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user319836
user319836
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To help you understand let me go through some MOSFET specs.

  1. You found \$R_{DS}(on)\$ in the datasheet. Just below that line is \$V_{GS}(th)\$ which has a minimum of 2V and a maximum of 4V. Therefore the voltage drop from gate to source must be greater than 4 volts to start conduction. There is no control over this value, when you buy a FET then \$V_{GS}\$ will be somewhere between those values. Based on your point 1, \$V_{GS} = 2\$V the minimum.
  2. If you look closely in the datasheet for \$R_{DS}(on)\$, you will see the conditions at which it was measured:\$V_{GS}=10\$V and \$I_{D}=28\$A. The low \$R_{DS}(on)\$ occurs only at high gate-source voltages.
  3. Looking at the chart you supplied, the solid lines represent a specific gate-source voltage. The lowest is 4.5V, the highest is 15 V. If we follow a vertical path along, say, the 1 volt \$V_{DS}\$ path, then the drain current increases as the gate-source voltage increases demonstrating that the drain-source resistance \$R_{DS}\$ decreases with increasing \$V_{GS}\$.

The transistor gets hot because \$R_{DS}(on)\$ is not small for the values of \$V_{GS}\$ that you are using. So yes, you could put heatsinks on and/or parallel several FETS to keep the temperature low.

If you adjust the potentiometer so there is 5V across the motor, the voltage from gate to ground must be 7V to 9V, \$V_{motor}+V_{GS}(th)\$. The 16W from your point 2 is the power dissipated by the motor. To find the power dissipated by the transistor (as heat) we need to know the voltage drop across it, namely $$V_{DS} = 12 - 10 = 2$$V Then calculate the FET power dissipation.$$P_{DS} = (20)(1.6) = 3.2W$$

Another way to explore the FET drive is putting the motor in the drain circuit as shown below.

EDIT: The 47k resistor will reduce the sensitivity of the potentiometer. It will also reduce the maximum voltage that you can apply. Experiment with different values. END EDIT

schematicschematic

simulate this circuitsimulate this circuit – Schematic created using CircuitLab

Cheers for exploring.

These are great circuits for introductory exploration of FETs. Except for on-off control they are not contenders for serious work. Too much power loss. A better way for speed control is though pulse width modulation (pwm). The complexity is greater but power efficiency is nearly perfect.

To help you understand let me go through some MOSFET specs.

  1. You found \$R_{DS}(on)\$ in the datasheet. Just below that line is \$V_{GS}(th)\$ which has a minimum of 2V and a maximum of 4V. Therefore the voltage drop from gate to source must be greater than 4 volts to start conduction. There is no control over this value, when you buy a FET then \$V_{GS}\$ will be somewhere between those values. Based on your point 1, \$V_{GS} = 2\$V the minimum.
  2. If you look closely in the datasheet for \$R_{DS}(on)\$, you will see the conditions at which it was measured:\$V_{GS}=10\$V and \$I_{D}=28\$A. The low \$R_{DS}(on)\$ occurs only at high gate-source voltages.
  3. Looking at the chart you supplied, the solid lines represent a specific gate-source voltage. The lowest is 4.5V, the highest is 15 V. If we follow a vertical path along, say, the 1 volt \$V_{DS}\$ path, then the drain current increases as the gate-source voltage increases demonstrating that the drain-source resistance \$R_{DS}\$ decreases with increasing \$V_{GS}\$.

The transistor gets hot because \$R_{DS}(on)\$ is not small for the values of \$V_{GS}\$ that you are using. So yes, you could put heatsinks on and/or parallel several FETS to keep the temperature low.

If you adjust the potentiometer so there is 5V across the motor, the voltage from gate to ground must be 7V to 9V, \$V_{motor}+V_{GS}(th)\$. The 16W from your point 2 is the power dissipated by the motor. To find the power dissipated by the transistor (as heat) we need to know the voltage drop across it, namely $$V_{DS} = 12 - 10 = 2$$V Then calculate the FET power dissipation.$$P_{DS} = (20)(1.6) = 3.2W$$

Another way to explore the FET drive is putting the motor in the drain circuit as shown below.

schematic

simulate this circuit – Schematic created using CircuitLab

Cheers for exploring.

These are great circuits for introductory exploration of FETs. Except for on-off control they are not contenders for serious work. Too much power loss. A better way for speed control is though pulse width modulation (pwm). The complexity is greater but power efficiency is nearly perfect.

To help you understand let me go through some MOSFET specs.

  1. You found \$R_{DS}(on)\$ in the datasheet. Just below that line is \$V_{GS}(th)\$ which has a minimum of 2V and a maximum of 4V. Therefore the voltage drop from gate to source must be greater than 4 volts to start conduction. There is no control over this value, when you buy a FET then \$V_{GS}\$ will be somewhere between those values. Based on your point 1, \$V_{GS} = 2\$V the minimum.
  2. If you look closely in the datasheet for \$R_{DS}(on)\$, you will see the conditions at which it was measured:\$V_{GS}=10\$V and \$I_{D}=28\$A. The low \$R_{DS}(on)\$ occurs only at high gate-source voltages.
  3. Looking at the chart you supplied, the solid lines represent a specific gate-source voltage. The lowest is 4.5V, the highest is 15 V. If we follow a vertical path along, say, the 1 volt \$V_{DS}\$ path, then the drain current increases as the gate-source voltage increases demonstrating that the drain-source resistance \$R_{DS}\$ decreases with increasing \$V_{GS}\$.

The transistor gets hot because \$R_{DS}(on)\$ is not small for the values of \$V_{GS}\$ that you are using. So yes, you could put heatsinks on and/or parallel several FETS to keep the temperature low.

If you adjust the potentiometer so there is 5V across the motor, the voltage from gate to ground must be 7V to 9V, \$V_{motor}+V_{GS}(th)\$. The 16W from your point 2 is the power dissipated by the motor. To find the power dissipated by the transistor (as heat) we need to know the voltage drop across it, namely $$V_{DS} = 12 - 10 = 2$$V Then calculate the FET power dissipation.$$P_{DS} = (20)(1.6) = 3.2W$$

Another way to explore the FET drive is putting the motor in the drain circuit as shown below.

EDIT: The 47k resistor will reduce the sensitivity of the potentiometer. It will also reduce the maximum voltage that you can apply. Experiment with different values. END EDIT

schematic

simulate this circuit – Schematic created using CircuitLab

Cheers for exploring.

These are great circuits for introductory exploration of FETs. Except for on-off control they are not contenders for serious work. Too much power loss. A better way for speed control is though pulse width modulation (pwm). The complexity is greater but power efficiency is nearly perfect.

Source Link
user319836
user319836
  • 17.1k
  • 3
  • 12
  • 43

To help you understand let me go through some MOSFET specs.

  1. You found \$R_{DS}(on)\$ in the datasheet. Just below that line is \$V_{GS}(th)\$ which has a minimum of 2V and a maximum of 4V. Therefore the voltage drop from gate to source must be greater than 4 volts to start conduction. There is no control over this value, when you buy a FET then \$V_{GS}\$ will be somewhere between those values. Based on your point 1, \$V_{GS} = 2\$V the minimum.
  2. If you look closely in the datasheet for \$R_{DS}(on)\$, you will see the conditions at which it was measured:\$V_{GS}=10\$V and \$I_{D}=28\$A. The low \$R_{DS}(on)\$ occurs only at high gate-source voltages.
  3. Looking at the chart you supplied, the solid lines represent a specific gate-source voltage. The lowest is 4.5V, the highest is 15 V. If we follow a vertical path along, say, the 1 volt \$V_{DS}\$ path, then the drain current increases as the gate-source voltage increases demonstrating that the drain-source resistance \$R_{DS}\$ decreases with increasing \$V_{GS}\$.

The transistor gets hot because \$R_{DS}(on)\$ is not small for the values of \$V_{GS}\$ that you are using. So yes, you could put heatsinks on and/or parallel several FETS to keep the temperature low.

If you adjust the potentiometer so there is 5V across the motor, the voltage from gate to ground must be 7V to 9V, \$V_{motor}+V_{GS}(th)\$. The 16W from your point 2 is the power dissipated by the motor. To find the power dissipated by the transistor (as heat) we need to know the voltage drop across it, namely $$V_{DS} = 12 - 10 = 2$$V Then calculate the FET power dissipation.$$P_{DS} = (20)(1.6) = 3.2W$$

Another way to explore the FET drive is putting the motor in the drain circuit as shown below.

schematic

simulate this circuit – Schematic created using CircuitLab

Cheers for exploring.

These are great circuits for introductory exploration of FETs. Except for on-off control they are not contenders for serious work. Too much power loss. A better way for speed control is though pulse width modulation (pwm). The complexity is greater but power efficiency is nearly perfect.