user17255: This is basically an off-shoot from https://arduino.stackexchange.com/questions/20569/unexpected-output-pin-behavior-was-relay-on-off-doesnt-loop. It's the same schematics and so on, however, I'm now wondering what the output voltage and current is. 
I've a 9V battery and the MOSFET transistor from the arduino starterpack (doesn't say the specs) connected to the 5V pin on the board. I need the following output current/voltage for it to activate a solenoid:

6V: 160 mA 
7V: 190 mA 
8V: 220 mA 
9V 240 mA 
10V :270 mA 
11V :300 mA
12V :320 mA

and I'm just not sure how to calculate it since I'm not sure about the output from the transistor and all the way to the solenoid.
The solenoid: http://www.electrokit.com/en/plastic-water-solenoid-valve-1-2-12v.51108


user4143: You don't need to "calculate" anything with a MOSFET.  You will be using it in the saturation zone. That means it is fully turned on and will allow through the current you need.
However, the gate of a MOSFET looks, to all intents and purposes, like a small capacitor. That means there will be a certain amount of inrush current to the gate when you turn it on. To limit that inrush current you need a small resistor (say 100-220Ω) on the gate.
When using a BJT, though, you need to ensure that you allow enough current through the base to allow enough current through the collector.  Divide the current you need through the collector by the Hfe (typically 100 for a small signal transistor) and that is the minimum current you need to allow through the base. Subtract a silicon junction voltage (0.6V) from the voltage of the IO pin (5V). Divide it by your base current calculated above, and that is the maximum resistor you can use on the base. 
And also your transistor symbol in your schematic is backwards.

17255: <p>This is basically an off-shoot from <a href="https://arduino.stackexchange.com/questions/20569/unexpected-output-pin-behavior-was-relay-on-off-doesnt-loop">https://arduino.stackexchange.com/questions/20569/unexpected-output-pin-behavior-was-relay-on-off-doesnt-loop</a>. It's the same schematics and so on, however, I'm now wondering what the output voltage and current is. </p>

<p>I've a 9V battery and the MOSFET transistor from the arduino starterpack (doesn't say the specs) connected to the 5V pin on the board. I need the following output current/voltage for it to activate a solenoid:</p>

<ul>
<li>6V: 160 mA </li>
<li>7V: 190 mA </li>
<li>8V: 220 mA </li>
<li>9V 240 mA </li>
<li>10V :270 mA </li>
<li>11V :300 mA</li>
<li>12V :320 mA</li>
</ul>

<p>and I'm just not sure how to calculate it since I'm not sure about the output from the transistor and all the way to the solenoid.</p>

<p>The solenoid: <a href="http://www.electrokit.com/en/plastic-water-solenoid-valve-1-2-12v.51108" rel="nofollow noreferrer">http://www.electrokit.com/en/plastic-water-solenoid-valve-1-2-12v.51108</a></p>

<p><a href="https://i.stack.imgur.com/ElqV9.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ElqV9.png" alt="It should be solenoid instead of relay in the schematics"></a></p>
 4143: <p>You don't need to "calculate" anything with a MOSFET.  You will be using it in the <em>saturation</em> zone. That means it is fully turned on and will allow through the current you need.</p>

<p>However, the gate of a MOSFET looks, to all intents and purposes, like a small capacitor. That means there will be a certain amount of <em>inrush</em> current to the gate when you turn it on. To limit that inrush current you need a small resistor (say 100-220Ω) on the gate.</p>

<p>When using a BJT, though, you need to ensure that you allow enough current through the base to allow enough current through the collector.  Divide the current you need through the collector by the <em>Hfe</em> (typically 100 for a small signal transistor) and that is the <em>minimum</em> current you need to allow through the base. Subtract a silicon junction voltage (0.6V) from the voltage of the IO pin (5V). Divide it by your base current calculated above, and that is the <em>maximum</em> resistor you can use on the base. </p>

<p>And also your transistor symbol in your schematic is backwards.</p>
