Week 14: 30-03-2020: Electronics¶

Demonstration of frequency instability¶

Frequency compensation¶

Download the poster: “Frequency compensation”

Design example: My First Voltage Amplifier¶

Design example “My First Voltage Amplifier”

Download the poster: myFirstVamp.pdf

Presentations

• myFirstVampPart-1.pdf
• myFirstVampPart-2.pdf

Videos

1. My first Voltage Amplifier, Part 1 (21:49)
2. My First Voltage Amplifier, part 2 (5:17)
3. My first Voltage Amplifier, Part 3 (3:55)
4. My first Voltage Amplifier, Part 4 (9:45)
5. My first Voltage Amplifier, Part 5 (4:57)
6. My first Voltage Amplifier, Part 6 (4:28)
7. My first Voltage Amplifier, Part 7 (5:49)

Downloads

• SLiCAP files
• View the SLiCAP output
• SIMetrix schematics
• SIMetrix magnitude plots
• SIMetrix step responses
• SIMetrix schematics
• TI OPA211 macro model

SLiCAP source files Chapter 12¶

Download the Chapter 12 SLiCAP files

Homework¶

1. If a feedback amplifier does not have an MFM characteristic, can it then always be compensated with the aid of one or more phantom zero’s? Motivate your answer.

   Answer

   • This is not true if we restrict ourselves to phantom zeros in the left half plane.
   • This is not true if we restrict ourselves to the application of passive phantom zeros, and if the source impedance, the load impedance and the feedback network do not establish an attenuation in the loop gain.

2. The loop gain of a negative feedback amplifier has a DC value of \(-10^4\) and two poles: one at \(-10\) kHz and one at \(-50\) MHz. The ideal gain of the amplifier equals its asymptotic gain and does not depend on frequency. Our aim is to give the servo function an MFM characteristic with the largest possible bandwidth.

   1. How large is this bandwidth?

      Answer

      • The first-order bandwidth would be: \(\left|(1-L_{DC})p_1\right| =\left| -10^8\right|=100\) MHz
      • The second-order bandwidth would be: \(\sqrt{\left|(1-L_{DC})p_1p_2\right|}=\sqrt{\left| 5\cdot 10^{15} \right|}=70.7\) MHz
      • Since the second-order bandwidth is less than the first-order bandwidth, the system has a second-order low-pass transfer with a bandwidth of 70.7MHz.

   2. Is frequency compensation possible with a phantom zero in the left half of the complex plane?

      Answer

      • The sum of the two dominant poles is about -50MHz, which is less than \(-\sqrt{2}\) times the achievable bandwidth. Hence, the poles are in not MFM positions and the system can be compensated with a passive phantom zero.

   3. If so, what should be the frequency of the zero?

      Answer

      • The frequency of the zero should be (calculation in MHz):
      \[z=-\frac{(70.7)^2}{100-50}=-100\text{MHz}\]

   4. Verify your results with SLiCAP.

      Answer

      You are invited to do so!

3. A passive feedback transimpedance amplifier converts the current of a signal source of a grounded capacitive source with a source capacitance of \(2\) pF into a voltage across a grounded resistive load with a resistance of \(2\) k \(\Omega\). The magnitude of the transimpedance should be \(50\) k \(\Omega\). The \(-3\) dB MFM bandwidth of the amplifier should at least be \(2\) MHz. An operational amplifier of which the specifications have been given below, is preferred as controller.

   1. The DC voltage gain equals \(0.5\times10^{6}\).
   2. The voltage gain has two poles, one at \(-200\) Hz and one at \(-100\) MHz.
   3. The DC output resistance equals \(50\Omega\).
   4. The input capacitance (between the inverting input and the ground) is \(3\) pF.

   Please answer the following questions, and verify your answer with SLiCAP.

   1. Can the required bandwidth be achieved with the preferred operational amplifier as controller?

      Answer

      • The poles of the loop gain are:

        \[p_1=-\frac{1}{2\pi R_f(C_i+C_s)}=-640\text{kHz}\]
        \[p_2=-200\]
        \[p_3=-100\cdot 10^6\]

      • The DC loop gain is:

        \[L_{DC}=-0.5\cdot 10^6 \frac{R_{\ell}}{R_{\ell}+R_o}=-488\cdot 10^3\]

      • The first-order bandwidth is:

        \[B_1=|(1-L_{DC})p_2| = 97.6\text{MHz}\]

      • The second-order bandwidth is:

        \[B_1=\sqrt{\left|(1-L_{DC})p_2p_1\right|} = 7.9\text{MHz}\]

      • The third pole at -100MHz is not dominant

      • Hence a bandwidth of 2MHz seems to be feasible with the given operational amplifier.

   2. If so, is frequency compensation required?

      Answer

      • Yes it is, the absolute value of the sum of the poles is less than \(\sqrt{2}\) times the achievable bandwidth.

   3. If so, give at least two ways for implementation of this compensation.

      Answer

      1. A capacitor in parallel with the feedback resistor
      2. A resistor between the source and the input of the amplifier
      3. An inductor in series with the load

   4. Implement the frequency compensation, using the most promising method.

      Answer

      • The most promising solution is a capacitor in parallel with the feedback resistor. This introduces a new non-dominant pole at a very high frequency.

      • The zero needs to be implemented at:

        \[z=-\frac{(7.9)^2}{7.9\sqrt{2}-0.64}=-5.9\text{MHz},\]

      • from which we obtain:

        \[C_{\mathrm{phz}}=\frac{1}{2\pi\times 5.9\cdot 10^6\times 50\cdot 10^3}=0.54\text{pF}.\]