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As seen in this LTSpice model, I am trying to control a white LED (3.6 Vf @ 20 mA) connected to a 7.5 voltage source with a 2N7002 N-channel MOSFET and a 5V control signal from an Arduino.

enter image description here

Given that my gate voltage is 5V from a micro-controller, I was expecting the MOSFET to basically act as a switch. By looking at the graph for the 2N7002, given that the desired current is 20mA and the Vgs is 5V, I was expecting a voltage drop across the transistor of close to zero such the source voltage was ~7.5V.

However, as seen in the simulation graph, the voltage across the transistor is in fact quite large, such that the source voltage is only ~3V (as opposed to the expected ~7.5V).

enter image description here

When I breadboarded this circuit, I got the same result, a source voltage of ~3V.

Can anyone explain why the source voltage of the MOSFET is much lower than expected? And can anyone please recommend a transistor that would effectively allow me in this situation to create a switch to drive the white LED with the use a 5V signal and a 7.5V supply power?

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  • \$\begingroup\$ Vgsth is about 2.1 to 2.5V @ Id=0.25mA, If assume when LED is forward biassed, 3.5V will appear across it and across 200Ohms resistor 20mA*200 =4V (actual voltage at a given forward current will get from LED characterstics). There is some problem with the circuit. COnnect the source to GND and design accordingly. \$\endgroup\$ – user19579 Jul 24 '13 at 5:31
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Place the load between the 7.5 Volt supply and the MOSFET drain, and you will get switch-like behavior.

schematic

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.

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    \$\begingroup\$ Then important point here is that the FET is controlled by the gate-to-source voltage, not the gate-to-ground voltage. \$\endgroup\$ – Peter Bennett Jul 24 '13 at 17:32
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Just to add to the previous comment, you can control your led with the arrangement you provided, but you'll need to use a P channel mosfet.

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  • \$\begingroup\$ Yep, but the OP should realize he'd need to use negative logic in this case, i.e. if the GPIO is 0 the LED will be on and vice versa. Or he could control a NPN BJT or NMOS that controls the PMOS to keep logic positive.. But now we are adding a lot of circuitry for a simple LED switch. \$\endgroup\$ – Jim Jun 6 '16 at 14:04

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