Is a Diode necessary in this DC Motor Control Circuit

Discussion in 'Analog & Mixed-Signal Design' started by free_state, Aug 26, 2018.

  1. free_state

    Thread Starter New Member

    Aug 26, 2018
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    Hey guys,

    Just wondering if I should put a diode or two in this circuit due to the two POTs. Basically I'm concerned about the position of the second POT (R_foot) having an impact on the motor speed when the first pot (R_dial) is engaged.

    upload_2018-8-26_19-47-6.png
     
  2. Alec_t

    AAC Fanatic!

    Sep 17, 2013
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    Welcome to AAC!
    Either put the motor in the drain path of the NFET or use a PFET instead. As is, the FET needs about 10V between gate and source to switch on fully, which doesn't leave much voltage across the motor. You will need a diode to protect the FET.
     
  3. Audioguru

    Expert

    Dec 20, 2007
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    A resistor in series with a motor reduces its starting power tremendously. You can slowly turn down its speed with your extremely simple circuit (when the motor is in series with the drain of the Mosfet) but it probably will not start running until you turn up the pot.
    Usually a PWM circuit is used to adjust the speed of a DC motor. Each pulse has full power for good starting.
     
  4. Audioguru

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    I forgot to say the switch should be at the gate of the Mosfet to select which pot. Then the pots will be completely separate.
    I said a resistor is in series with the motor. The Mosfet is the resistor.
     
  5. crutschow

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    Mar 14, 2008
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    That will cause a very non-linear motor speed versus pot position.
    You will get no motor voltage until the pot output reaches the MOSFET threshold voltage, and then it will take only a few tenths of a volt more to reach full motor voltage.

    Better would be to use a Sziklai pair in place of the MOSFET.
    That way the maximum motor voltage is only about 1V below the supply voltage.
    For that you should use 10kΩ pot instead of a 100kΩ.
    Below is an LTspice simulation of the basic circuit.

    Note that to avoid interaction between the two pots the switch should be put at the output of the two pots, not the input.

    upload_2018-8-26_8-15-33.png

    But the best is to use a PWM circuit as AG suggested.
     
  6. ScottWang

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    Aug 23, 2012
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    The motor was moved to the Drain and two VR100K will be as in parallel, so moved it to the Gate(or you can in series two diodes for each VR100K), the 1M resistor is added to avoid during the switch is switching from R-Dial to R_Foot still keep a low level and not open state, you can ignore the 1M resistor and then it will be as your circuit when you put the switch on the left side.

    upload_2018-8-26_19-47-6_free_state_ScottWang.png
     
  7. ebp

    Well-Known Member

    Feb 8, 2018
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    Putting the motor in the drain makes the motor vastly less controllable. In the source the voltage would at least be fairly consistent versus the pot setting, though the maximum motor voltage would be around 8 volts. The IRF540 has a minimum transconductance of 8.7 S (spec'd at high current, but that's all we have for a numeric value). The motor requires 0.12 A maximum. This means a delta Vgs of only about 14 mV above the gate-source threshold voltage will change the motor current from zero to maximum. For a typical single-turn pot this would be about 0.3 degrees of rotation with 12 volts on the pot.

    The diodes in the pot circuits contribute nothing.

    A FET is simply not appropriate for open-loop control. Crutschow's circuit is much superior.
     
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  8. ScottWang

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    Using the MOSFET and adjust its Vgs to control the speed, the control voltage range will be too narrow, for a MOSFET that it is use as the ne555 pwm motor controller more easy to control the speed, but I didn't suggest it, because I just follow what the TS did.

    Please recheck the two VR100K from the Vg. of MOSFET.
     
  9. ebp

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    Feb 8, 2018
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    "I just follow what the TS did."

    The TS had the circuit configured as a source follower (common drain). In that configuration the voltage across the motor would be something around 3 to 4 volts less than the gate voltage. That would not be ideal, and I assume not as intended. If the threshold voltage were 4 V it would mean that the "bottom" 90 degrees of pot rotation would be "dead", but the top 180 degrees would give reasonably linear control of the motor voltage. If a logic level FET with exceptionally low threshold voltage were used, performance would be improved but Crutschow's circuit would likely still be superior.

    "Please recheck the two VR100K from the Vg. of MOSFET."

    Yes. And ... ? The original configuration would not work as intended, but the diodes in your version do nothing of any value when the wipers of the pots are switched to the FET's gate. The gate to source resistor is necessary in that configuration to keep the FET off when switching between the pots.
     
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  10. ScottWang

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    You still didn't get why I did it, they were used to avoid the two VR100K in parallel and affecting each other, because the pin2 of two VR100K are connected together and another pin probably pin3 are connected to ground.
     
  11. ebp

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    Feb 8, 2018
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    I see your circuit at #6 and I see absolutely no reason for the diodes. There is nothing whatever wrong with connecting the ends of two pots in parallel if the wipers are not connected together, so no, I don't get why you did it. If the wipers are connected together the diodes still won't do anything of value. If the wipers are connected together then there is no configuration of diodes to the ends of the pots that will prevent their interaction.
     
  12. free_state

    Thread Starter New Member

    Aug 26, 2018
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    Thanks for the replies everyone! I went ahead and built a simple version of the circuit just using one POT and ebp is correct, across the motor I had about 9V (max) and yea the first 90 degrees or so of the pot did nothing. I have taken on board crutschow's circuit although I'm quite new to this so trying to figure out how it works.

    The reason I went with this design is the fact I have to make a number of them, I thought making an analog circuit would certainly be cheaper in the long run than running a digital PWM circuit, the circuit will also see up to 10amps. As I said I'm new to this so I may have been very wrong =/. Thanks again!

    upload_2018-8-27_8-45-33.png
     
  13. MaxHeadRoom

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    Jul 18, 2013
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    Also the motor appears to have a gear box?
    If so, there is a certain amount of torque required to be overcome due to this load before the motor starts to turn.
    Max.
     
  14. free_state

    Thread Starter New Member

    Aug 26, 2018
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    The motor seems to start fine, even at low voltage.
     
  15. MaxHeadRoom

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    Jul 18, 2013
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    Referring to the reason for the comment 'the first 90 degrees or so of the pot did nothing. '!
    Max,
     
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  16. crutschow

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    That significantly changes the design as compared to your circuit schematic, which only shows 120mA
    Not for that current level.
    For that high current you should use PWM, otherwise the maximum transistor dissipation will be about 30W and it would need a large heatsink.
    The PWM circuit will be a lot cheaper than the cost of the heatsink, as well as being a lot more efficient.

    Below is an example, simple PWM circuit using a cheap 555 timer IC.

    Pot U2 controls the PWM duty-cycle and thus the motor speed from <1% minimum to >99% maximum.
    The output is shown for a 10% and 90% pot setting.
    In your circuit you would have two pots, of course.

    The N-MOSFET can be any with a 20V or greater voltage rating, and a 10mΩ or less on-resistance to avoid having to heatsink the transistor.

    Its PWM frequency as shown is about 1kHz, but that can be changed by using a different value for C3, if that frequency makes too much motor noise.

    upload_2018-8-27_0-32-49.png
     
    Last edited: Aug 27, 2018
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  17. Alec_t

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    Sep 17, 2013
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    I agree. The OP's circuit has a number of problems. PWM would be much better.
     
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