E84 Lab 6: Design of a transimpedance photodiode amplifier

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1 E84 Lab 6: Design of a transimpedance photodiode amplifier E84 Fall 2017 Due: 11/14/17 Overview: In this lab you will study the design of a transimpedance amplifier based on an opamp. Then you will design a transimpedance amplifier to meet a specification. You will simulate if your design works. What to turn in: 1. Your circuit schematic with the parts labeled. 2. Your parts list within the budget and vendor/source of said parts ( to Prof. Bryce). 3. Any paper calculations or design thoughts. 4. A transient SPICE simulation of your design at 10 MHz driven by the current/capacitance model of the photodiode. 5. An AC SPICE simulation of your design showing it meets the AC specs. Feedback: Let us know: What went well in this assignment? Were there points of confusion? How long did you need to work on this assignment?

2 Your task for this lab is to design a transimpedance amplifier around the SFH2701 Si pin photodiode. It should meet the following specifications: Gain 20,000 Volts/Amp 3 db Bandwidth 18 MHz, with < 1 db gain peaking Coupling DC Output impedance 50 Ohm External power input 6-9 V DC Cost of BOM 10 dollars exclusive of PCB PCB 1 square inches, 2 layers Noise Minimize You should design your amplifier on paper first, then simulate it. You are free to select any parts you wish from any vendor, you may need to use other versions of SPICE or import models if you wish to use parts not available from Linear Technologies. You must submit your part list, and calculations/simulations by the due date. The next lab is dependent on this lab as you will design a PCB for your circuit. A point of emphasis in the specification list is that you are allowed only a single power input to the system 6-9 V DC (you do not get to choose the voltage in this range you only know that it is bounded to this range). You must accommodate this range of input voltages without damaging your circuit. This may require you to select a LDO or other power management solution. To minimize the noise of the system the specification of this power management system may be important. The rest of this document is intended to help you design your transimpedance amplifier.

3 Consider the simplest of all opamp based transimpedance amplifiers: Here D1 is a photodiode and creates a current proportional to the amount of light hitting it. Consulting our photodiode s datasheet we see that SFH2701 will produce 0.5 A/W for 780 nm light. The relative response of the photodiode as a function of wavelength is provide in the datasheet and is shown at left. This shape is typical of Si photodiodes and has largely to do with the band structure of Si convolved with the number of photons per watt as a function of wavelength. When we measure the response of the circuit in lab 9 we will use 850 nm light. Let s consider U1 an ideal opamp. If this is the case then the current produced by the photodiode will flow through R1. Thus the output voltage will be V out = I D R 1. The transimpedance gain of this circuit would be R1 V/A. Thus to meet the specification of a gain of 20,000 V/A we could simply use a 20 kohm resistor. If there was such a thing as an ideal opamp we would indeed be done at this point. However there is of course no such thing as an ideal opamp. A more complete analysis of the circuit above is complicated by the existence of the diode. The simplest model we might apply to replace the diode would be:

4 As the photodiode provides a current we have modeled it as an ideal current source. Of course it is not ideal but given the voltage across it should be close to 0 due to the action of the opamp, a shunt resistor to model the small loss is likely not needed. What is needed however is the fact that the photodiode has capacitance. Indeed this is critically important. This capacitance sets a limit on the bandwidth of the non-ideal circuit. Referring to the datasheet there are two limits on the speed of the photodiode. Both have plots: The capacitance will limit the speed of the overall circuit by interacting with other parts of the system. The switching time can be an intrinsic property of the semiconductor device depending on its operation. Both of these figures are plotted as a function of reverse bias on the photodiode. The rise and fall times decrease as more reverse bias is applied to the diode because this increases an electric field in the photodiode that removes charge from the active region of the semiconductor. With 0 V of bias across the photodiode from the figure we might reasonable estimate the rise and fall times are on the order of 10s of ns. t rise = 0.22/BW This formula comes from t rise being a first order response. The bandwidth for the specification is 18 MHz. This means we can tolerate up to 12 ns of rise time. Any reverse bias greater than 1V would accomplish this. This is a constraint on our design, we must reverse bias the diode by more than 1 V to meet the specification.

5 We also see the capacitance of the photodiode decreases as a function of reverse bias. At 0 V the capacitance is nominally 3 pf, at 1 V it is less than 2 pf, the asymptotic value conservatively looks like 1.8 pf. To understand the basics of our circuit for now we will neglect the need to reverse bias the diode and presume we have provided this function. From the datasheet we have learned that we must consider the value of C1 between 1.8 pf and 3 pf. If we were to use a breadboard to construct this circuit the breadboard itself would add 2 pf for two pins next to each other! These capacitances might seem small but if you want to go fast they are important! Returning to the transimpedance amplifier circuit we can solve for what we need to learn by considering the non-ideal properties of the opamp. Let s consider the input impedance of the opamp still to be infinite but for the gain to be both finite and a function of frequency: A = A(w) = A 0 /(1 + j w w 0 ). This models the opamp s gain as being limited by single simple pole. This is often very close to correct as designers will often intentionally put a so called dominant pole in the transfer function of an opamp to make designing with it simpler. The value of w 0 can be determined by either the -3 db bandwidth or the gain bandwidth product (GBP). We would like to solve for V out /I d which is the transimpedance gain as a function of frequency. By superposition if we consider only the diode current source than we see that C1 and R1 are in parallel. Let the current contribution from this source in C1 be I Cd and let the current in R1 from this source be I Rd. Then: Z C1 I Rd = I d Z C1 + R 1 I Cd = I d R 1 Z C1 + R 1 Now we shall consider the current source off. By definition of the voltage source and the value of the voltage at the V p : V out = A (V p V m ) = A V m For this source the components are in series so the current from the voltage source is:

6 V out I s = Z C1 + R 1 It follows that: V m = (I s + I Cd ) Z C1 V out = A (I s + I Cd ) Z C1 = A ( V out + R 1I d ) Z R 1 + Z C1 R 1 + Z C1 C1 Solving for V out /I d : V out I d = A R 1 Z C R 1 + Z C + AZ C Rearranging to put it in a better form: V out I d = R R 1 + Z C1 A Z C1 Notice that in the limit of A being very large this reduces to R 1 as we would expect. Meanwhile if the capacitor is very large Z C1 will be small and the result will no longer be R 1! If we define: β = jwr 1 C 1 Then the transfer function reduces to: V out = R 1 I d Aβ Recall that: A = A j w w 0 The function has two interacting poles. If the phase at the negative terminal of the opamp becomes -180 degrees while the gain is still above 0 db the system will oscillate at that frequency. This can be solved by placing a capacitor in parallel with R 1. We leave it as an exercise to replace R 1 with the effective impedance of the resistor and capacitor in parallel. The analysis is otherwise the same. Formula in hand let s look at some possible real life numbers using the LT6230 on a breadboard. The specifications for the device over the normal temperature range begin on page 4. If we use the LT6230 on a breadboard and we reverse bias the diode a few volts we might reasonably estimate C 1 = = 11.7 pf (2 pf from the diode, 2 from the breadboard and 7.7 pf from the opamp (see differential capacitance)). We need 20,000 Volts/Amp so we would choose R 1 = 20,000. Next we look at datasheet (page 4) and see that the large signal open loop gain reported is 78,000 V/V (A 0 ) while the GBP is >150

MHz (minimum). This implies that w 0 = 12083 Hz (if we want consistency with a dominate pole, which it is not actually in the plots, but is useful in this generic analysis as a first estimate).

7 MHz (minimum). This implies that w 0 = Hz (if we want consistency with a dominate pole, which it is not actually in the plots, but is useful in this generic analysis as a first estimate). We can now plot the magnitude of the transfer function. Sadly this design has two major problems. First it falls short and of the required -3 db bandwidth only reaching 15 MHz. Second it has a very bad gain peaking problem. The gain peaking problem could lead to oscillation of the circuit; it also adds a lot of noise to the broadband output for input tones that are at lower frequencies. Both are bad. To fix this problem we must add a capacitor in parallel with the feedback resistor, placing the pole to the left of this gain peak. This will reduce the bandwidth of the circuit even more. These calculations are fine as far as they go, but a full SPICE based simulation is warranted. An LTSpice simulation a LT6230 on a breadboard (2 pf), with the diode at 2.5V reverse bias (2 pf), and a 20 kohm feedback resistor results in the simulated result shown at left. The gain peak occurs 7 MHz rather than 10 MHz in the simple model, while the -3 db bandwidth is only 10 MHz. Getting rid of the 2 pf from the breadboard breadboard and moving to a PCB would help increasing the -3 db bandwidth to 12 MHz, but it is still not enough. Another opamp/design is needed. A rule of thumb for calculating the approximate

8 position of the 3dB bandwidth is: f 3dB f RC GBW. Here f 2 RC = 1/(2πRC). Using this rule we arrive at 8.4 MHz for R = 20,000, C = 12 pf and GPB = 215 MHz. This is a reasonable estimate but the simulation is critical to see the full behavior including the exact gain peaking. The core idea behind the TIA: The simple rule points to the core purpose of using an opamp as a transimpedance amplifier. One could convert the current from the photodiode to a voltage by simply adding a resistor to the photodiode: This circuit and the equivalent current model clearly show that we would get the needed I/V conversion, but the -3 db bandwidth would be f RC = 1/(2πRC), only 2 MHz for the 20 kohm and 4 pf. How does the opamp speed up the circuit? Intuitively it does so by reducing the voltage swing across the capacitor. Notice how the diode s effective capacitor is attached across two points which should be almost the same voltage. If there is no voltage difference then the passive component effectively does not exist. This is the key concept as to why the opamp is helpful in speeding up the circuit. The ability of the opamp to make the capacitance vanish depends critically on how small it can make the voltage swing across the diode capacitance. This of course depends on how large the gain of the opamp is. The larger the gain (and thus the larger the GBP), the better the opamp will do in terms of speed. The downside to all of this magic is gain peaking which must be compensated (damped out) to keep the system stable and low noise.

Design: As was mentioned the circuit analyzed above does not reverse bias the photodiode. Your task is now to come up with a circuit to meet the specifications.

9 Design: As was mentioned the circuit analyzed above does not reverse bias the photodiode. Your task is now to come up with a circuit to meet the specifications. There are many design choices in front of you. Should you use single supply or dual supply? If you use a dual supply how will you create the negative rail? Should you use one stage or several? How will you supply low noise power to your amplifier? How will you apply the reverse bias to the photodiode and how much can you apply? These and more interplay in how to accomplish the specifications within budget. Be creative, and enjoy exploring different options. It is wise to explore multiple circuit configurations and parts options in each configuration. Finding parts: Finding parts is one of the major skills to get out of this lab. The design you choose will be determined by what parts you can find and what specifications they have. You will need to be able to read the part s datasheets and parse the needed information to determine if the parts are suitable for your design. It is helpful to have settled on a configuration or topology for your circuit before looking for parts but you may find you cannot implement one topology and have to choose another. Part selection is always complicated by costs and availability/lead times. For this design you have a small budget. There are many distributors of electronic parts. If you know the exact part you want octopart.com is a good website for finding the lowest cost distributor and who has the part in stock. If you do not know what parts exist you are left with 3 options: internet search, distributor search and vendor search. Internet search is normally the worst of the 3. While there are other great distributors available a major contributing factor of Digi-key s success (digikey.com) is their excellent cataloging and thus good parametric search tools. If the parameter you need to filter parts with is listed in their menus you can quickly find parts across vendors. As good as Digi-Key is for finding things never overlook vendor s websites. They have far more detailed specifications available in their parametric tools.

For instance here is Linear s category options: Texas Instruments, Linear Technology, and Analog Devices are the most likely sources of opamps to design this circuit.

10 For instance here is Linear s category options: Texas Instruments, Linear Technology, and Analog Devices are the most likely sources of opamps to design this circuit. Depending on your design you may need to select one or more power management ICs as well, such as LDOs and/or charge pumps or other switch mode converters. Be sure to select a part available in a package that you can solder. To meet this requirement it should be a leaded package (e.g. SOT-23, SOT-223, SOIC-x, etc.). Good luck.

EE320L Electronics I. Laboratory. Laboratory Exercise #2. Basic Op-Amp Circuits. Angsuman Roy. Department of Electrical and Computer Engineering

EE320L Electronics I. Laboratory. Laboratory Exercise #2. Basic Op-Amp Circuits. Angsuman Roy. Department of Electrical and Computer Engineering EE320L Electronics I Laboratory Laboratory Exercise #2 Basic Op-Amp Circuits By Angsuman Roy Department of Electrical and Computer Engineering University of Nevada, Las Vegas Objective: The purpose of