Electronics > Projects, Designs, and Technical Stuff
Visible Light Oscilloscope Probe
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StillTrying:
Looking at the light shape of light sources is an interesting idea, I wish I'd thought of it. :)
Zero999:
For high speed, high intensity applications, such as monitoring very bright pulses from a flash lamp, just use a photodiode and resistor. I just used a 50 Ohm terminator, at the end of a co-axial cable connected to an oscilloscope. Here's a schematic of the set up I used to drive a high power LED, with very high current pulses and sense them with a photodiode.

https://www.eevblog.com/forum/projects/ultra-short-ultra-fast-led-flash/msg2247069/#msg2247069
graybeard:
not1xor1:
I'm not an EE and have not much experience, so I wonder if there any particular reason to put the switch on the negative of the battery and to connect capacitors from the virtual 2.5V ground to both the positive and the negative supply rails.

In single supply circuits I'm used to put a switch on the positive rail and capacitors between the virtual ground and the negative and between the positive and the negative of the battery rails (close to the supply terminal of the opamps).
I realize your circuit replicates the usual schematic of real ground circuits. So I wonder if I'm wrong or if that makes any difference.  :-//

I think batteries have low noise, but with a noisy single supply voltage, wouldn't capacitors from the positive rail to the shunt regulator increase the virtual ground noise?
graybeard:

--- Quote from: not1xor1 on December 04, 2019, 06:02:21 am ---In single supply circuits I'm used to put a switch on the positive rail and capacitors between the virtual ground and the negative and between the positive and the negative of the battery rails (close to the supply terminal of the opamps).
I realize your circuit replicates the usual schematic of real ground circuits. So I wonder if I'm wrong or if that makes any difference.  :-//

--- End quote ---


The battery and the switch are in series, the order of components in a series circuit does not matter, only the orientation (of some components) needs to be maintained.  Thus it makes no difference where the switch is, if you like it better on the positive battery terminal you can put it there and it will function the same.


--- Quote from: not1xor1 on December 04, 2019, 06:02:21 am ---I think batteries have low noise, but with a noisy single supply voltage, wouldn't capacitors from the positive rail to the shunt regulator increase the virtual ground noise?

--- End quote ---

The capacitors are there to provide a low impedance return path at high frequencies for the signals flowing onto the voltage reference node through R3 & R5 since the impedance of the shunt bandgap reference U3 increases with frequency do to the roll of of it's internal feedback.  The battery starts as a low impedance, but that increases as it discharges.

I did not really pay that much attention to the decoupling capacitors.  I just put them in and only paid attention to the 10µf limit of capacitance in parallel with U3.  I suspect the circuit would work fine C1 through C4 removed.   If I was going to mass produce this I might look into removing them to decrease the parts cost.

The simple answer to your noise question is yes, but the real answer is more subtle.  You would need to look at the impedance of U3 vs frequency and factor that with capacitors, the sensitivity of the resistivity of Q1 to VCE changes, the impedance of the battery vs frequency and discharge state, and the power supply rejection and balance that against the high frequency current return paths.  I did not do that analysis since I did not think it was important in this case.  However it might be in another case.


Both of your questions are good questions, thank you for asking.
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