Photometer : Current To Voltage Converter

hgmjr

Joined Jan 28, 2005
9,029
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
 

hgmjr

Joined Jan 28, 2005
9,029
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
 

hgmjr

Joined Jan 28, 2005
9,029
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
 

Thread Starter

dickcruz

Joined Jan 7, 2007
20
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
 

Thread Starter

dickcruz

Joined Jan 7, 2007
20
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.
 

beenthere

Joined Apr 20, 2004
15,819
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).
 

Thread Starter

dickcruz

Joined Jan 7, 2007
20
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).
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
 
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