Here is a slightly different idea. The inductors are place in a asymmetric bridge. This allows the energy in the inductor to be recovered and returned to the supply. This allows testing of high current inductors with a small bench supply. The bench supply only has to supply the circuit losses.
Of course, one could make much simpler circuit by leaving just the bottom mosfet and putting the sense resistor between mosfet source and ground. The inductor under test would then lie between positive supply and the mosfet drain. The main disadvantage would then be that the fet gate charging spike would also be measured. But then one could put a diode from mosfet drain to the cap bank, to recuperate the stored energy back to use (think it as a boost converter boosting its own supply voltage!). Also, the gate driver could be simpler then. With this configuration, one could use more elaborate high-side current sense amplifier, to avoid measuring mosfet gate current.
Hi,
The circuit that Jahonen proposed looks like this:
This circuit does not recover the energy stored in the inductor. The energy is dissipated in the winding resistance of the inductor and the diode.
The duty-cycle should be low, to allow the inductor current to reset and control dissipation in the diode.
The MOSFET driver required is a little bit different than a normal half-bridge driver.
For my circuit to work properly, both of the MOSFETs in the bridge have to be ON at the same time. Normally you would not want this condition and the circuit would have dead time between the top and bottom MOSFETs.
The other issue with MOSFET drivers like the IR2011 is that they require the circuit to be switching to provide a charge pump action to drive the top MOSFET. In this application, it is desireable to be able to do a single pulse or low repetition frequency.
I chose a P Channel MOSFET for the top MOSFET. This allows me to drive it with the level shifter M3 and Q1. As originally drawn there are some limitations. The supply voltage must be high enough to turn on the MOSFETs but less than Vgs max (typically 20V).
I deliberately chosen not to drive the MOSFETs very hard, that is why I have 22 ohm series resistor on their gates.
A 555 will provide enough gate drive for this application.
I would also recommend that you place a 51 ohm resistor in series with the output from the sense resistor. Make the connection to the scope with 50 Ohm coax and terminate the scope input with a 50 Ohm resistor.
This will divide the signal amplitude by 2, but you will get much cleaner waveforms.
IR2011 does not have any cross-conduction protection or integrated dead time so you can actually turn both outputs on simultaneously, and it should work fine as posted by BravoV. Only thing is that it needs a path to charge the bootstrap capacitor. Otherwise, the high side driver fails to turn on due to UVLO. Also, if you don't use four mosfets (i.e. full bridge configuration), and two IR2011s, then make sure that the diodes used for energy return are beefy enough, they should be able to withstand the maximum current in the inductor.
I am trying to use generic parts, that people should have in their junk boxes or that easy to get.
I am going to try an build the circuit over the weekend.
Like this ?
Ok, I'm a bit confused here, especially the decaying route pointed by the dotted arrow, once the energy in the inductor depleted, isn't that same path will charge the inductor again if we don't control the timing precisely ?
Also about the charge for the bootstrap capacitor, on my previous topology using diode for the decaying route instead of mosfets, any idea how to solve it without getting the circuit grows overly complicated ?
In this first picture I have 'zoomed out' to show the more of the cycle. The upper trace is the output from J1. The sharp edge is caused when the MOSFET Q2 is turned off, it is no longer possible to measure the inductor current.
The voltage across the inductor is the difference between the TP3 and TP4 waveforms. Again, this shows how the energy is recovered.
The current draw from the 10V power supply from this test was 0.058A, which is 0.58W and this includes all the power consumption in the gate drive circuitry.
During the decaying period, once the inductor's voltage (minus diodes drop out) is equal to power supply rail, the inductor will self dissipate the rest of it's own excess energy through the ringing as observed at the scope shot, am I correct ?
Just curious and purely academical here, curious in ball park number on how much percentage of the energy left in percentage of total only at that ringing period ?
Still learning, hope you're not bored to death entertaining my noobness.
I'm aware of the energy required to charge those big caps when the circuit was turned on for the 1st time, but lets ignore this initial state, from this result that this circuit draws only 58 ma at 10 volt is pretty low isn't it ? This is valuable info, so I don't need a beefy power supply.
Varying the Supply voltage.
As I have designed the circuit, it is all dc coupled. I also designed the circuit using commonly available parts.
I have given some thought to using transformer coupled gate drive. This means using an non-junk box part, I could probably find some thing on Digikey. If I did this it would be really easy to separate the supplies.
Stay Tuned.......