# Notes can be found as interactive webpage at

9: Op-Amps, Comparators & Charge Sharing


# 03/29: Operational Amplifier and Comparator #

## Dependent Sources #

As briefly mentioned before, the voltage and current sources covered so far have been independent sources, meaning they have a constant, fixed value. However, dependent sources also generate currents and voltages but their output depends on some other “controlling” current .$i_x$ or voltage .$v_x$ in the circuit.

VCVS: Voltage-controlled voltage source

CCCS: Current-controlled current source

VCCS: Voltage-controlled current source

• .$I = f_b (v_x)$
• Units: Siemens, .$S$; .$\Omega^{-1}, \mho$
• E.x. Transconductance amplifier with conductance .$b$
• Input, output impedance: .$\infty, \infty$

CCVS: Current-controlled voltage source

• .$V = f_d (i_x)$
• Unit: Ohms
• E.x. Transresistor with resistance .$d$
• Input, output impedance: .$0, 0$
$^1$The proportionality constant between dependent and independent variables is unit/dimension-less if they are both currents (or both voltages)

## Op-Amp #

• An operational amplifier (op-amp) is a device that transforms a small voltage difference into a very large voltage difference.
• Two input terminals with potentials:
• $U_+$ (non-inverting) and
• $U_-$ (inverting)
• Signal voltage = $\Delta U = U_+ - U_-$
• Two power supply terminals
• $V_{DD}$ (positive power supply) and
• $V_{SS}$ (negative power supply)
• One output terminal with potential:
• $U_{out} \in [V_{SS}, V_{DD}]$

• We can model the op-amp behavior with the two VCVS circuit to the right
• Notice that no current is drawn from either power supply terminals so the behavior of the circuit shouldn’t change
• $A$ is the internal signal gain
• Unit-less quantity
• The larger $A$, the sharper/quicker the transition between $V_{SS}$ and $V_{DD}$

• With KVL we can find $U_{out}$:
• Except we can’t have $U_{out} > V_{DD}$ or less than $V_{SS}$ so we end up clipping $U_{out}$ – this is called railing $$U_{out} = V_{SS} + \frac{V_{DD} - V_{SS}}{2} + A \Delta U$$ $$\dots = \frac{V_{DD} + V_{SS}}{2} + A \Delta U$$ $$\dots = \begin{dcases} V_{SS} & \frac{V_{DD} + V_{SS}}{2} + A \Delta U < V_{SS} \\ V_{DD} & \frac{V_{DD} + V_{SS}}{2} + A \Delta U > V_{DD} \\ \frac{V_{DD} + V_{SS}}{2} + A \Delta U & \text{otherwise} \\ \end{dcases}$$
• Rail-determined offset: $\dfrac{V_{DD} + V_{SS}}{2}$
• $y$-intercept in $V_{out}, \Delta U$ graphs
• Zero in NFB, that is $$U_+ = U_-$$
• Signal gain: $A \Delta U = A (U_+ - U_-)$
• $A$ is the slope of the zone between rails
• $A \approx \infty$ is a vertical line

### Op-Amp as a Comparator #

• The sign of the output will indicate which input voltage is larger.
• We often want to know whether $V_?$ is smaller or larger than $V_{ref}$ so we’ll set
• $V_- = V_{ref}$ and $V_+ = V_?$
• The blue region is the (super narrow, and in the ideal case, perfectly vertical) linear region where the input actually has “space” to be amplified without clipping
• For the vast majority of realistic inputs, the outputs sits at one of the rails (here, $V_{SS} = β2V$ and $V_{DD} = 3V$)
• In this way, the sign of the output directly gives the relationship between $V_{?}$ and $V_{ref}$
• Note that with zero input difference, the output voltage ($y$-intercept) sits at the rail-offset: $\left(0, \dfrac{V_{DD} + V_{SS}}{2} \right) = (0, 0.5)$

## Comparator #

• Comparator: Enables us to take a wonky digital signal (i.e voltage) and convert it to a binary output signal
• Optimized for binary output and speed
• Used in devices that measure and digitize analog signals i.e ADCs

$$U_{out} = \begin{cases} V_{DD} & U_+ > U_- \\ V_{SS} & U_+ < U_- \\ \end{cases}$$

### Capacitive Touchscreen with Comparator #

#### AC #

See last weeks notes: 8: AC Source

• Because they are on the same node, we can observe $U_+ = V_C$ and $U_β = V_{ref}$
• The power sources are $V_{DD} = 3.3V$ and $V_{SS} = 0$, thus we can write the following:

$$V_{out} = \begin{cases} 3.3V & V_C > V_{ref} \\ 0V & V_C < V_{ref} \\ \end{cases}$$

• So how do we choose $V_{ref}$? We want our comparator to be high (on, $V_{out} = V_{DD}$) depending on touch (or lack thereof)
• Thus we should set $V_{ref}$ to be between the peak of $V_C$ with a finger and the peak of $V_C$ without a finger
• To be fairly robust to noise/error, $V_{ref}$ should be exactly between the two peaks: $$V_{ref} = \frac{V_{C, touch} + V_{C, no\ touch}}{2} = \dots = \frac{1}{2} \frac{C_{eq} + \Delta C}{C_{eq} + C_{ref} + \Delta C}\frac{C_{eq}}{C_{eq}+C_{ref}}$$

When there is a finger present, $V_C$ is always less than $V_{ref}$ so the output voltage is 0

Now every period of the current source cycle, $T$, we can measure if there is a finger touching this pixel

#### DC #

We can also use a DC voltage source too and end up with the same equation

# 03/31: Charge Sharing and Negative Feedback #

## Charge Sharing #

• Floating node: Node out of or into which no charge can flow – virtual ground
• We apply charge sharing to these nodes
• If the sum of charges is 0 on a floating node, the sum of charges on that floating node at steady state will stay 0.
• Examples:
• Nodes connected only to capacitor plates
• Nodes connected only to op-amp inputs or comparator inputs
• Notation:
• In each phase, the circuit has a different configuration by opening/closing different switches.
• Generally, a switch labeled $\phi_i (i = 1,2,3, \dots)$ will be closed in phase $i$ and open in all other phases.
• Switches labeled $\phi_1$ will be closed in phase 1 and open in phase 2,
• Switches labeled $\phi_2$ will be open in phase 1 and closed in phase 2.

### Algorithm #

#### Step 1 #

Label the voltages across all the capacitors. Choose whichever direction (polarity) you want for each capacitor - this means you can mark any one of the plates with the β+β sign, and then you can mark the other plate with the βββ sign. Make sure you stay consistent with this polarity across phases.

#### Step 2 #

Draw the equivalent circuit during both phases. For this problem, phase 1: $\phi_1$ closed, $\phi_2$ open, and phase 2: $\phi_1$ open, $\phi_2$ closed. Also, label all node voltages on the circuit for both phases. No need to try and maintain the same names, since certain nodes of the phase 1 circuit might be merged or split in phase 2.

$\phi_1$

$\phi_2$

#### Step 3 #

Identify all βfloatingβ nodes in your circuit during phase 2. You can identify those nodes as the nodes connected only to capacitor plates, op-amp inputs or comparator inputs. These will be the nodes where we apply charge sharing.

In this case the only node that is floating during phase 2 is node $u_3$. (Node $u_2$ is connected to the voltage source, i.e. $u_2 = V_S$, and the 3rd node is the ground node)

#### Step 4 #

For steps 4-6 we will examine each phase 2 floating node individually. Pick a floating node from the ones you found in step 3 and identify all capacitor plates connected to that node during phase 2. Then, calculate the charge on each of these plates in the steady state of phase 1. To do so, identify all nodes in your circuit during phase 1. Label all node voltages, and write the voltages across each capacitor as functions of node voltages (step 2 should help you with that). Do this according to the polarities you have selected. Then the charge is found as $Q = CV_C$ (where $V_C$ is the voltage across a capacitor).

• Careful: The plate marked with the βββ sign will have $Q = -CV_C$ and the plate marked with the β+β sign will have $Q = CV_C$ stored onto them.
• Careful: We assume here that you know all node voltages during phase 1. If you donβt, before starting this procedure calculate the node voltages you need using one of the previously introduced circuit analysis techniques (most likely KVL).

Looking at our single floating node we can see that the β+β plates of $C_1$ and $C_2$ are connected to it during phase 2. Letβs calculate the charge on these plates connected to the floating node $u_3$ during phase $\phi_1$, denoted $Q_{u3}^{\phi 1}$

$$Q_{u3}^{\phi 1} = V_1^{\phi 1} C_1 + V_2^{\phi 1} C_2$$ $$\dots = (V_S - 0)C_1 + 0$$ $$\dots = V_S C_1$$

#### Step 5 #

Find the total charge on each of the floating nodes in the steady state of phase 2 as a function of node voltages using the same process as Step 4. Make sure you kept the polarity same and pay attention to the sign of each plate.

$$Q_{u3}^{\phi 2} = V_1^{\phi 2} C_1 + V_2^{\phi 2} C_2$$ $$\dots = (u_3 - u_2)C_1 + (u_3 - 0)C_2$$ $$\dots = (u_3 - V_S)C_1 + u_3 C_2$$

#### Step 6 #

Apply conservation of charge. The charge in the steady state of phase 1 we calculated in Step 4, $Q_{u3}^{\phi 1}$, is the initial charge of phase 1. Charge must be conserved at a floating node, so the steady state charge of phase 2, $Q_{u3}^{\phi 2}$, should equal the initial charge $Q_{u3}^{\phi 1}$. We can then write in terms of the unknown(s).

$$Q_{u3}^{\phi 1} = Q_{u3}^{\phi 2}$$ $$V_S C_1 = (u_3 - V_S)C_1 + u_3 C_2$$ $$u_3 = \frac{2C_1}{C_1 + C_2}V_S$$

#### Step 7 #

Repeat steps 4-6 for every floating node. This will give you one equation per floating node (i.e. if you have $m$ floating nodes you will have $m$ equations). You can then solve the system of equations to find the node voltages during phase 2 (unknowns).

In this problem we did not go through step 7 since we only had one floating node during phase 2. This means we have only one unknown node voltage ($u_3$) for which we solved using our single equation from Step 6. Example 2 on page 4 has multiple floating nodes.

### Important Notes #

1. A node that is floating during phase 2 is not necessarily floating during phase 1. In fact, it could have been two separate nodes during phase 1 depending on how your switches are configured. We only care about the floating nodes during phase 2 since those are the nodes with unknown voltages. What about floating nodes during phase 1? You should be able to calculate the voltage of these nodes based on the initial condition of the circuit and the circuit techniques that you have learned so far (most likely KVL).
2. When handling charge sharing problems you should avoid using any parallel or series capacitance formulae that you have learned. Any simplification of the circuit might lead to mistakes since the circuit in different phases is configured differently and some capacitors placed in series during one phase may be in parallel or even not connected at all in another phase!