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I am just confused about understanding a simple concept.

For the following circuit:

enter image description here

Say the blue line acts as a short for whatever reason. If the LED has a forward voltage drop of 23 V and forward current of 30mA (LED contains everything it needs for this example (as in resistor is inside just for this example))

If we don't have a resistor and just a wire connecting the MCU to the mosfet, when the short occurs will the current still remain at 30 mA? As in the current would not become infinite/very large? Essentially nothing will change except the current will be continuously entering the MCU pin and could potentially blow teh MCU if the 30mA exceeded its rating?

In short: Essentially I just want to know that if a short occurred over the mosfet and say the MCU pin was low for that state, will the current remain the same at 30mA due to the limiting resistor of the LED. As in that's all that would happen (and obviously that the MCU will always have current going into it)

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  • \$\begingroup\$ Sorry the LED contains the resistor inside (but yeah I understand it needs the Resistor) \$\endgroup\$ – Student May 8 at 13:45
  • \$\begingroup\$ The LED has a forward voltage of 24V? You'll need more than 24V supply then. It is only 30mA of current if the current limiting resistor for it is selected to give 30mA. You need to add a resistor to it otherwise the LED will likely be ruined. What is the voltage from the MCU? Once you have these factors, redraw the circuit without the MOSFET and with a voltage source instead of the MCU. You should be able to solve it from there. \$\endgroup\$ – MCG May 8 at 13:45
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    \$\begingroup\$ Not drawing the resistor is confusing! Just include it in the schematic please. \$\endgroup\$ – Bimpelrekkie May 8 at 13:46
  • \$\begingroup\$ @Student if the resistor is internal, draw it as an external resistor in your schematic \$\endgroup\$ – MCG May 8 at 13:46
  • \$\begingroup\$ @MCG Essentially i just want to know that if a short occurred over the mosfet and say the MCU pin was low for that state, will the current remain the same at 30mA due to the limiting resistor of the LED. As in thats all that would happen (and obviously that the MCU will always have current going into it) \$\endgroup\$ – Student May 8 at 13:49
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With the shorts in place this is likely to happen:

I'm assuming that the MCU does not run on 24 V but on a different supply of 5 V.

I'm also assuming that the MOSFET is broken and that it behaves as an "open" on all pins.

Then the output of the MCU will be "pulled up" by the LED + resistor + shorts.

That will forward bias one of the ESD diodes inside the MCU.

These ESD diodes are present in practically any IC.

This is then the situation:

schematic

simulate this circuit – Schematic created using CircuitLab

This assumes that the 5 V supply is such that it stays at 5 V.

It is possible that the 5 V supply cannot "absorb" the extra 24 mA and will increase in value. For example to 10 V. That could potentionally damage your MCU!

If the MOSFET is not broken then it will act as a "MOSFET diode" and limit the voltage at the drain to a few volts (slightly more than the Vt of the MOSFET).

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  • \$\begingroup\$ Ok thanks, just a quick question why is it 24 mA? But thanks for the explanation, it makes a lot of sense. \$\endgroup\$ – Student May 8 at 14:02
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    \$\begingroup\$ why ~24 mA? Voltage across R1: 24 V - 3 V (D1) - 0.6 V - 5 V = 15.4 V / 680 ohm = 23 mA. In my 24 mA calculation I forgot about D2. \$\endgroup\$ – Bimpelrekkie May 8 at 14:06
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    \$\begingroup\$ Yes if the MOSFET is still working then it will act like a diode but dropping Vt (couple of volts). So not "straight to ground" but with some voltage across it. \$\endgroup\$ – Bimpelrekkie May 8 at 14:08
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    \$\begingroup\$ I just look at the circuit, analyze it and predict what happens. If you can't follow that then maybe you need more circuit analysis experience. That requires practice and more practice. I've been doing that (analyzing how circuits work) for more than 30 years so no worries if you don't "see" the solution so quickly. \$\endgroup\$ – Bimpelrekkie May 8 at 14:21
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    \$\begingroup\$ @Student I'll try to predict what Bimpelrekkie actually meant. D2 will be forward biased and D3 will be reverse biased. Considering that BUF1 and the rest impedances are much higher than D2. Thus, 99% of the current will flow towards D2. \$\endgroup\$ – Unknown123 May 8 at 15:37
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This is your initial scenario with 30mA LED current. V1 is the MCU output:

enter image description here

And this is if you short those nodes:

enter image description here

So LED current decreases and the MCU out might burn or gets damaged because it gets shorted to ground. But that current passes through MOSFET drain and source also will damage the MOSFET. So eventually the LED will also be off.

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  • \$\begingroup\$ 2287A? I don't think your simulation is providing useful results. \$\endgroup\$ – Elliot Alderson May 8 at 13:59
  • \$\begingroup\$ Yes I didn't add 50 ohm output impedance for MCU just to show the effect. \$\endgroup\$ – user16307 May 8 at 14:01
  • \$\begingroup\$ I am a little confused on why the current goes to ~2.28kA? \$\endgroup\$ – Student May 8 at 14:03
  • \$\begingroup\$ @Student Because the simulation is not realistic. Ignore those results. \$\endgroup\$ – Elliot Alderson May 8 at 14:03
  • \$\begingroup\$ You can add 50 Ohm output resistance in series with V1. \$\endgroup\$ – user16307 May 8 at 14:04

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