Why not use a diode instead of a resistor for current sense circuits?Why use an input resistor in this current sense circuit?Stepper current measurement puzzleLM311 Current Sensing SwitchCurrent sensing in H bridge for Stall detection-Purpose of DiodeMeasuring current - shunt resistor or current transducer or something elseA question on current sensing through a very low resistorSimple Op-Amp setup for current sense has a strange working rangeCurrent drawn by a motor driverCurrent sense amplifier + op-amp buffer + ADC: Measuring down to 0 with a single supplyWhat is the predictable difference of the voltage drop between the V+ and V- lines on a shunt resistor?

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Why not use a diode instead of a resistor for current sense circuits?


Why use an input resistor in this current sense circuit?Stepper current measurement puzzleLM311 Current Sensing SwitchCurrent sensing in H bridge for Stall detection-Purpose of DiodeMeasuring current - shunt resistor or current transducer or something elseA question on current sensing through a very low resistorSimple Op-Amp setup for current sense has a strange working rangeCurrent drawn by a motor driverCurrent sense amplifier + op-amp buffer + ADC: Measuring down to 0 with a single supplyWhat is the predictable difference of the voltage drop between the V+ and V- lines on a shunt resistor?






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;








1












$begingroup$


We're building a motor-drive circuit using an L6206 dual H-bridge circuit. The bottom side of its H drivers come out of a pin called $V_sense$, intended to be connected to ground or to a low-side current-sense circuit. I'm expecting to drive motors with PWM up to an amp or two.



I want a logic sense line to indicate whether there's any current flowing (say $>10mA$), or not. It's essentially detecting the operating presence of the (removable) motor.



I think I can do this by comparing the voltage drop across a Schottky diode with an approx $0.25V$ reference. I'm thinking of using a divider to create a weak $0.25V$ voltage source from my $5V$ power supply, a $2A$ Schottky diode (eg SB240, $V_f=0.5V$ at $2A$) with a small $25Omega$ bypass resistor to eat any leakage, and a comparator (eg LM324 op-amp, or LM339) to create the $5V$ logic result.





schematic





simulate this circuit – Schematic created using CircuitLab



Wherever I have looked up designing current sense circuits, there is always a low-value power resistor to create the voltage drop. I haven't found any that use a diode. I think that a Schottky diode is perfect: its voltage drop will clamp off nicely so it won't waste heat, it's cheap and simple.



Perhaps everyone else is interested in measuring the current level, where I just want to know if there is a current at all? Or, they only care about being close to a maximum current, where I want a minimum?



Why don't I see any diode-based current sensors?



Note: my logic output essentially goes into a latch, so I don't care that the signal will flap around with the PWM.










share|improve this question











$endgroup$







  • 5




    $begingroup$
    It’ll certainly waste more heat than a low impedance resistive current shunt and amplifier and, it won’t be linear. Sounds a bad idea to me.
    $endgroup$
    – Andy aka
    8 hours ago










  • $begingroup$
    You could reduce the bypass resistor from 25 to about 2 ohms (using a cheap comparator whose offset voltage is 10mV, sensing your 10mA decision point). Yes, the diode will clamp on big currents - its a design decision: is diode cost worth the efficiency increase?
    $endgroup$
    – glen_geek
    7 hours ago






  • 1




    $begingroup$
    @Andyaka Ok, I think I don't want it to be linear - I want it to snap over at around 10mA (give or take) and stay there all the way up. If I use a little $0.2Omega$, it means I need to detect a 2mV gap for my 10mA trigger: might that be quite challenging with noisy power PWM and motors just next door? and it still drops $800mW$ at the max 2A, about same as the diode. I can make the resistor lower, but the sensitivity requirements get harder...
    $endgroup$
    – SusanW
    7 hours ago


















1












$begingroup$


We're building a motor-drive circuit using an L6206 dual H-bridge circuit. The bottom side of its H drivers come out of a pin called $V_sense$, intended to be connected to ground or to a low-side current-sense circuit. I'm expecting to drive motors with PWM up to an amp or two.



I want a logic sense line to indicate whether there's any current flowing (say $>10mA$), or not. It's essentially detecting the operating presence of the (removable) motor.



I think I can do this by comparing the voltage drop across a Schottky diode with an approx $0.25V$ reference. I'm thinking of using a divider to create a weak $0.25V$ voltage source from my $5V$ power supply, a $2A$ Schottky diode (eg SB240, $V_f=0.5V$ at $2A$) with a small $25Omega$ bypass resistor to eat any leakage, and a comparator (eg LM324 op-amp, or LM339) to create the $5V$ logic result.





schematic





simulate this circuit – Schematic created using CircuitLab



Wherever I have looked up designing current sense circuits, there is always a low-value power resistor to create the voltage drop. I haven't found any that use a diode. I think that a Schottky diode is perfect: its voltage drop will clamp off nicely so it won't waste heat, it's cheap and simple.



Perhaps everyone else is interested in measuring the current level, where I just want to know if there is a current at all? Or, they only care about being close to a maximum current, where I want a minimum?



Why don't I see any diode-based current sensors?



Note: my logic output essentially goes into a latch, so I don't care that the signal will flap around with the PWM.










share|improve this question











$endgroup$







  • 5




    $begingroup$
    It’ll certainly waste more heat than a low impedance resistive current shunt and amplifier and, it won’t be linear. Sounds a bad idea to me.
    $endgroup$
    – Andy aka
    8 hours ago










  • $begingroup$
    You could reduce the bypass resistor from 25 to about 2 ohms (using a cheap comparator whose offset voltage is 10mV, sensing your 10mA decision point). Yes, the diode will clamp on big currents - its a design decision: is diode cost worth the efficiency increase?
    $endgroup$
    – glen_geek
    7 hours ago






  • 1




    $begingroup$
    @Andyaka Ok, I think I don't want it to be linear - I want it to snap over at around 10mA (give or take) and stay there all the way up. If I use a little $0.2Omega$, it means I need to detect a 2mV gap for my 10mA trigger: might that be quite challenging with noisy power PWM and motors just next door? and it still drops $800mW$ at the max 2A, about same as the diode. I can make the resistor lower, but the sensitivity requirements get harder...
    $endgroup$
    – SusanW
    7 hours ago














1












1








1


1



$begingroup$


We're building a motor-drive circuit using an L6206 dual H-bridge circuit. The bottom side of its H drivers come out of a pin called $V_sense$, intended to be connected to ground or to a low-side current-sense circuit. I'm expecting to drive motors with PWM up to an amp or two.



I want a logic sense line to indicate whether there's any current flowing (say $>10mA$), or not. It's essentially detecting the operating presence of the (removable) motor.



I think I can do this by comparing the voltage drop across a Schottky diode with an approx $0.25V$ reference. I'm thinking of using a divider to create a weak $0.25V$ voltage source from my $5V$ power supply, a $2A$ Schottky diode (eg SB240, $V_f=0.5V$ at $2A$) with a small $25Omega$ bypass resistor to eat any leakage, and a comparator (eg LM324 op-amp, or LM339) to create the $5V$ logic result.





schematic





simulate this circuit – Schematic created using CircuitLab



Wherever I have looked up designing current sense circuits, there is always a low-value power resistor to create the voltage drop. I haven't found any that use a diode. I think that a Schottky diode is perfect: its voltage drop will clamp off nicely so it won't waste heat, it's cheap and simple.



Perhaps everyone else is interested in measuring the current level, where I just want to know if there is a current at all? Or, they only care about being close to a maximum current, where I want a minimum?



Why don't I see any diode-based current sensors?



Note: my logic output essentially goes into a latch, so I don't care that the signal will flap around with the PWM.










share|improve this question











$endgroup$




We're building a motor-drive circuit using an L6206 dual H-bridge circuit. The bottom side of its H drivers come out of a pin called $V_sense$, intended to be connected to ground or to a low-side current-sense circuit. I'm expecting to drive motors with PWM up to an amp or two.



I want a logic sense line to indicate whether there's any current flowing (say $>10mA$), or not. It's essentially detecting the operating presence of the (removable) motor.



I think I can do this by comparing the voltage drop across a Schottky diode with an approx $0.25V$ reference. I'm thinking of using a divider to create a weak $0.25V$ voltage source from my $5V$ power supply, a $2A$ Schottky diode (eg SB240, $V_f=0.5V$ at $2A$) with a small $25Omega$ bypass resistor to eat any leakage, and a comparator (eg LM324 op-amp, or LM339) to create the $5V$ logic result.





schematic





simulate this circuit – Schematic created using CircuitLab



Wherever I have looked up designing current sense circuits, there is always a low-value power resistor to create the voltage drop. I haven't found any that use a diode. I think that a Schottky diode is perfect: its voltage drop will clamp off nicely so it won't waste heat, it's cheap and simple.



Perhaps everyone else is interested in measuring the current level, where I just want to know if there is a current at all? Or, they only care about being close to a maximum current, where I want a minimum?



Why don't I see any diode-based current sensors?



Note: my logic output essentially goes into a latch, so I don't care that the signal will flap around with the PWM.







current-measurement






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 7 hours ago







SusanW

















asked 8 hours ago









SusanWSusanW

15510 bronze badges




15510 bronze badges







  • 5




    $begingroup$
    It’ll certainly waste more heat than a low impedance resistive current shunt and amplifier and, it won’t be linear. Sounds a bad idea to me.
    $endgroup$
    – Andy aka
    8 hours ago










  • $begingroup$
    You could reduce the bypass resistor from 25 to about 2 ohms (using a cheap comparator whose offset voltage is 10mV, sensing your 10mA decision point). Yes, the diode will clamp on big currents - its a design decision: is diode cost worth the efficiency increase?
    $endgroup$
    – glen_geek
    7 hours ago






  • 1




    $begingroup$
    @Andyaka Ok, I think I don't want it to be linear - I want it to snap over at around 10mA (give or take) and stay there all the way up. If I use a little $0.2Omega$, it means I need to detect a 2mV gap for my 10mA trigger: might that be quite challenging with noisy power PWM and motors just next door? and it still drops $800mW$ at the max 2A, about same as the diode. I can make the resistor lower, but the sensitivity requirements get harder...
    $endgroup$
    – SusanW
    7 hours ago













  • 5




    $begingroup$
    It’ll certainly waste more heat than a low impedance resistive current shunt and amplifier and, it won’t be linear. Sounds a bad idea to me.
    $endgroup$
    – Andy aka
    8 hours ago










  • $begingroup$
    You could reduce the bypass resistor from 25 to about 2 ohms (using a cheap comparator whose offset voltage is 10mV, sensing your 10mA decision point). Yes, the diode will clamp on big currents - its a design decision: is diode cost worth the efficiency increase?
    $endgroup$
    – glen_geek
    7 hours ago






  • 1




    $begingroup$
    @Andyaka Ok, I think I don't want it to be linear - I want it to snap over at around 10mA (give or take) and stay there all the way up. If I use a little $0.2Omega$, it means I need to detect a 2mV gap for my 10mA trigger: might that be quite challenging with noisy power PWM and motors just next door? and it still drops $800mW$ at the max 2A, about same as the diode. I can make the resistor lower, but the sensitivity requirements get harder...
    $endgroup$
    – SusanW
    7 hours ago








5




5




$begingroup$
It’ll certainly waste more heat than a low impedance resistive current shunt and amplifier and, it won’t be linear. Sounds a bad idea to me.
$endgroup$
– Andy aka
8 hours ago




$begingroup$
It’ll certainly waste more heat than a low impedance resistive current shunt and amplifier and, it won’t be linear. Sounds a bad idea to me.
$endgroup$
– Andy aka
8 hours ago












$begingroup$
You could reduce the bypass resistor from 25 to about 2 ohms (using a cheap comparator whose offset voltage is 10mV, sensing your 10mA decision point). Yes, the diode will clamp on big currents - its a design decision: is diode cost worth the efficiency increase?
$endgroup$
– glen_geek
7 hours ago




$begingroup$
You could reduce the bypass resistor from 25 to about 2 ohms (using a cheap comparator whose offset voltage is 10mV, sensing your 10mA decision point). Yes, the diode will clamp on big currents - its a design decision: is diode cost worth the efficiency increase?
$endgroup$
– glen_geek
7 hours ago




1




1




$begingroup$
@Andyaka Ok, I think I don't want it to be linear - I want it to snap over at around 10mA (give or take) and stay there all the way up. If I use a little $0.2Omega$, it means I need to detect a 2mV gap for my 10mA trigger: might that be quite challenging with noisy power PWM and motors just next door? and it still drops $800mW$ at the max 2A, about same as the diode. I can make the resistor lower, but the sensitivity requirements get harder...
$endgroup$
– SusanW
7 hours ago





$begingroup$
@Andyaka Ok, I think I don't want it to be linear - I want it to snap over at around 10mA (give or take) and stay there all the way up. If I use a little $0.2Omega$, it means I need to detect a 2mV gap for my 10mA trigger: might that be quite challenging with noisy power PWM and motors just next door? and it still drops $800mW$ at the max 2A, about same as the diode. I can make the resistor lower, but the sensitivity requirements get harder...
$endgroup$
– SusanW
7 hours ago











2 Answers
2






active

oldest

votes


















2












$begingroup$

You can choose the resistor so that it consumes the same power, or less if you want to. At 2 A the diode will consume 1 W, so "equal" resistor would be 0.25 Ω. At smaller current, say 1 A, the diode will consume roughly 0.5 W because the voltage doesn't drop that much. There the resistor only consumes 0.25 W, which is significantly less.



The diode is not accurate. Its $V_f$ varies especially with temperature. Junction to ambient thermal resistance is 45°C so you can expect that much heating with the maximum current. And if you have some heating from outside (direct sunlight for example) you can easily reach 100°C temperature for the junction which drops $V_f$ to half from the original. A decent resistor drifts something like 50 ppm/°C, so at temperature rise from 25°C to 100°C the resistance changes only 0.375%



With a resistor you do need an amplifier, because at low current the voltage is also low. But there are circuits made for just that, so implementation is not difficult. Search for e.g. current sense amplifier. Some of those even have a built in comparator so the number of components might even be the same as with the diode + comparator solution. And while you are adding an amplifier, you can drop the resistor size, and get away with a lot less power loss than with the diode.



So the answer to your question is, diode based solutions are not accurate, and consumer more power compared to shunt resistor solutions.






share|improve this answer











$endgroup$












  • $begingroup$
    Your point about the temperature is excellent - I'd thought about whether the heat was excessive but I'd carelessly assumed the effect on $V_f$ to be negligible. And I notice that some current sense amps discuss PWM immunity - interesting. Ok, brilliant - thanks, good answer. I'll leave it a couple of days to give others a voice, but ... yep, thanks :-)
    $endgroup$
    – SusanW
    6 hours ago


















1












$begingroup$

Change the idea. You can insert an active low current pulse circuit to detect if there's a load connected. If the voltage depends enough on the pulses, there's no load.






share|improve this answer









$endgroup$















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    2 Answers
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    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    You can choose the resistor so that it consumes the same power, or less if you want to. At 2 A the diode will consume 1 W, so "equal" resistor would be 0.25 Ω. At smaller current, say 1 A, the diode will consume roughly 0.5 W because the voltage doesn't drop that much. There the resistor only consumes 0.25 W, which is significantly less.



    The diode is not accurate. Its $V_f$ varies especially with temperature. Junction to ambient thermal resistance is 45°C so you can expect that much heating with the maximum current. And if you have some heating from outside (direct sunlight for example) you can easily reach 100°C temperature for the junction which drops $V_f$ to half from the original. A decent resistor drifts something like 50 ppm/°C, so at temperature rise from 25°C to 100°C the resistance changes only 0.375%



    With a resistor you do need an amplifier, because at low current the voltage is also low. But there are circuits made for just that, so implementation is not difficult. Search for e.g. current sense amplifier. Some of those even have a built in comparator so the number of components might even be the same as with the diode + comparator solution. And while you are adding an amplifier, you can drop the resistor size, and get away with a lot less power loss than with the diode.



    So the answer to your question is, diode based solutions are not accurate, and consumer more power compared to shunt resistor solutions.






    share|improve this answer











    $endgroup$












    • $begingroup$
      Your point about the temperature is excellent - I'd thought about whether the heat was excessive but I'd carelessly assumed the effect on $V_f$ to be negligible. And I notice that some current sense amps discuss PWM immunity - interesting. Ok, brilliant - thanks, good answer. I'll leave it a couple of days to give others a voice, but ... yep, thanks :-)
      $endgroup$
      – SusanW
      6 hours ago















    2












    $begingroup$

    You can choose the resistor so that it consumes the same power, or less if you want to. At 2 A the diode will consume 1 W, so "equal" resistor would be 0.25 Ω. At smaller current, say 1 A, the diode will consume roughly 0.5 W because the voltage doesn't drop that much. There the resistor only consumes 0.25 W, which is significantly less.



    The diode is not accurate. Its $V_f$ varies especially with temperature. Junction to ambient thermal resistance is 45°C so you can expect that much heating with the maximum current. And if you have some heating from outside (direct sunlight for example) you can easily reach 100°C temperature for the junction which drops $V_f$ to half from the original. A decent resistor drifts something like 50 ppm/°C, so at temperature rise from 25°C to 100°C the resistance changes only 0.375%



    With a resistor you do need an amplifier, because at low current the voltage is also low. But there are circuits made for just that, so implementation is not difficult. Search for e.g. current sense amplifier. Some of those even have a built in comparator so the number of components might even be the same as with the diode + comparator solution. And while you are adding an amplifier, you can drop the resistor size, and get away with a lot less power loss than with the diode.



    So the answer to your question is, diode based solutions are not accurate, and consumer more power compared to shunt resistor solutions.






    share|improve this answer











    $endgroup$












    • $begingroup$
      Your point about the temperature is excellent - I'd thought about whether the heat was excessive but I'd carelessly assumed the effect on $V_f$ to be negligible. And I notice that some current sense amps discuss PWM immunity - interesting. Ok, brilliant - thanks, good answer. I'll leave it a couple of days to give others a voice, but ... yep, thanks :-)
      $endgroup$
      – SusanW
      6 hours ago













    2












    2








    2





    $begingroup$

    You can choose the resistor so that it consumes the same power, or less if you want to. At 2 A the diode will consume 1 W, so "equal" resistor would be 0.25 Ω. At smaller current, say 1 A, the diode will consume roughly 0.5 W because the voltage doesn't drop that much. There the resistor only consumes 0.25 W, which is significantly less.



    The diode is not accurate. Its $V_f$ varies especially with temperature. Junction to ambient thermal resistance is 45°C so you can expect that much heating with the maximum current. And if you have some heating from outside (direct sunlight for example) you can easily reach 100°C temperature for the junction which drops $V_f$ to half from the original. A decent resistor drifts something like 50 ppm/°C, so at temperature rise from 25°C to 100°C the resistance changes only 0.375%



    With a resistor you do need an amplifier, because at low current the voltage is also low. But there are circuits made for just that, so implementation is not difficult. Search for e.g. current sense amplifier. Some of those even have a built in comparator so the number of components might even be the same as with the diode + comparator solution. And while you are adding an amplifier, you can drop the resistor size, and get away with a lot less power loss than with the diode.



    So the answer to your question is, diode based solutions are not accurate, and consumer more power compared to shunt resistor solutions.






    share|improve this answer











    $endgroup$



    You can choose the resistor so that it consumes the same power, or less if you want to. At 2 A the diode will consume 1 W, so "equal" resistor would be 0.25 Ω. At smaller current, say 1 A, the diode will consume roughly 0.5 W because the voltage doesn't drop that much. There the resistor only consumes 0.25 W, which is significantly less.



    The diode is not accurate. Its $V_f$ varies especially with temperature. Junction to ambient thermal resistance is 45°C so you can expect that much heating with the maximum current. And if you have some heating from outside (direct sunlight for example) you can easily reach 100°C temperature for the junction which drops $V_f$ to half from the original. A decent resistor drifts something like 50 ppm/°C, so at temperature rise from 25°C to 100°C the resistance changes only 0.375%



    With a resistor you do need an amplifier, because at low current the voltage is also low. But there are circuits made for just that, so implementation is not difficult. Search for e.g. current sense amplifier. Some of those even have a built in comparator so the number of components might even be the same as with the diode + comparator solution. And while you are adding an amplifier, you can drop the resistor size, and get away with a lot less power loss than with the diode.



    So the answer to your question is, diode based solutions are not accurate, and consumer more power compared to shunt resistor solutions.







    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited 6 hours ago









    Huisman

    4,3522 gold badges5 silver badges29 bronze badges




    4,3522 gold badges5 silver badges29 bronze badges










    answered 6 hours ago









    TemeVTemeV

    7581 silver badge9 bronze badges




    7581 silver badge9 bronze badges











    • $begingroup$
      Your point about the temperature is excellent - I'd thought about whether the heat was excessive but I'd carelessly assumed the effect on $V_f$ to be negligible. And I notice that some current sense amps discuss PWM immunity - interesting. Ok, brilliant - thanks, good answer. I'll leave it a couple of days to give others a voice, but ... yep, thanks :-)
      $endgroup$
      – SusanW
      6 hours ago
















    • $begingroup$
      Your point about the temperature is excellent - I'd thought about whether the heat was excessive but I'd carelessly assumed the effect on $V_f$ to be negligible. And I notice that some current sense amps discuss PWM immunity - interesting. Ok, brilliant - thanks, good answer. I'll leave it a couple of days to give others a voice, but ... yep, thanks :-)
      $endgroup$
      – SusanW
      6 hours ago















    $begingroup$
    Your point about the temperature is excellent - I'd thought about whether the heat was excessive but I'd carelessly assumed the effect on $V_f$ to be negligible. And I notice that some current sense amps discuss PWM immunity - interesting. Ok, brilliant - thanks, good answer. I'll leave it a couple of days to give others a voice, but ... yep, thanks :-)
    $endgroup$
    – SusanW
    6 hours ago




    $begingroup$
    Your point about the temperature is excellent - I'd thought about whether the heat was excessive but I'd carelessly assumed the effect on $V_f$ to be negligible. And I notice that some current sense amps discuss PWM immunity - interesting. Ok, brilliant - thanks, good answer. I'll leave it a couple of days to give others a voice, but ... yep, thanks :-)
    $endgroup$
    – SusanW
    6 hours ago













    1












    $begingroup$

    Change the idea. You can insert an active low current pulse circuit to detect if there's a load connected. If the voltage depends enough on the pulses, there's no load.






    share|improve this answer









    $endgroup$

















      1












      $begingroup$

      Change the idea. You can insert an active low current pulse circuit to detect if there's a load connected. If the voltage depends enough on the pulses, there's no load.






      share|improve this answer









      $endgroup$















        1












        1








        1





        $begingroup$

        Change the idea. You can insert an active low current pulse circuit to detect if there's a load connected. If the voltage depends enough on the pulses, there's no load.






        share|improve this answer









        $endgroup$



        Change the idea. You can insert an active low current pulse circuit to detect if there's a load connected. If the voltage depends enough on the pulses, there's no load.







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        answered 4 hours ago









        user287001user287001

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