Transimpedence amp project exercise - Page 1
Forum section: electronics / projects

author: Silente
author_rank: Newbie
timestamp: January 20, 2020, 09:26:54 pm
content: I'm trying to understand how poor I am in analog design, to improve, and I wanted to make a sample project, all the way (on paper), of a transimpedance amplifier, which takes the input signal from a photodiode illuminated by the 'external. The specifications I want to respect are:1. power of the incident radiation, when there is a signal of -30 dBm,2. diode responsiveness = 0.8 A / W,3. peak output voltage of at least 5 mV,4. 50 Ohm output impedance,5. input impedance < 50 Ohm,6. band > 100 MHz.The mental path I take immediately is: I need a transimpedance that earns at least 5mV/(0.8A/W*1uW) = 6250, it must accept a current signal (so I will use a common base at the beginning to obtain a low Z_in which I can then convert to voltage and amplify (therefore a common emitter stage also). I also think of a possible cascode on the common emitter in the case that, after having feedbacked it, the gain and bandwidth are not sufficient to cover the specifications. Finally, being the required output impedance low enough, I think a common collector buffer would be useful.Hoping I didn't say stupid things and I didn't miss fundamental considerations, I could start thinking about a circuit for the small signal, which I will see later on how to polarize well (*):I'd like to understand if I have made heavy mistakes so far that I don't notice, so I can correct myself and try to continue the project.The feedback that I thought of using, at a first cause-effect-cause-effect analysis, is actually a negative feedback.I would also like to know how to make a few more considerations on the possible choice of pnp transistors instead of npn, which I have not really considered now.Please don't give me ready circuits, I would like to get there on my own, albeit slowly.Thanks a lot to anyone who wants to help me.(*) surely the polarization will insert components that will then influence the signal circuit again, but in any case starting from the latter and then worrying about polarization seems to me the best way to proceed, at least I hope.Thank you.PS: is there a way to insert formulas (latex)?

author: moffy
author_rank: Super Contributor
timestamp: January 20, 2020, 10:54:49 pm
content: Q1 is only pulling current out of the base of Q2, won't work, you need a collector resistor connected to the collector of Q1. Iin same problem, no biasing. Everything is connected to gnd where is your power supply and biasing of the transistors? You would need a collector resistor connected to Q3 also. Doesn't look like the most efficient topology. Why the less than 50 ohm input impedance, is it because of the diode capacitance? An opamp though intrinsically high impedance, with proper feedback has a very low impedance at the -ve input (inverting configuration).

author: RoGeorge
author_rank: Super Contributor
timestamp: January 20, 2020, 11:23:51 pm
content: Quote from: Silente on January 20, 2020, 09:26:54 pmLatex enclosed by backslash dollar will render it inline, when enclosed by double dollar will render centered, on a new line.Example:Code:From Faraday's induction law, the induced voltage \$V\$ in an \$N\$ turns coil will be:$$V = -N\frac{\vartriangle \Phi}{\vartriangle t}$$where \$\Phi\$ is the magnetic fluxFrom Faraday's induction law, the induced voltage \$V\$ in an \$N\$ turns coil will be:$$V = -N\frac{\vartriangle \Phi}{\vartriangle t}$$where \$\Phi\$ is the magnetic fluxThe Latex render will be seen only after posting the message, not at preview.

author: Silente
author_rank: Newbie
timestamp: January 21, 2020, 06:31:07 am
content: Quote from: moffy on January 20, 2020, 10:54:49 pmYes, this is because:Quote from: Silente on January 20, 2020, 09:26:54 pmpolarize=bias, sorry for my english.Quote from: moffy on January 20, 2020, 10:54:49 pmSame reason just explained.Quote from: moffy on January 20, 2020, 10:54:49 pmI can't reply to this, it is a requirement of the exercise. Anyway, yes, I think that it's for that reason.So, could the small-signal topology be this?Thank you.

author: moffy
author_rank: Super Contributor
timestamp: January 21, 2020, 09:33:10 am
content: You need to include biasing before you can do small signal analysis. e.g. Emitter resistance of Q1 is (at room temp) 26/(Ie(ma)). So how can you know what your input impedance is without the biasing current? Your feedback values will also tend to reduce the input impedance.You also need to put some kind of values to the resistors, and some sort of type to the transistors. You said 100MHz BW, the transistor capacitance is very much going to limit that and interfere with feedback stability. Try using LTSpice to play with your topology and get a feel for what happens with different values.That is the best place to start. You'll need some good HF transistors with an ft better than or equal to 1GHz.

author: magic
author_rank: Super Contributor
timestamp: January 21, 2020, 09:40:40 am
content: Yes, Q1 would benefit from some bias current in parallel with the diode to bring its emitter input resistance and bandwidth to reasonable numbers.

author: StillTrying
author_rank: Super Contributor
timestamp: January 21, 2020, 11:07:00 am
content: "Try using LTSpice"Yep. Ive not managed to add RF transistors to LT, but the already there 2N2369 is high speed.

author: magic
author_rank: Super Contributor
timestamp: January 21, 2020, 11:29:21 am
content: Come on, models are everywhere and adding them is trivial. Which part you have a problem with?

author: Kleinstein
author_rank: Super Contributor
timestamp: January 21, 2020, 02:53:18 pm
content: For the proposed circuit, the output impedance would not be 50 Ohms, but usually much smaller. The feedback from the output reduces the impedance to less than Re+Impedance of the transistor in parallel. This is of cause easy to fix with the usual series termination.For the input part the transistor in base configuration will have low impedance. So part of the current will flow to the transistor and only a part will flow through Rf. So the transimpedance gain would be smaller than 1/Rf.For low current from the photo-diode, Q1 would also add extra current noise.

author: Marco
author_rank: Super Contributor
timestamp: January 21, 2020, 03:13:41 pm
content: Quote from: Silente on January 21, 2020, 06:31:07 amIt's just a simplified circuit, you haven't substituted the transistors yet.If you use biasing resistors they do belong in both the simplified circuit and the small signal model. If you use PNP/PMOS current sources then use current sources in the simplified circuit and just drop them in the small signal model. Also just draw the power rail in the simplified circuit.

author: awallin
author_rank: Frequent Contributor
timestamp: January 21, 2020, 07:09:15 pm
content: Quote from: Silente on January 20, 2020, 09:26:54 pmyou need to specify the photodiode size, which goes hand in hand with the photodiode capacitance, to fully specify the problem.For 100 MHz you need small-ish photodiodes, maybe 1mm diameter or smaller, with capacitance 10 pF or lower.like these maybehttps://www.hamamatsu.com/resources/pdf/ssd/s5971_etc_kpin1025e.pdf

author: Silente
author_rank: Newbie
timestamp: January 21, 2020, 08:57:25 pm
content: Thank you to all.I'll try to refine the schematic after your tips, and I will show calculations and simulations that I will do.I'm downloading LTSpice.Thank you again for now.

author: Marco
author_rank: Super Contributor
timestamp: January 21, 2020, 10:14:58 pm
content: Personally for a toy circuit like this I wouldn't use LTSpice ... can't stand the interface. Simetrix comparatively easier to work with for me and the free version is powerful enough for this.

author: magic
author_rank: Super Contributor
timestamp: January 21, 2020, 10:59:14 pm
content: Quote from: StillTrying on January 21, 2020, 12:39:09 pmThat's technically off topic, but since OP may run into the same gotcha:The parameters that show up in "pick new transistor" are only informative and don't affect simulation. Many 3rd party models omit them, but they still work.