# Photometer : Current To Voltage Converter

Discussion in 'The Projects Forum' started by dickcruz, Jan 29, 2007.

1. ### dickcruz Thread Starter Member

Jan 7, 2007
20
0
I need to know what the input current can be I have a 400mAh, 4.8v | 300mAh, 4.8V. would this suffice?

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2. ### dickcruz Thread Starter Member

Jan 7, 2007
20
0
what would be the output range?

3. ### hgmjr Moderator

Jan 28, 2005
9,030
214
Is your question, "what is the estimated current consumed by this circuit?"

I am assuming you are trying to decide whether you can get away with using a 4.8V battery that is rated 400 mAh and one that is rated at 300 mAh.

hgmjr

4. ### hgmjr Moderator

Jan 28, 2005
9,030
214
Both the total current consumption and the output level is going to depend on the nature of the load you intend to connect the circuit to.

Can you tell us what that load consists of?

hgmjr

5. ### dickcruz Thread Starter Member

Jan 7, 2007
20
0
the load can be a resistor right?
im new to this.

6. ### dickcruz Thread Starter Member

Jan 7, 2007
20
0
could i use a battery with these ratings? which do you think is better?

7. ### dickcruz Thread Starter Member

Jan 7, 2007
20
0
could you help me ? i just want to use this to measure the intensity of light. i donno what my load would be.

8. ### hgmjr Moderator

Jan 28, 2005
9,030
214
What are going to feed the output of this circuit to? A microcontroller, perhaps?

hgmjr

9. ### dickcruz Thread Starter Member

Jan 7, 2007
20
0
could it just be a volt meter?

Jan 7, 2007
20
0

11. ### hgmjr Moderator

Jan 28, 2005
9,030
214
An off-the-shelf DVM would introduce very little in the way of loading so I would guess-timate the current consumption to be around 4 ma. That means that you could expect in the neighborhood of 100 hours of continuous operation from the 400 mAh battery and around 75 hours of continuous operation from the 300 mAh battery.

I would opt for the 400 mAh battery for the greatest battery life.

hgmjr

12. ### dickcruz Thread Starter Member

Jan 7, 2007
20
0
Yeah, i got that. thanks. now if i were to drive a load circuit like a bulb or something, where would the terminals have to be connected?
And, i can use the volt meter without having to bias the voltage right cause thats all i want to do.
Thanks

13. ### dickcruz Thread Starter Member

Jan 7, 2007
20
0
Finally,

just one question, the current between the photodiode and the inverting input of the opamp, would it be in the direction of the photdiode or the opamp.
and this current would be alternating right?

thanks for all the help.

14. ### beenthere Retired Moderator

Apr 20, 2004
15,815
283
Hi,

The photodiode is marked for polarity, which indicates the direction of current.

Why would the current be AC in nature? You'll see some variance if measuring the output from a fluorescent tube, and some ripple if a tungsten bulb, but the output from the photodiode has to be DC (think about it).

15. ### dickcruz Thread Starter Member

Jan 7, 2007
20
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ok so the direction is from the photodiode to the inverting terminal.

but if the photodiode is DC, then the blocking cap will not let it pass through. only if its ac will it be let through right?

So effectively what i assumed was that the current would be AC and then the transistor would act as a rectifier and supply the DC to the inverting input. and the opamp would act as a difference amp and give V+ - V-. just correct me everywhere ive been wrong.
Thanks for all the help

16. ### dickcruz Thread Starter Member

Jan 7, 2007
20
0
ok sorry,i wast taling about the output current .i was talking about the current b/w the cap and the photodiode.