The load is a vehicle damper. I'm assuming it is inductive. My problem is, if I replace the load with a 4R4 load resistor, I can measure the current going through the resistor (placed in series with it) absolutely fine. My power supply shows around the same current that my current meter is displaying. If I attach the damper load to it instead of the resistive load, my current measurement is showing 1.8A but the power supply source is showing ~1A current being used. As I cannot possibly have more current going through the load than I am supplying, there appears to be an issue! I have verified my current measurement meter by using 2 different types of meter. The power supply display is also correct. The PWM signal when using both types of load is also identical. Any suggestions would be most welcome !
Yes, you can. Inductors store energy in their magnetic field, which is later returned, in this case to flow through D8. With a resistive load all the power supplied is converted to heat, with an inductive load it is stored then returned, flows in D8 and the resistance of the inductor and eventually ends up as heat. It does not surprise me that you measure more current as you are kind of measuring the same current twice.
I was hoping that D8 would effectively discharge the energy from the magnetic field each time the mosfet is 'high'. Is there a solution to this or should I look at faster reacting flyback diodes? Thanks
OK, so you are measuring the current into the inductor when the MOSFET is conducting and you are measuring the current into D7 D8 from the inductor when the MOSFET is not conducting. Both these currents will be in the same direction so will be added together.
I don't know what you mean by 'improve'. As far as I can tell the circuit is behaving as expected.
And so it should be if the power supply is any good!
I think the problem is you are not seeing what you thought you would see and you are having trouble getting your head around it. I think what you are seeing is exactly what you should see.
This is perfectly normal in a switched inductive circuit - currents circulate round the diode and the load that are in addition to the power supply current - the load sees both currents and is thus larger (on average, as seen by a meter).
The principle of operation of buck and boost converters relies on this fact.
I'm only interested in the current being used by the inductor. If I remove D7 it should reduce the current being measured? I don't understand why current will flow through D7 when the MOSFET is not conducting unless the anode is more positive than the cathode, which it can't be as the anode is grounded?
I have also attached a current shunt from the MOSFET source to ground. The voltage drop across the current shunt is fed into an Arduino. I'm using a display to show the current. This current is very close to what the power supply is displaying.
Because its an inductor. Inductors actively resist changes in current so when the MOSFET turns off the current in the inductor keeps flowing but through the diode instead, gradually falling due to losses until it decays away to nothing or the MOSFET turns on again.
If you switch off current to an inductor without providing an alternative path for the current the inductor will generate whatever voltage is necessary to force the current to stay flowing, perhaps thousands of volts - which is why the diode is needed.
Ok, I have a 3A 75mV current shunt going from ground to source. Why would this current be matching the current provided by the power supply (1A), but the current being read by my fluke is 1.8A at A1- to the inductor negative? I think they should be matching?
No...
The source - ground current is the supply current, the current at A1 is the supply current plus the current from the inductor when the MOSFET is off.
Have you taken into account that my original reply #8 had a typo so was misleading? I believe it's correct now.
We've answered this multiple times now - the average current through the inductor is not the same as the average current through the MOSFET. The peak currents are the same, but the MOSFET is off some of the time so its average current is less. The inductor current may never be falling to zero for instance.
The average current through the inductor is the sum of the average currents through the MOSFET and through the diode.
This is how buck converters can have output current higher than input current.
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