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An Op Amp Tutorial
(Based on material in the book Introduction to Electroacoustics and Audio Amplifier Design, Second Edition - Revised Printing, by W. Marshall Leach, Jr., c published by Kendall/Hunt, ° 2001.) An op amp has two inputs and one output. The circuit is designed so that the output voltage is proportional to the difference between the two input voltages. In general, an op amp can bemodeled as a three-stage circuit as shown in Fig. 1. The non-inverting input is vI1 . The inverting input is vI2 . The input stage is a differential amplifier (Q1 and Q2 ) with a current mirror load (Q3 −Q5 ). The diff amp tail supply is the dc current source IQ . The second stage is a high-gain stage having an inverting or negative gain. A capacitor connects the output of this stage to its input. Thiscapacitor is called the compensating capacitor. Other names for it are lag capacitor, Miller capacitor, and pole-splitting capacitor. It sets the bandwidth of the circuit to a value so that the op amp is stable, i.e. so that it does not oscillate. The output stage is a unity-gain stage which provides the current gain to drive the load.

Figure 1: Op amp model. If we assume that Q1 and Q2 arematched, that Q3 and Q4 are matched, that base currents can be neglected, and that the Early effect can be neglected, we can write the following equation for iO1 : iO1 = iC1 − iC3 = iC1 − iC4 = iC1 − iC2 But iC1 + iC2 = IQ and iC1 = IQ/2 + ic1 . Thus we obtain iO1 = 2iC1 − IQ = 2ic1 (2) (1)

1

Open-Loop Transfer Function
We wish to solve for the transfer function for Vo /Vid , where Vid is thedifference voltage between the two op amp inputs. First, we solve for the current Ic1 as a function of Vid . For the diff amp, let us assume that the transistors are matched, that IE1 = IE2 = IQ /2, the Early effect can be neglected, and the base currents are zero. In this case, the small-signal ac emitter equivalent circuit of the diff amp is the circuit given in Fig. 2(a). In this circuit, re1 andre2 are the intrinsic emitter resistances given by VT 2VT re1 = re2 = re = = (3) IE IQ Note that the dc tail supply IQ does not appear in this circuit because it is not an ac source. From the emitter equivalent circuit, it follows that Vid (4) Ic1 = Ie1 = 2 (re + RE ) where Ic1 = Ie1 because we have assumed zero base currents.

Figure 2: Circuit for calculating Ie1 . (b) Circuit for calculatingVo . Figure 2(b) shows the equivalent circuit which we use to calculate Vo . We assume that Req is the effective load resistance on the current 2Ic1. In this case, the current which flows through the compensating capacitor Cc is given by Vo1 Vid Vo2 Io1 = 2Ic1 + = − (5) Req re + RE KReq where we have used Eq. (4) and the relation Vo1 = −Vo2/K. The voltage Vo2 is given by · ¸ −Vo2 Vid 1 Io1 Vo2 = +(6) Vo2 = Vo1 + − Cc s K re + RE KReq Cc s If we assume that the output stage has a gain that is approximately unity, then Vo ' Vo2. Let G (s) = Vo /Vid . It follows from Eq. (6) that G (s) is given by Vo Vo2 KReq 1 G (s) = ' = × (7) Vid Vid re + RE 1 + (1 + K) Req Cc s This is of the form A G (s) = (8) 1 + s/ω 1 where A and ω 1 are given by KReq 1 A= ω 1 = 2πf1 = (9) re + RE (1 + K) Req Cc 2 Figure 3: Asymptotic Bode magnitude plots. (a) Without feedback. (b) With feedback.

Gain Bandwidth Product
The asymptotic Bode magnitude plot for |G (jω)| is shown in Fig. 3(a). Above the pole frequency ω 1 , the plot has a slope of −1 dec/dec or −20 dB/dec. The frequency at which |G (jω)| = 1 is called the unity-gain frequency or the gain-bandwidth product. It is labeled ω x in the figure andis given by ω x = 2πfx = Aω 1 = K 1 1 ' 1 + K re + RE Cc (re + RE ) Cc (10)

where the approximation holds for K À 1. It follows that an alternate expression for G (s) is A G (s) = (11) 1 + sA/ω x For maximum bandwidth, fx should be as large as possible. However, if fx is too large, the op amp can oscillate. A value of 1 MHz is typical for general purpose op amps. Example 1 An op amp is to be...
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