ECEN4827/5827 lecture notes Two Stage Op-Amp: DC Solution (Part 1) Objectives in this segment of the course are to: 1. Review necessary CMOS technology and device characteristics 2. Understand the internal configuration of a two-stage CMOS op-amp. 3. Perform analysis and determine the DC solution for the two-stage CMOS op-amp Figure 1 shows a basic two-stage CMOS op-amp configuration. The resistance R provides biasing for the entire operational amplifier. M1 and M2 form a differential pair and thus the input of the first gain stage of the op amp. The current mirror formed by M8 and M5 supplies the differential pair with bias current IB1. The input differential pair is actively loaded with the current mirror formed by M3 and M4. Node 1 forms the output of the first stage of the op amp. The second stage consists of M6 which is a common-source amplifier actively loaded with the transistor M7. It should be noted that M7 does not provide biasing for M6, and that M6 is biased from the gate side. Figure 1: Two-stage CMOS op-amp. The inputs are set to zero to determine the nominal DC bias operating point.Numerical Example: Given: Process parameters: µnCox = 60 µA/V2 and µpCox = 20 µA/V2, Vtn (The threshold voltage of NMOS devices) = +1V and Vtp (The threshold voltage of PMOS devices) = −1V. Supply voltages: VDD = +5V and –VSS = −5V Device aspect ratios: (W/L)8 = 1, (W/L)1 = (W/L)2 = 100, (W/L)3 = (W/L)4 = 20. Dc bias specifications: IR=1µA, IB1=10µA, IB2=100µA DC Solution The first step in finding the DC solution is to determine the value of the resistance R so that the bias current IR equals the specified value (1µA in the numerical example): VDD = R*IR + VGS8 – VSS, (1) which yields R = (VDD + VSS – VGS8)/IR (2) While the values of VDD, VSS, and IR are known, the voltage VGS8 is not. In order to solve for VGS8, we observe that the device M8 operates in the active/saturation region since the gate of M8 is shorted to the drain of M8, and therefore the following conditions are met: 1. VGS8 ≥ VTN 2. VTN ≥ VGD8 Since M8 is in the active/saturation region, we have ID8 = IR = (µnCox/2)(W/L)8(VGS8 - VTN ) 2 (3) which can be solved for VGS8, tnGSVKIVR+=88/ , where K8 = 1/2* µnCox*(W/L)8. (4) For the values in the numerical example, VGS8 = 1.2 V The value of VGS8 is then substituted into (2) and the value of R is determined to be 8.8 MΩ. The value of the gate voltage VG8 can be found as:VG8 = −VSS + VGS8 = −5 + 1.2 = −3.8V (5) Assuming that M5 and M7 are in the active/saturation region, the specifications for the DC currents IB1 and IB2, together with the property of the current mirror can be used to find the aspect ratios of M5 and M8: RBIILWLW185)/()/(= = 10 and RBIILWLW287)/()/(= = 100. (6) The gates of M1 and M2 are grounded to obtain the nominal DC solution, thus VGS1 = VGS2. Also, ID1 + ID2 = IB1. Since M1 and M2 are identical NMOS devices having the same VGS voltages, assuming that M1 and M2 are in the active/saturation region, the bias current IB1 divides equally between M1 and M2, ID1 = ID2 = 5µA. (7) At this point we can check whether the device M5 is in the active/saturation region. The drain voltage of M5 is the source voltage of M1 and M2, i.e., the negative of the gate to source voltages of M1 and M2. VD5 = −VGS1 = −VGS2 = −(Vt,n + 11 / KID ), where K1= 1/2* µnCox*(W/L)1 (8) The drain voltage of M5 is calculated to be -1.04V, which is much larger than its gate voltage −3.8V. Thus, M5 is indeed in the active/saturation region. Now continuing with the PMOS devices M3 and M4, M3 is in saturation as its gate and drain terminals are connected together. Thus, the source to gate voltage of M3 can be calculated as VSG3 = |Vtp| + 33 / KID where K3= 1/2* µpCox*(W/L)3 = 200µA/V2 (9) For the numerical example, we get: VSG3 = 1.16V (10) The drain voltage of M1 is VD1 = VDD – VSG3 = 3.84V which shows that M1 is easily in active/saturation region. The calculation of the drain voltage of M4 is done in the next lecture.Comments: • The circuit utilizes two power supplies, which are ±5V in the numerical example. The values of ±2.5V or even lower supply voltages are also typical. Single supply voltages (VSS = 0) are also in common use, with VDD as low as 2.5 V in battery-powered electronic systems. • An NMOS transistor operates in the active/saturation region if the gate-to-source voltage is greater than the threshold voltage Vtn and if the gate-to-drain voltage is less than the threshold voltage. The active/saturation region of operation is most commonly utilized in analog circuits. • A PMOS transistor operates in the active/saturation region if the source-to-gate voltage is greater than the absolute value of the threshold voltage |Vtp|, and if the drain-to-gate voltage is less than the absolute value of the threshold voltage. • In the active/saturation region, the effects of the drain to source voltage on the current of the device is neglected in the analysis so far. In other words, the effect of channel length modulation is not