mosfetoperational-amplifier
I need to analyze a schematic and I am having trouble with this part:
The thing is I don't get at all the utility of the N-channel MOSFET at the output of the op-amp. Could anyone explain the purpose of this component?
Because I think, the conversion would be done even without this transistor.
Place the load between the 7.5 Volt supply and the MOSFET drain, and you will get switch-like behavior.
simulate this circuit – Schematic created using CircuitLab
In your current layout, the source is floating depending on current through the LED and resistor. Hence, Vgs is not 0 to 5 Volts as you assumed, but much lower, depending on where Source is floated to at a point in time.
What is needed is for the "switch" to conduct when given a positive signal from the Arduino, thus pulling its Drain node to ground (or close to it), thereby expressing the desired ~7 Volts across the LED + R3.
See here -> Digikey has the BSS138 and they can be obtained from them relatively quickly.
They do not seem to have through hole versions.
Once you play a little you will find that the SOT23 smd pkg is not too hard to hand solder - and you can add wire leads if needed.
That level shifter circuit is trickier and more critical of part characteristics than most simple circuits are.
The BSS138 is an "N Channel MOSFFET". It has a much lower turn on voltage (Vgson) than most similar MOSFETs and this is important in this circuit.
Alternative parts in through hole are available but they are less liable to be available to you than a BSS138. eg see Digikey for TN0702 or maybe the LP0701 - but BSS138 is available from the same supplier. If you can learn to handle the SOT23 pkg a lot more parts will be usable.
Best Answer
This circuit converts a voltage in a current, as you can see in the transfer function.
The transistor isn't relevant in calculating the output current, which is only dependent on the input voltage and R1.
From the circuit you can find that:
$$ V_{in-} = V_{SS} + I_{OUT} \cdot R_1 $$
But if the Opamp is in the high-gain region, you'll also have that (ideally):
$$ V_{in-} = V_{in+} = Vin $$
Therefore you can compare the right term of both equations and obtain:
$$ Vin = V_{SS} + I_{OUT} \cdot R_1 $$ $$ I_{OUT} = \frac{Vin}{R_1} $$
The transistor is meant to drive the output current depending on the gate voltage. Think about it this way: the Opamp will do what's needed to make its input equals, and this willimply supplying a voltage so that R1*Iout is equal Vin. The relation between Iout and Vo(opamp) will be set by the transistor.
So the transistor will perform the real V-I conversion, creating a feedback loop with the op-amp.