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Transimpedence amp project exercise

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Silente:
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  \$\Omega\$, 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)?

moffy:
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).

RoGeorge:

--- Quote from: Silente on January 20, 2020, 09:26:54 pm ---PS: is there a way to insert formulas (latex)?

--- End quote ---

Latex 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 flux

--- End 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 flux

The Latex render will be seen only after posting the message, not at preview.

Silente:

--- Quote from: moffy on January 20, 2020, 10:54:49 pm ---Iin same problem, no biasing.
--- End quote ---
Yes, this is because:


--- Quote from: Silente on January 20, 2020, 09:26:54 pm ---I could start thinking about a circuit for the small signal, which I will see later on how to polarize well (*):

--- End quote ---
polarize=bias, sorry for my english.


--- Quote from: moffy on January 20, 2020, 10:54:49 pm ---Everything is connected to gnd where is your power supply and biasing of the transistors?

--- End quote ---
Same reason just explained.


--- Quote from: moffy on January 20, 2020, 10:54:49 pm ---Why the less than 50 ohm input impedance, is it because of the diode capacitance?

--- End quote ---
I 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.

moffy:
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.

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