop gains in order to see the effect that closed loop gain has on output response. o Here we show an inverting amplifier circuit in a gain of -99V/. the input is a small signal step of 20mVpp. Based on the gain, the output should be approximately 2 Vpp. Does the output respond as a small signal or as a large signal? ●到日前为止,大多数讨论都基于电压跟随器,或者叫缓冲器。现在我们来看看不同的闭环电路结构对输岀响应的影响。 ●这里显示的是增益为9yV∧的反相放大电路,对其输入一个幅值为20mV峰峰值的小阶跃信号,根据増益的设置,输岀屯压应该为2ν峰峰值。此吋的输出响应应该按小信号计算还是大信号计算呢? m NOs 0 GBW 1OMHZ 10CkHz 100 2. 0.35 9%(2843us1 L「02316.0 r sIm 3.285 20u 30u 40u TEXAS INSTRUMENTS o This is the simulated output response of the circuit from the previous slide o Notice that the output signals rise and fall in an exponential way which indicates a small-signal response o Also, if we use the small signal rise time formula we can see that the calculated rise time of 3. 5us is very close to the simulated rise time of 3. 28us o Thus, the amplifier responds to the 20mv input step as a small signal, and we can conclude that the amplitude of the input signal determines the behavior of the op amp the output signal amplitude does not determine whether the response is small or large signal ●这是上一页幻灯片所示电路的输出响应仿真注意刭输岀信号按指数形式上升或这下降,即表现为小信号输入响 ●如果我们使用小信号输入时的上升时间计算公式,可以求得其上升时间为 35us,这一结果与仿真得到的3.28u5上升时间非常接近 ●因此,尽管有2V的输出,此时运放仍按照小信号输入方式对20mv阶跃输入进行响应。这样我们可以得到一个结论:即运放的输入信号决定了运放的响应情况,无论输岀信号嶇債大还是小,输岀都无法决定运放的响应 VouT G=:10 G=-10 VIN Time(200 nsdi Time(200 ns/div) POSITIVE OVERLOAD RECOVERY NEGATIVE OVERLOAD RECOVERY Hi TEXAs INSTRUMENTS o Up to this point, we have considered the small-signal and large-signal response of signals that are inside the linear range of the amplifier In other words, we have not violated the input common mode range or the output swing limitations Now we l discuss what happens when we drive signals beyond the op amp's output swing limitations. Driving the output beyond its linear range is called overloading the output. Once the overload condition is corrected there is a time delay before the overloaded output can recover, called TOVERLOAD, and return to its normal linear behavior. In the next few slides, I' ll explain overload recovery and show how it relates to small and large signal response ●到目前为止,我们仅仅考虑了运放工作在线性放大区域时对输入的响应,换言之,我们还没有考虑高输入共模信号的影响和输出电压摆幅限制的影响。 ●下面我们讨论当输出信号超过运放的输岀摆幅范围时公发生什么。运放的输出电压超过线性输出范围的现象被称作输出幅度过载失真。输出电压从失真域回到线性放大区时会有一个过度时间,称为过载恢复时间, Overload 后面几页幻灯片将解释过载恢复及其对小信号和大信号响应的影响。 Q1 Q1 Normal QZ Norm Q2 Q2 Overload Q1 Overload Condition Condition Drain to Source voltage (v) TEXAS INSTRUMENTS o The overload condition occurs when an amplifier is driven beyond its output swing limits. e In this example if the output is driven very close to the positive supply (v+), the output transistor Q1 becomes saturated, while transistor Q2 is nearly cutoff. Also notice that the transistor Q1 is in the ohmic region of its operating curve as opposed to the active region where it normally operates. Overload recovery is the time required for all the internal transistors in the output stage to transition from an abnormal state, whether saturated or cutoff to a normal state ●当运放将输出电压驱动到其摆幅限制范围以外时,就公发生幅度过载现象。 ●图中,当输岀信号非常接近正电源轨Ⅵ时,晶体管Q1将会出现饱和,将不再工作在放大区,而进入其可变电阻区,同时Q2将逼近其截止工作区区。过压恢复就是运放输出级的品体管从其饱和区或者截止区回到线性放大区时所需要的时间。

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