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I have an Aglient power supply Model number N5747A, and I want to use its current sinking ability to become an active load.

This is to test a DC/DC converter (5VDC Output), Gaia MGDS-35-H-C, I want to load the output of the DC/DC to 75% load.

Requirement of Electronic Load I am using an active load because I need to vary the load from 1% to 100% in 15 steps. An active load will be smoother and more accurate.

According to the datasheet 75% load is around 5.25A. My idea is to hook up the Agilent power supply and set the current limit to 5.25A on the output of the DC/DC converter.

My question is, what kind of "fixed" resistance would I need, would it be just to protect the Agilent supply and DC/DC converter? I assume I would need some kind of power resistor.

The power supply will create a current loop and force the DC/DC converter to draw 75% of its maximum current. Am I correct in my understanding of this?

Below is a drawing for 100% load (7A) on the output of the DC/DC converter

Quick Drawing

Then the fixed resistance I chose was to maintain a safe operating voltage for the current sink. Example: 60V current source, to maintain 7A's of current through the loop a 4 Ohm resistor will operate the current source in the middle of its operating voltage range of around (28V - 5V )= 23V. If you chose an 8 Ohm resistor for example, then you'll be operating really close to the supply's limits (7A * 8 Ohms = 56V-5=51V). Also the fixed resistor limit in cause the supply goes to 0V (a dead short), so at least 5V/7A = 1 Ohm minimum (rounded up for safety) to put a fixed cap on the current that will flow.

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    \$\begingroup\$ What in the datasheet or manual for the N5747A makes you think it has a "current sinking ability"? I read the datasheet and I didn't see anything to suggest that. But I didn't read the manual. \$\endgroup\$
    – The Photon
    Commented Sep 26, 2013 at 3:50
  • \$\begingroup\$ Instead of the N5747A, have a look at the 6063B "electronic load". \$\endgroup\$
    – The Photon
    Commented Sep 26, 2013 at 4:07
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    \$\begingroup\$ Another source for electronic loads is Kepco: kepcopower.com The BOP series are 4-quadrant supplies. The EL series are electronic loads specifically. \$\endgroup\$
    – The Photon
    Commented Sep 26, 2013 at 17:24
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    \$\begingroup\$ Assuming only one (or none) of the supplies ties its negative terminal to earth ground, what you drew will not immediately blow up. I haven't thought through all the failure modes, though. Also be aware the resistor is dissipating 200 W. You will want to think carefully about what kind of resistors to use and how to heat-sink them. The Agilent supply will also need to have a terminal disconnect feature (or you need to add a switch) to be able to turn the system off (I=0) programatically. \$\endgroup\$
    – The Photon
    Commented Sep 26, 2013 at 21:28
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    \$\begingroup\$ For example, if both of them had their (-) terminal tied to ground, the Agilent supply would be short circuited. It probably wouldn't literally blow up, but it'd probably be shut down by their protection circuits. \$\endgroup\$
    – The Photon
    Commented Sep 26, 2013 at 22:00

1 Answer 1

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Most power supplies do not have "current sinking" capability.

When discussing this, we talk about the four quadrants of the I-V space:

enter image description here

Where the response curve of the supply is plotted with I on the x-axis and V on the y-axis. For example an ideal voltage supply would be characterised by a horizontal line at constant voltage, no matter what the current.

Most power supplies can operate only in the I quadrant. This means that current is going out of the more positive terminal of the supply. Some can also (or you can flip the leads to work in) the III quadrant. In these quadrants, the supply is delivering power to its load.

Only specially designed supplies can operate in the II and IV quadrants. These are called "4 quadrant supplies". An example of this is the BOP series supplies from Kepco.

There are also "electronic loads" designed to operate in the II and IV quadrants, such as the Agilent N6063B.

However, it's not obvious from what you said about the goals of your test that you couldn't simply connect a (or a combination of) power resistors to the supply being tested instead of spending a lot of money on an electronic load.

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  • \$\begingroup\$ The electronic Load will be easier for the test because the requirements are to ramp the load from minimum load all the way up to 100% load in 15 steps. The Agilent supply or any electronic load would be smoother and easier to control than switching in and out power resistors. \$\endgroup\$ Commented Sep 26, 2013 at 20:46

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