Transistor / Photodiode Circuit Switch Analysis

[Jamessamuel]

Homework Statement

Firstly, this is more of a schoolwork issue, not coursework or a set task. It is more to do with understanding the circuit seen in the image provided:

I am trying to understand why this circuit does what it is supposed to; activate the lamp when a certain light intensity is incident on the photodiode.

Homework Equations

The Attempt at a Solution

My logic is that whether or not the transistor will allow a current to flow depends on the voltage difference between the gate and source nodes into the transistor (horizontal wire and lower vertical wire). As i believe these points and the photodiode lie in parallel, the potential difference between gate and source is the potential difference induced in the photodiode. So, my belief is that the following will take place:
1. intensity of light increases.
2. photocurrent increases proportionally.
3. voltage increases as the current increases (not sure if proportional, but it is a positive relationship). Since the photodiode is connected to ground, the potential just after the diode is the potential at the gate.
4. when the voltage induced by the photodiode is large enough, an inversion layer forms in the MOSFET, current flows and the lamp is lit etc.

The book talks about the resistor and how the potential difference across it relates to the Gate-Source voltage we need to know about. However as you can see i don't even consider the resistor. Am i right still?

P.S i have another small query. why do we need a larger, 6V supply voltage overlooking the whole circuit?

Guidance appreciated as ever.

Regards,

James.

 

[gneill]

Your item (3) needs a bit more thought. Think of the photodiode as a current source whose current is proportional to the light intensity incident upon it. The more light, the more current it will generate. In the dark it will generate almost no current. Now, what is the path of this current? What established the potential at the MOSFET gate?

As for the 6 V supply, something is required to power the circuit. After all, there's a 6 V lamp to light.

 

[Jamessamuel]

I think the current all goes into the resistor... at least i think i can make that assumption.
Now this is my solution; the potential at the gate is 6 - IR where I = photocurrent.
as I gets larger (more light) the potential grows smaller. but, if it starts to get dark, the current drops. We subtract a smaller figure from the 6, and the voltage, relative to its previous smaller state, rises eventually above the functional voltage threshold. Current flows into the MOSFET, the lamp lights.

in response to your answer about the 6V presence, is the reason that we need it evident from the equation i wrote above? the 6V boosts the sum to a higher number so we can hit the threshold voltage? I mean, the current is supplied by the diode so under any other circumstance, If someone told me we don't need it I'd believe them.

 

[QuantumQuest]

You have to figure out what happens to the current produced in the photodiode. When overcoming the small potential difference of the transistor between drain - gate, what happens to the current output at the drain. When there is no light is there a potential difference on lamp regarding 6V? What happens as the current increases at drain regarding resistor?

 

[Jamessamuel]

im confused now; will the voltage difference between drain and gate influence when the lamp lights? and due to parallel configurations, is the voltage across the resistor the sum of the lamp and the drain-gate voltage difference?

 

[gneill]

im confused now; will the voltage difference between drain and gate influence when the lamp lights?

The potential between the Gate and Source is the influence that controls the Drain current.

and due to parallel configurations, is the voltage across the resistor the sum of the lamp and the drain-gate voltage difference?

Yes, but remember that Gate current is essentially zero for practical purposes, and might as well be an open circuit.

 

[QuantumQuest]

im confused now; will the voltage difference between drain and gate influence when the lamp lights? and due to parallel configurations, is the voltage across the resistor the sum of the lamp and the drain-gate voltage difference?

First, sorry for the typo:

When overcoming the small potential difference of the transistor between drain - gate

I meant between source - gate. This voltage controls the conductivity of the channel from the source to the drain.
Now, what I just wanted to point out, is that you must follow the path of various currents in order to figure out voltages. That way we did it a long ago at a technical school for electronics I was - although mainly for BJTs back then, but the idea remains the same. It is of no help to try to figure out how things work by a high point of view. This may sound difficult at the beginning, but pays well back. The best way I think, is to create a diagram with pen and paper and draw currents and whatever is of help to you on it. Taking into account how the transistor functions in general and in the context of your specific setting, you can draw safe conclusions. For the second question, things work as gneill points out.