Recom R12-150B Boost Converter Output Changes with BK 1787B Current Limit Settings

Hi,
I am using some Recom R12-150B DC boost converters to output around their maximum output of 200 V.I have the strange problem that if I turn on my power supply to the boost converter (a BK precision 1787B Programmable DC power supply), and I put the power supply at 10 V and current limited to 0.2 A, the boost converter will output about 126V while drawing only 0.69 V and the full 0.2 A current from the power supply. If, however, I set the voltage from the power supply to 10V and the current limit to 0.4 V or greater, suddenly 9.98 V is being drawn from the power supply and a much smaller current of only 0.04 A, and the output voltage from the boost converters jumps to 203 V. This second result seems like the proper operation of the boost converter, but why is it being affected by the current limit I set considering the current drawn is so small?The resistive load on the boost converter is 5 GΩ.I'll include the datasheet for the boost converters and a schematic of my circuit. Apologies if the schematic is a bit poor; I am a novice with Eagle. (Note: the converters have a quirk where each pin has a duplicate. Apparently, according to the manufacturer, they're connected to their duplicates internally, but I connected them on my PCB before I was sure.)Any guidance you have would be greatly appreciated.Best,Mike

I don't have a lot of design experience with boost converters but I suspect that it has to do with the current required to get the converter started.  There are capacitors that need to be charged up which takes some current.  The voltage and current limits on a bench supply are just that, limits.  If your load exceeds the current limit, the power supply will  enforce that limit by lowering its voltage however far it needs to to keep the current from exceeding that limit.  If the load allows the voltage to exceed the set voltage limit the PS will enforce that limit by lowering its current however far it needs to to keep the voltage at the limit.Another way of looking at this is, depending on which limit is reached, the power supply operates either in constant voltage mode or constant current mode.  In constant current mode it adjusts the voltage (within the set limit) as needed to maintain that current.  In constant voltage mode, it adjusts the current (within the set limit) as needed to maintain that voltage.  Generally, you don't use the power supply to provide a specific current and voltage at the same time.  You set the voltage and the load determines the current or you set the current and the load determines the voltage.In your case, until the boost converter gets up and running, it requires more than 0.2 A to get it there.   Once there, the voltage limit is engaged and the current ends up at 0.04 A as determined by the load.Doubling up of pins is probably not a quirk.  Since pins can be quite small, usually it is to provide more current capability to that node  inside the part.  It is appropriate to connect to both pins on your board so as to balance the current between the two pins and not over stress just one pin or the other.  It might also help to point out that the power being drawn from the supply when failing to start is only 0.69 V * 0.2 A = 0.138 W.   When running, it is 9.98 V * 0.04 A = 0.399 W.  When the voltage is low (like during startup) the current must be high by the same amount to deliver the same power.  When the converter is running, try turning the PS voltage up and down a little (say from 9 to 11 V) and look at what happens to the current draw from the PS.  It should go down as voltage goes up assuming power drawn is about the same.  Please also note that as you put load on the converter, at 2o3 V, the load increase from the supply should be about 20 times that because its voltage is about 20 times lower (10 V vs. 203 V).  Thus a 10 mA load at 203 V should increase the current draw from the PS by about 200 mA or 0.20 A to 0.24 A.
Now that I have glanced at the datasheet, you should see the output drop below 203 V as you load it up as per their power limit curves so the current drawn from the PS will be lower than I just said, accordingly.I also now see that you have two converters connected (for double the voltage I expect) which can also double the current required to get started, or did you only have one converter installed for this test?  Also, when you say you have a 5 GΩ load on the 203 V, you are basically saying there is no load applied, right?  The outputs are open?  How did you arrive at 5 GΩ?Best.  -JD

Thank you, JD. That does make quite a bit of sense. The two converters are there to amplify a positive and negative supply rail for an amplifier. Also, the load isn't in the diagram because it's actually on the amplifier output.
I appreciate your help.
Best,Mike

You are welcome, Mike.  My pleasure. -