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Sorry in advance if I get the terms wrong here.

And I have a potentially (no pun intended) simply stupid question here.

In a MOSFET if I have 12 Volts on Source (the potential that drives the circuit I need to switch).

And I put 5V on the Gate (the pin that switch the MOSFET on) via a 10 Ohm resistor. This pin is on a 100k resistor to ground, so the MOSFET should be off unless signal is on.

The query, isn't there only a 7V potential difference now between the source to drain. OR is it the full 12V that drives the circuit? Since the 5V delivered to gate pin is always there when the MOSFET is switched on.

Is there a difference if it's different sources of the different voltages, or if it's the same voltage source (via voltage limiter/divider). They will have the same ground in both cases.

This particular use case is a RAMPS 1.4 board and the pin D9 (Edit: wrong pin, D8 is the correct one, you should never rely on wet tissue memory for specifics) MOSFET I use to heat the bed. And the driving (gate?) pin can take the 5V to turn the MOSFET on either from USB or from the RAMPS power input (via Arduino voltage limiter).

This question is mainly for my knowledge base and not the cause of the problem I'm facing. I suspect a bad capacitor in the PSU is causing my power issues.

Edit: The MOSFET is IRL3034 Mouser and the load is 1.2 Ohms, And the gate voltage is +5V through a voltage divider at 10 Ohms and 100K to ground. So basically 4.6V. Ramps 1.4 schematic It's gate pin D8, and PS1/PS2 bed outputs, that my question specifically is about. The other outputs are connected through two capacitors in parallel to ground, so that complicates matters a tiny bit.

For my question this is irrelevant, since I am asking if the gate voltage supplied wouldn't interfere with the potential difference, because they share a common ground. If you have 5V on one side of a circuit and 12V on the other side, wouldn't that mean the circuit only have 7 "effective" volts? Just like voltage drops across resistors in series.

Maybe a better question is: if I inject 5V between two resistors (loads) in series that's driven by 12V, would the 5V Volts create a difference in the ampere across the resistor between 12V+ and 5V+? And isn't this analogues to my MOSFET question?

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  • \$\begingroup\$ Which kind of MOSFET? There are several types. \$\endgroup\$ Commented Aug 28 at 7:28
  • \$\begingroup\$ It's the voltage from gate to source that determines how much the MOSFET is turned on, and that can vary a lot between different types of MOSFET and between different MOSFETs of the same type. So we really need to see a schematic diagram (draw on a piece of paper and photograph it if it's easier) and component details to help. \$\endgroup\$ Commented Aug 28 at 8:49
  • \$\begingroup\$ Please specify MOSFET part number. Link to datasheet if you have it. | Please specifi RAMPS bed load resistance and voltage if known and RAMPS manual. ||||. Gate drive supply and MOSFET and RAMPS power supply MUST share a common ground. || When gate of an N Channel MOSFET is driven adquately high relative to ground (5V or more typically) the Drain voltage will drop from about V+ to about ground. MOSFET has resistance Rdson for a given gate voltage and drain voltage = IR = Iload x Rdson. . \$\endgroup\$ Commented Aug 28 at 14:22
  • \$\begingroup\$ It seems the wall of text does not decribe a circuit detailed enough to grasp what is happening. For example, with that N-type FET, it makes no sense to have 12V on Source terminal and 5V on Gate terminal, as you would need something like 14V on Gate to turn it on. Perhaps you are using the FET incorrectly, so please draw a diagram with FET, 5V and 12V batteries as supplies and put resistors as load and gate dividers. It also makes no sense that 10 ohm and 100k divide 5.0V down to 4.6V, as it should be more like 4.9995V. \$\endgroup\$ Commented Aug 28 at 18:43
  • \$\begingroup\$ That's an N-Channel MOSFET. Putting 12 V on the source (with, presumably, the load connected between the drain and ground?) makes no sense as it's (a) backwards and (b) the body diode will power the load all the time regardless of gate voltage. \$\endgroup\$ Commented Aug 28 at 18:45

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So this part of the RAMPS schematic is the actual part you're talking about, and it's the correct way to use an N-Channel MOSFET to switch a load without relying on special drivers or additional supply rails to control the gate. It's known as a low-side switch because it's the low side of the load that is switched rather than the high side:

Portion of RAMPS schematic showing P$1 and P$2 output driver

You seem to be confused about which terminal of the MOSFET is which. The Source pin is the one connected to GND, not to +12 V. The load is connected between +12 V and the Drain pin, so that when the MOSFET turns on it applies 12 V between P$1 and P$2. The terminal names can be a little confusing; "source" refers to the terminal where charge carriers enter the device and "drain" refers to the terminal where they leave it. This is especially annoying as we often tell people to just consider conventional current flow and not worry about electrons or holes.

The MOSFET is controlled by the gate-source voltage, Vgs, which in this case is the same as the gate voltage because the source is grounded. When D8 is set high by the CPU, Vgs is high enough to turn the MOSFET on. A common misunderstanding is that a MOSFET operates like a relay, and turns fully on when the Vgs threshold voltage is reached whereas in reality the threshold voltage is only usually specified for a drain current of about 250 uA. Looking at the datasheet for the IRLB8034 shows you how Vgs affects the relationship between drain current and drain-source voltage; it's important to make sure your MOSFET is operating such that the product of the two (i.e. the power dissipated in the MOSFET) is not going to lead to a thermal problem:

enter image description here

You haven't said what your actual problem is but one thing you need to be very careful with is the wiring to your load. If it has a resistance of 1.2 Ohms it's going to be drawing 10 Amps from your supply, and that can cause issues such as "ground bounce" if you're not careful; this is where the current through the ground wires causes a voltage drop that can be enough to stop the circuit working properly and may show as the load switching rapidly on and off or the processor resetting itself or locking up.

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  • \$\begingroup\$ Thanks for correcting my terminology. I do find the terms contra intuitive. This isn't a problem as much as a question regarding whether the voltage level on D8 would/could produce a difference in the amperage draw over the 12V load, other than a too low voltage not turning it on at all. And ground bounce should be mitigated by R15 if I understand the wikipedia entry correctly. \$\endgroup\$ Commented Aug 28 at 21:31
  • \$\begingroup\$ I know that they are not on the std diagram, but a reverse gate zener would do no harm. Nor would a reverse polarity load diode. I imagine the RAMPS heater is largely resistive, but Murphy loves "is largely". \$\endgroup\$ Commented Aug 29 at 12:57
  • \$\begingroup\$ It is a little confusing; "source" refers to the terminal where charge carriers enter the device and drain refers to the terminal where they leave it. Especially annoying as we often tell people to just consider conventional current flow and not worry about electrons or holes. \$\endgroup\$ Commented Aug 29 at 13:32
  • \$\begingroup\$ I've updated the answer to show the effect of changes in Vgs \$\endgroup\$ Commented Aug 29 at 13:50

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