ive the basic current mirror below. What would the value of W/L of M2 in order for to have an output vice its Iref (current reference)? 40 VDD IREF 3 M₂ out W

5. For the following circuit with ideal OPAM and positive saturation voltage Vsat+ = 9V and negative saturation voltage Vsat-=-8V. Find: a. It is a function of V1, in active region and positive and negative saturation regions. b. The current iA in active region and in positive and negative saturation regions. Remember that no current enters the inverting and non-inverting terminals, and also V+ is different from V- when the OPAM is saturated. 10ΚΩ w 1ΚΩ www Va we+ 1A 1ΚΩ 1ΚΩ w 10kQ

In which region BJT is working?

QI Ref RG RM A1 Vou VOUT Digital A DACI Input, DIN VDAC Figure QI Figure QI shows the circuit used to condition the output of the n- bit Digital to Analog Converter "DACI" having a digital input value of Dw. DACI may be assumed to have a voltage output with zero source resistance coded in "straight binary" such that • For a binary input of Dy = 000...000 the output voltage vaac is 0 V • For a binary input of Dw = 111..111 the output voltage vaac is Vn Where Vg is a positive voltage reference. Neglecting Ry and assuming that opamp Al is ideal, show that the output voltage v, is (а) OUT R. for D, - 000000 R (i) Vay- ref (ii) Vor for D,-111..111 Hence choose values of Ro and R to give an output voltage range -V Svar S (iii) Re-draw Figure Q1 to show all sources of DC opamp error and say why they have the signs you have given them. (b) Hence derive an expression for the emor in vug due to opamp DC input cur. and voltage errors including both bias and offset currents. (c) If the two…

9. A 6-bit DAC has an input of 1001012 and uses a 10.0 V reference. i. Find the output voltage produced. ii. Specify the conversion resolutions

Determine the input resistance Rin as indicated in the figure. Ignore resistor ro of the transistor and capacitors  are large.   A. Rin = R1//R2//R3    B. Rin = R2//R3 (//R4//R5)    C. Rin = R2//R3 //   ( r π + [1+gm r π ]  (//R4//R5) )    D. Rin = R2//R3 //  r π  (//R4//R5) )