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

 

what would be the output range?

 

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

 

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

 

the load can be a resistor right?
im new to this.

 

could i use a battery with these ratings? which do you think is better?

 

could you help me ? i just want to use this to measure the intensity of light. i donno what my load would be.

 

What are going to feed the output of this circuit to? A microcontroller, perhaps?

hgmjr

 

could it just be a volt meter?

 

Could someone please talk to me about the battery?

 

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

 

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

 

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.

 

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).

 

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

 

ok sorry,i wast taling about the output current .i was talking about the current b/w the cap and the photodiode.