Microelectronic Circuits

Mon Jan 17, 2022 [Tue Apr 19, 2022]

Lec 1 #

Op Amp is the closest we can get to Ideal Amplifier

Lec 2 #

Superposition Theorem #

used when there are multiple independent sources
For each independent source, form a subckt with all other independent srcs set to zero. All voltage sources are shorted, all current sources are open.
Find the response to that independent source acting alone for each subckt
Total response is sum of individual responses.

Lec 3 #

Capacitor #

\[I = C \frac{dV}{dt}\] \[X_c = \frac{1}{j\omega C}\]

At dc, \(\omega = 0\), open ckt
At high freq ac, \(\omega \approx \infty\), short ckt

Inductor #

\[V = L \frac{dI}{dt}\] \[X_L = j\omega L\]

At dc, \(\omega = 0\), short ckt
At high freq ac, \(\omega \approx \infty\), open ckt

Amplifiers #

\(v_i \sim \mu V\), \(v_o \sim mV\)

Linear Amplifier #

\[v_o (t) = Av_i(t)\]

A: a constant independent of \(\omega\)
If \(v_i = a \sin \omega t\), then \(v_o = Aa \sin(\omega t + \phi)\), only amplitude changes. Don’t worry about phase in this class. Either in-phase or out of phase.
Non-linear amp: \(v_o(t) = f(v_i(t))\)

Gain #

\[A_p = \frac{v_oi_o}{v_ii_i} = A_VA_I\]

Decibels #

V-Gain (dB) = \(20 \log_{10} |A_v|\) dB
I-Gain (dB) = \(20 \log_{10} |A_I|\) dB
Power Gain (dB) = \(10 \log_{10} |A_p|\) dB

Amp Power Supply #

\[P_{DC} = V_{CC}I_{CC}+V_{EE}+I_{EE} \] \[\underbrace{P_{DC}}_{\text{Power from supply}} + \underbrace{P_I}_{\text{i/p power in the signal}} = P_{\text{load}} + P_{\text{diss}} \] \[\eta = \frac{P_L}{P_{DC}}, \ P_I \ll P_{DC}\]

Amplifier Saturation #

Linear amplification happens only over a range of inputs. Outside this range, no amplification occurs, distortion may happen.
Output voltage is limited to \(\{L_-,L_+\}\) so \[\frac{L_-}{A_v} \le v_i \le \frac{L_+}{A_v}\]

Lec 4 #

Naming Convention #

DC: Capital letters, \(V_O, I_C\)
AC: Lowercase, \(v_c, i_c\)
Total current = DC + AC so \(i_C(t) = I_C+i_c(t)\).
DC power supplies: \(V_{CC} , V_{DD}\).

Voltage Amplifier Model #

\(v_i, v_o, A_v\)

For ideal volt amp, \(R_{in} = \infty\), \(R_{out} = R_o = 0\)
only a fraction of the source signal v s actually reaches the input terminals of the amplifier
Thus, net gain < \(A_v\)

Amplifier with Load and Signal #

Measurement #

Open ckt voltage gain
\[R_{in} = \frac{v_i}{i_{in}}\]
\[R_o = v_o/i_o\]

Amplifier Types #

Cascade Amps #

Lec 5 #

MOSFET #

Neglect body effect, was only an issue in older technologies.
Gate current is zero, i.e. \(I_G = 0\). Only gate voltage matters.
Threshold voltage is at which inversion occurs, i.e. drain becomes source

MOSFET Params #

Design Params: W, L (\(\frac{W}{L}\)); Less flexibility to change L
Process Params: \( V_{DD}, V_T, T_{ox}, N_{sub} \)
Process transconductance parameter \(k_n’ = \mu_nC_{ox}\)
MOSFET transconductance parameter \(k_n = \mu_nC_{ox}(W/L)\)

Lec 6 #

Channel Length Modulation #

At some point in saturation, in pinchoff: as we increase \(V_D\), \(\Delta L\) increases which reduces effective channel length from \(L \text{ to } L-\Delta L\)
We can model this non-ideality using the channel length modulation factor \(\lambda \propto \Delta L/ L\): \[I_{DS} = \frac{1}{2}\mu_{neff}C_{ox}\frac{W}{L}(V_{GS}-V_T)^2(1+\lambda V_{DS})\]
\(\frac{\partial I_{DS}}{\partial V_{DS}} = \lambda I_{DSAT}\)
- Thus, resistance \[r_0 = \frac{1}{\lambda I_{DSAT}}\]
In terms of Early effect: \(\lambda = \frac{1}{V_A}\) where \(V_A\) is the Early voltage.
- \[V_A = V_A’ \cdot L\]

Transconductance #

\[g_m = \frac{\partial I_{DS}}{\partial V_{GS}}\] \[g_m = k_n’(W/L)\overbrace{(V_{GS}-V_T)}^{overdrive}\] \[g_m = \sqrt{2\mu_nC_{ox}(W/L)I_D} \] \[g_m = \frac{2I_D}{V_{GS}-V_T}\]

Body effect #

When source and body are not tied together
Not used in this class

Long Signal Models #

Lec 7 #

Small Signal Model #

\(I_{ds} = I_{DS}+i_{ds}\): Amplitude of \(i_{ds}\) is very smol.
- Small signal parameters depend on the biasing point we choose

Linearisation of I-V characters around a bias point #

At a bias point: \(I_{DS}, V_{GS}, V_{DS}\)
Since \(I_{DS} = f(V_{GS}, V_{DS}) \), and we think of the small signal \(i_{ds}\) as a variation in the dc signal, we can say

\[i_{ds} = \Delta I_{DS} = \frac{\partial f}{\partial V_{GS}}v_{gs}+\frac{\partial f}{\partial V_{DS}}v_{ds}\] \[i_{ds} = \frac{\partial I_{DS}}{\partial V_{GS}}v_{gs}+\frac{\partial I_{DS}}{\partial V_{DS}}v_{ds}\]

Assume an NMOS in saturation: \[I_{DS} = \frac{1}{2}k_n(V_{GS}-V_T)^2(1+\lambda V_{DS})\]

\[i_{ds} = \frac{\partial I_{DS}}{\partial V_{GS}}v_{gs}+\frac{\partial I_{DS}}{\partial V_{DS}}v_{ds}\] \[i_{ds} = g_mv_{gs}+\frac{v_{ds}}{r_0} \text{ where } \ r_0 = \frac{1}{\lambda I_D}\] \(i_{ds} = g_mv_{gs}\) when no channel-length modulation (a): No channel length modulation (b): With channel length modulation

Lec 8 #

Transistor Amplifier #

\[A = \frac{v_o}{v_i} = -g_mR_D\]

Negative gain implies inverting amplifier, i.e. \(180^\circ\) phase between input and output.
When channel-length modulation is accounted for:

\[A_v = \frac{v_o}{v_i} = -g_m(r_0||R_D)\]

CMOS Amplifier #

Common Source (CS)
Common Gate (CG)
Common Drain (CD) or Source Follower

CS Amplifier #

\[A_v = -\sqrt{2k_n’(W/L)I_D}R_D\]

For high gain, we need to increase \(I_D, R_D\) but that makes it harder to keep \(M_1\) in saturation

Lec 9 #

CS Amp #

For high gain, increase \(R_D\) till infinity so that it becomes an ideal current source
So \(A_v = -g_mr_o\), \(R_{on} = \infty\), \(R_{out} = r_o\)
- For an ideal MOS, \(r_o = \infty\) so gain is infinite
We can get an ideal current source by using another MOS in saturation

\[A_v = -g_{m1}(r_{o1}||r_{o2})\]
- As \(Q_1\) is converting the input voltage to output current, it is doing the transconductance

Lec 10 #

CS Stage with Source Resistance (Degeneration) #

We use \(R_S\) to control the magnitude of the signal \(v_{gs}\) and thereby ensure that \(v_{gs}\) does not become too large and cause unacceptably high nonlinear distortion.
But \(R_S\) will bring stability to gain at the cost of reducing it.
\[v_{gs} = \frac{v_{in}}{1+g_mR_S}\]
- \( v_o = - g_mv_{gs}R_D\)
\[A_v = \frac{v_o}{v_{in}} = \frac{-g_mR_D}{1+g_mR_S} = - \frac{R_D}{1/g_m + R_S} = -G_MR_D\]
- If a load resistance \(R_L\) is connected at the output, replace \(R_D\) by \(R_D || R_L\)
- The factor \((1+g_mR_s)\) is the amount of negative feedback introduced by \(R_s\)
Voltage gain from gate to drain = \(- \frac{\text{Total resistance in drain}}{\text{Total resistance in source}}\)
\[R_{out} = r_o + (1+g_mr_o)R_S\]
\(R_{in} = \infty\)

Lec 11 #

Coupling Capacitor #

In DC Analysis, treated as open
For small signal ac analysis, shorted
So, it only lets the ac component of the signal through.
For large capacitance, area of cap needs to be high which is not possible in integrated ckts, so we have to use differential amps

Lec 12 #

Common Gate Amplifier #

\(R_{in} = \frac{1}{g_m}\)
\(A_{v} = g_mR_D\)
\(R_{o} = R_D\)
The CS configuration suffers from a limitation on its high-frequency response
- Combining the CS amp with a CG amp can extend the bandwidth considerably

Lec 13 #

Source Follower (Common-Drain) #

\[A_v = \frac{g_m(r_o||R_L)}{1+g_m(r_o||R_L)} = \frac{r_o||R_L}{\frac{1}{g_m}+(r_o||R_L)} \approx 1\]
- Thus, used as a voltage buffer.
\(R_{in} = \infty\)
\(R_{out} = \frac{1}{g_m}\)
\(R_L, (W/L), V_{GS}(\text{or } V_{ov})\) are the parameters to calculate for designing a circuit

Problems with Resistors #

Occupy too much space on the SoC
Generate heat which heats up all the components around it and \(n_i^2 \propto e^{-1/T}\)
- We test the design at high temperatures to check for the worst case
Speed up transistor aging and \(I_D\) starts dropping

Thus, other than discrete designs, resistors aren’t used for biasing. We use current source biasing in IC designs.

Lec 14 #

Cascode Amplifier #

Cascoding: using a CG xtor to provide current buffering for the output of a CS/CE xtor

\(R_{in} = \infty\) (gate)
Comparing with a source-degenerated CS amp, \(R_s = r_{o1}\) so \[R_o = (g_{m2}r_{o2}) \ r_{o1}\]
\[A_v = -g_{m1}R_o = -(g_{m1}r_{o1})(g_{m2}r_{o2})\]

Effective Transconductance #

\(A_v = - G_m R_{out}\)

To calculate the effective transconductance \(G_m\), ground the output and measure \(i_{out}\) as a function of \(v_{in}\)

\[G_m = \frac{i_{out}}{v_{in}}\]

To calculate \(R_{out}\), short the input voltage and measure \(v_{out}\) as a function of \(i_{out}\) \[R_{out} = \frac{v_{out}}{i_{out}}\]

Cascode Current Source #

For a gain of \(A_0^2\), the load \(R_L\) must be the same order as \(R_o\) of the cascode amplifier
\(Q_3\) raises the output resistance of the current source \(Q_4\)

Cascode Amp with Cascode i-Src Load #

\[A_v = -g_{m1}[R_{on} || R_{op}]\]

If all transistors are identical \[A_v = -\frac{1}{2}(g_mr_o)^2\]

Lec 15 #

Current Mirrors #

\(I_{REF} = I_{D1} = \frac{1}{2}k_n’(W/L)_1(V_{GS}-V_{Tn})^2\)
\(I_O = I_{D2} = \frac{1}{2}k_n’(W/L)_2(V_{GS}-V_{Tn})^2\)
\[\frac{I_O}{I_{REF}} = \frac{(W/L)_2}{(W/L)_1}\]
- In case of identical xtors, the circuit becomes a current mirror

Effect of \(V_O\) on \(I_O\) #

To ensure \(M_2\) is saturated, \(V_O \ge V_{GS}- V_{Tn}\) \(V_O \ge V_{OV}\)

Channel Length Modulation #

\(R_O := \frac{\Delta V_O}{\Delta I_O} = r_{o2} = \frac{V_{A2}}{I_O}\) \[I_O = \frac{(W/L)_2}{(W/L)_1}I_{REF}(1+\frac{V_O-V_{GS}}{V_{A2}}\]

Current Steering #

Xtors \(M_1, M_2, M_3\) form a two-output current mirror.

\[I_2 = I_{REF}\frac{(W/L)_2}{(W/L)_1}\] \[I_3 = I_{REF}\frac{(W/L)_3}{(W/L)_1}\]

\(I_3\) is fed to the input of a i-mirror formed by pMOS xtors \(M_4, M_5\)

\[I_5 = I_4 \frac{(W/L)_5}{(W/L)_4}\] \[V_{D5} \le V_{DD} - |V_{OV5}|\]

While \(M_2\) pulls its current \(I_2\) from a circuit (not shown), \(M_5\) pushes its current \(I_5\) into a circuit (not shown). Thus \(M_5\) is appropriately called a current source, whereas \(M_2\) should more properly be called a current sink. In an IC, both current sources and current sinks are usually needed.

Small Signal of Current Mirror #

\[A_{i} = \frac{g_{m2}}{g_{m1}} = \frac{(W/L)_2}{(W/L)_1}\]

Lec 16 (Mane Arc Begins) #

\[A_v = \frac{\text{Output Resistance}}{\text{Resistance seen through source}}\]

MOS Differential Pair #

Two matched xtors whose sources are joined together and biased by a current source.

Common-Mode Input Voltage #

When equal voltages are applied at gate, i.e. \(v_{G1} = v_{G2} = V_{CM}\)
As xtors are matched, \(i_{D1} = i_{D2} = I/2\)
- \[\frac{I}{2} = \frac{1}{2}k_n’(W/L)V_{OV}^2\]
- \[V_{OV}= \sqrt{\frac{I}{k_n’(W/L)}}\]
\[v_{D1} = v_{D2} = V_{DD} - (I/2)R_D\]
- Thus, diff in output voltage will be zero, i.e. differential pair does not respond to common-mode inputs.
Since both xtors must remain in saturation, there would be a range over which the diff-pair operates properly,
- \[V_{CM_{max}} = min(V_{DD}, V_{DD} - (I_{SS}/2)R_D + V_T)\]
- \[V_{CM_{min}} = -V_{SS} + V_{CS} + V_T + V_{OV}\]
- \[V_{CM_{min}} = V_{ISS} + V_{GS_1} = V_{ISS} + V_{OV} + V_T \]

Differential Input Voltage #

For all the current \(I_{SS}\) to flow through \(M_1\), the current through \(M_2\) must be zero which happens at cutoff, i.e. \(V_{GS_2} = V_T\)
- \(V_{in_1}-V_{GS_1}+V_{GS_2}-V_{in_2}=0\)
- \(2V_{in_1}= V_{GS_1}- V_{GS_2}\)
- \(2V_{in_1}= V_{GS_1}- V_T = V_{OV_1}\)
- Thus, \(v_{id_{max}} = \sqrt{2}V_{OV}\)
Must be operated between the range:

\[-\sqrt{2}V_{OV} \le v_{id} \le \sqrt{2}V_{OV}\]

\[\\ \text{offline modo hiatus} \\ \]

Frequency Response #

Thus far, we have been assuming that our amplifiers are operating in the middle-frequency band or midband, where the gain is almost constant.
However, at lower frequencies, the magnitude of the amplifier gain falls off.
- Coupling and bypass caps no longer have low impedances
\(r_o\) is neglected, as has a negligible effect in discrete-circuit amps