# Arduino controlled switch for 8V 5A

So, my problem is very simple: I need to use an Arduino to control whether or not my circuit is open or not.

However, I need a low resistance (less than 1 Ohm, hopefully) switch (Relay, Transistor, I don't really know what to look for) that lets a 4A current through.

I have a 8V battery that I need to connect to a 2 Ohm (can be tweaked slightly according to the switch's resistance) load at certain times, but so far I've fried everything I tried.

I used the N-channel MDF5N50B transistor with these characteristics and a few relays that I was 99% sure would fry, and fry they did.

What should I look for in the component?

I searched some electromagnetic relays but they all seem to require a higher current than the Arduino provided one to function.

It's ridiculous that I already took 3 electronic courses in uni and for all the theory on transistors I still have no idea what to look for.

• (1) What type of circuit do you want to switch on and off? I'm asking for the sake of context. (2) About electromagnetic relays. You seem to have found a relay which requires more current into its coil than a microcontroller pin can provide. You can switch the relay coil current with a transistor (BJT or MOSFET), and control that transistor with a digital pin of a microcontroller. That's a normal practice. Don't forget the flyback diode, of course. – Nick Alexeev Jun 9 at 17:50
• Errm. The resistance of most heating filaments I am aware changes with temperature. What are the chances yours pulls more than 4A when first powered on? That is, does it maybe draw more than 4A before it gets hot? – JRE Jun 9 at 20:35
• Or, to rephrase JRE's comment, is the filament 2Ω when it's hot or cold? On a different note, what voltage did you apply to the gate of MDF5N50B when you turned it on (and it fried)? – Nick Alexeev Jun 9 at 20:43
• @André Notice that the ratings in the MDF5N50B datasheet are conditional on the +10V gate voltage (Vgs). +5V at the gate is not enough to turn on that MOSFET. That's true for majority of MOSFETs. There are, however, MOSFETs which are designed to work with low gate voltages; so called logic-level MOSFETs. As an aside, it looks like the MDF5N50B is made for switching high voltage. 500V is quite high for a MOSFET. (Did you salvage your MOSFET out of some AC to DC power supply?) At the same time, you don't care about high voltage for what you're trying to do; you've got +8V. – Nick Alexeev Jun 9 at 22:01
• @André That one has a similar Vgs (gate voltage) issue. Read-up: Is MOSFET gate threshold voltage a limit or minimal “Full-on” switching voltage?, MOSFET as a Switch. – Nick Alexeev Jun 9 at 23:24

This is really more of a (really!) long comment and not so much a direct answer to your original question. However, this information is relevant to any answer that would involve using a MOSFET as an ON|OFF switching element to drive the load directly, or to drive the coil of a relay.

The MOSFET's gate threshold voltage specification, $$\V_{GS(th)}\$$ or $$\V_{TH}\$$, identifies the gate-source voltage that causes the MOSFET to just begin to conduct current in its drain-source path. This is the point where the MOSFET starts the transition out of CUTOFF mode and into saturation mode (small signal amplification mode).

(n.b. A MOSFET's saturation mode corresponds to a BJT's forward active (small signal amplification) mode. MOSFET saturation does NOT mean "fully ON".)

If you look at data sheets for various MOSFETs you'll probably find that the manufacturer specifies the threshold conditions as $$\I_{D}=250\,\mu A\;@\; V_{TH}\$$. A current of 250 uA is miniscule, and the MOSFET is nowhere near its fully ON state (a.k.a., the MOSFET's ohmic or triode or linear mode of operation).

So you must ensure your gate-source voltage is well above $$\V_{GS(th)}\$$ to ensure your drive the MOSFET fully into its linear mode of operation--i.e., out of cutoff, and all the way through its saturation region (small signal amplification mode), and into its fully ON linear mode of operation.

Okay, then what exactly do I mean by well above $$\V_{GS(th)}\$$? This is a complex question to answer so rather than give a rigorous mathematical solution I'll give you a "back of the envelope" approach you can use as a starting point for your designs.

If you look at a MOSFET's data sheet you'll usually find a graph figure that plots $$\V_{GS}\$$, the MOSFET's gate-source voltage, against $$\Q_C\$$, the total charge on the MOSFET's gate (usually with units of nanocoulombs, nC). Figure 1 (below) is an example of such a plot from the data sheet for an International Rectifier IRL530N NMOS transistor. I added the colored content seen on the figure.

A MOSFET's gate acts like a variable capacitance as the MOSFET is turning ON|OFF, and the voltage on the gate is a function of the total charge on the gate; the "QVC" relationship: $$\Q = V \cdot C\$$, or $$\V=Q/C\$$. On Figure 1, the red circle labeled "A" corresponds to $$\V_{GS(th)}\$$, the point where there is just enough charge on the gate to cause the MOSFET to start conducting current in its drain-source path. The horizontal, green dotted oval region labeled "B" corresponds to the Miller Plateau; this is the region where the gate capacitance is changing as the transistor's conduction increases through its saturation (small signal amplification) mode of operation. The blue oval labeled "C" corresponds to the MOSFET's linear region where the transistor is fully ON.

Figure 1. Plot of VGS vs. QG for an IRL530N NMOS (reference).

Looking at Figure 1 you can get a good estimate of the gate-source voltage that's required to put enough charge onto the MOSFET's gate to put the MOSFET into its linear mode of operation (its fully ON state). For example, if the Miller Plateau is occurring at around $$\V_{GS}=3.3\,V\$$, then you want to ensure your gate driver circuit applies, say, $$\V_{GS}\ge 4\,V\$$ to ensure the MOSFET is driven fully into its linear mode.

And, of course, the really important factor here is the gate charge. You want to push and pull charge onto/from the gate as quickly as possible when operating the MOSFET as a switch, and voltage alone isn't going to do this; you need a gate drive circuit that can supply both voltage and current. For example, a microcontroller's digital output pin sources and sinks a fairly small amount of current; consequently, a digital output pin needs a relatively long time to put enough charge onto the MOSFET's gate to fully turn on the MOSFET (analogous to filling a swimming pool with a garden hose; it'll get the job done but it'll take a long time). Ditto for pulling charge off the gate to turn the MOSFET off. This is why special "gate driver" ICs exist:

microcontroller --> gate driver IC --> MOSFET gate


The gate driver can source/sink much higher currents compared to the microcontroller's digital output pin, and thereby ensure rapid charging and discharging of the MOSFET's gate (e.g., the sound suppression system on NASA's mobile launch platform vs. a garden hose). IOW, the gate driver IC's higher rate of charge transfer $$\dQ/dt\$$ reduces the time required to traverse the Miller region, thereby shortening the MOSFET's ON|OFF switching times.