hw09sol

# hw09sol - ECE 315 Homework 9 Solution Fall 2007 1(Design of...

This preview shows pages 1–3. Sign up to view the full content.

1 ECE 315 Homework 9 Solution Fall 2007 1. (Design of cascode differential amplifier) (a) Sketch the circuit diagram of an active-load MOS differential amplifier where the input transistors are cascoded, and a cascode current mirror is used for the active load. (8 pts) (b) If all of your NMOS transistors are operated at an overdrive voltage of V OV and all NMOS and PMOS transistors have the same early voltage of | V A |, derive the gain expression as: A d = 2(V A /V OV ) 2 . Evaluate the gain for V OV =0.25V and V A =20V. (8 pts) We found in the last homework the expression for a differential amplifier with active load using the small signal model. We can apply the same results here noting now that the output resistance is the parallel combination of the pMOS and nMOS cascode stacks. The transconductance G M will be the same transconductance of a cascode amplifier, which is just g m,n G M = g m,n = OV D V I 2 g m,p = g m,n = g m R o = [(2r o,p + g m,p r o,p 2 ) || (2r o,n + g m,n r o,n 2 )] D OV A D A OV D o m I V V I V V I r g 2 2 2 2 2 2 = = (since g m r o >> 1) A d =G M R o = 2 2 2 OV A V V . v in + v in - V C V bias V DD v out + v out -

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
2 At V OV =0.25V and V A =20V, A d = 12,800. This is a large value even BJT hard to match, but you do need a large V DD in order to make the five-transistor stack saturated. For sure whenever you work with a high-gain system, everything else has to be precise to avoid gain saturation. 2. (Frequency response of amplifiers) For an NMOSFET in an amplifier setup with C gs =0.5pF, C gd =0.1pF, C db =0.1pF, C L =1pF (including C db ), g m =5mA/V, r o = 20k Ω , and R sig = R L = 20k Ω , (a) In the CS amplifier of Fig. 6.20, use the Miller theorem to find the quasi-static gain A M and the 3-dB corner frequency f H . (8 pts) The quasi-static gain for this simple first-order (single pole) system is just A M = -g m (r o || R L ) or A M = -50. The Miller Theorem allows us to substitute the feedback capacitor, C gd , with two separate capacitors from the gate and the drain, both going to ground. The Miller Theorem estimates the equivalent capacitance on the gate as C in,miller = C gd [1 + g m (r o || R L )] = 5.1 pF The equivalent capacitance at the output at the output is given by C out,miller = C gd [1 + (g m (r o || R L )) -1 ] C gd Therefore, the total input capacitance is C in = C in,miller + C gs = 5.6 pF And we obtain the 3-dB corner frequency: f H = in sig C R π 2 1 = 1.42 MHz (b) Repeat (a) with the open-circuit time constant method. Is the answer different? Give the percentage contribution to τ H by each of the three capacitances ( C gs , C gd and C L ). Give also the gain-bandwidth product. (8 pts) The open-circuit time constant method of finding the 3-dB frequency can be done by: 1) Find the equivalent resistance seen by each capacitor using Thevenin/Norton equivalent circuits (note that all other capacitors become open circuits, which is the origin of the name; don’t forget
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 15

hw09sol - ECE 315 Homework 9 Solution Fall 2007 1(Design of...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online