Current Amplifiers

[MrDropsy]

Figured if i ground op amp inputs, id reduce noise. I also put capacitors across the rails to ground. This significantly reduces me noise.

That and using a 12 V battery rather than a power supply.

[Genesis92]

the oscillation you see in pic 1 is the sinewave from your AC bulb
1/0.020 = 50Hz
the oscillation on the pic 2 is a capacitive coupling of the sensitive input and high gain amplier with your finger. Your body create a big antenna
as explained in other topic 1M is a very high gain!

To give you an idea my laser harp sensor "see" scattered laser light at 3 meters with only 220K of feedback in my I/V converter!

the LM324 in not rail to rail input/output and even if they said "low voltage" it's for biased input not from groud to V+

it's for AC signal with V+/2 bias at the input and the output will move around the Vsupply/2. this is why we must use AC coupling capacitors in that case.
for your application you must use a rail to rail in/out amplifier or +8v/-2v supply voltage to get à +5V swing voltage at the output.

look this circuit:


R3 and R4 makes a voltage divider to VCC/2 for the V+ input. it's a single voltage application. and you have AC coupling capacitors (C1 and C2) at the input and output to remove the DC bias voltage

from LM324 datasheet:
Large Output Voltage Swing 0V to V+ − 1.5V
so you can get the ground but only 3.5V max with +5v supply. over this value the signal is deformed

[MrDropsy]

Thank you genesis92 for your input. First im just experimenting to see if it response to light at all. I have reached where it clearly does respond to light.

The difference between dark and very bright gives me a voltage swing but only time will tell whether that swing is large enough.

The problem with arduino is that its only 0-5V - surely that limits the range i possibly achieve.

Do you know if i put in a negative voltage through the arduino?

i am aware there are voltage converters - so i can put in +5 and obtain -5. So that way i can have -5 to +5.

Im not sure if this is a wise thing to do.

[jremington]

Do you know if i put in a negative voltage through the arduino?

No, that will damage the Arduino.

[MrDropsy]

Im putting through the signal from my circuit into arduino. But only using 0 to 5V gives me a range of around 0-700,

How can i maximise the full range 0-1024 when i cant even get more 5v for the board

[Genesis92]

You have asked same question Here
use rail to rail in/out operational amplifier instead of the standard LM324 to get 0 -> 5V amplifier output

[MrDropsy]

rail in/out confuses me. doesnt that just mean theres a V+ and V- pin on chip, which is what the Lm324 has.

ill be trying the Tl072, that also has V+ and V- pins

[MrDropsy]

The MCP6L02 you mentioned - is that possible to get as a through hole component or is it only a surface.

[Genesis92]

from the other topic: LMC6482 is a good choice too and in DIP package
and from Ih55t user (last post)
I've had great success with LM6134 from TI. It's designed to run on a single 5v supply (can go down to less than 2V), and both the inputs and outputs are rail to rail. You should still be able to get free samples from TI.
unfortunately I've got an update from TI last week and now only companies and schools can get sample.

they all have V+ and V- but some can work with only one supply rail and other must have negative supply to have their output reaching the ground (0v) or going negative.
rail to rail output amplifer can be powered with only 5v an the output can go between the grounf and 5V with a little voltage drop near 50mV

[MrDropsy]

yes i was just looking at that one in my local electronics store. If you think i can use it for high intensity photodiode applications, and i will make use of the full 0-1024 , i will purchase it.

[Genesis92]

you will not get exactly 1024 but near this value 1010/1015 I think based on 50mV drop.

about high intensity it depends of the I/V resistor value

[MrDropsy]

Yes the closer i can get to the maximum range, the better. Im still trying to figure out the resistor value. I understand that its ohms law - but im not sure how to apply it.

V= IR

is my voltage just 5V?

and Current just whatever i get from the photodiode... which is in the range of a few microamps from memory.

R = V/I

EDIT:

i guess the problem is that i dont know what current im getting out of the diode
so 5/3 uA

R= 1.6 M ohm

that doesnt make sense to me

[Genesis92]

use the spec of the photodiode to know the current value or if you have a multimeter with low mA range try to connect the photodiode like a current source. expose the photodiode to the highest light level and note the max current value.
after that the R value can be found by 5/imax = R
example Imax = 1mA
5/0.001 = 5000 -> 5K

[MrDropsy]

I see, thank you Genesis92, i suspected as much.

Ill try to find the current im getting at the maximum level of light.

The data sheet does show a correlation between lux and short circuit current.

so if i can figure out how many lux i have, ill be able to see how many mili amps etc.

Makes a lot of sense now ! Thank you

Ill let you know how it goes. Ill have to do it tomorrow at work.

[Genesis92]

yes photodiode datasheet is not easy because the amount of current depend of the photodiode response (with its "sensitive surface" size) and wavelenght
BPW21 is a good choice for visible wavelenghts
here you have a luxmeter example if that can help you
link

schematic diagram
but this one uses a differential measure with gain and offset adjustment (to remove the dark + bias current errors). it must be powered by two supply voltage (positive and negative)