0
\$\begingroup\$

I have a EMCO AG15P-5 DC-DC converter (datasheet) that takes 0-5V in and outputs up to 1.5kV (proportional to input voltage). I need to operate the converter at 500V output. I've done testing with a precision voltage source and determined that an input voltage of 1.14V gives 500V output. Under full load, the converter uses ~20mA, and this value would be constant in my application. Given the converter's proportional output, the input voltage needs to remain as stable as possible. Moreover, the precision of the voltage source would ideally keep it steady in the 10s of millivolts. So now I need to figure out how to provide stable 1.14V @ 20mA to power this sucker.

My first thought was to use an LM317 with a divider network. I think this wont work considering the 1.25V Vref of the LM317. I've searched digikey for regulators and found this regulator, but am not sure if buck regulators are the best for this, considering stability

\$\endgroup\$
5
  • \$\begingroup\$ Have you considered a voltage divider from the high voltage side and closing the entire loop to remove all drifts except your reference on your primary side? \$\endgroup\$
    – winny
    Commented Jun 1, 2022 at 15:07
  • \$\begingroup\$ @winny I have no considered manipulating the output side. Could you elaborate? \$\endgroup\$
    – earl
    Commented Jun 1, 2022 at 15:09
  • \$\begingroup\$ A voltage divider on the high voltage side, with low voltage output going to an opamp and reference, or perhaps just a TL431 if you are creative, to constantly adjust to your desired 500 V. \$\endgroup\$
    – winny
    Commented Jun 1, 2022 at 15:11
  • \$\begingroup\$ To keep a 500 V output steady to some 10s of millivolts may be difficult and would require additional low pass filtering with high voltage capacitors. Output ripple is about 0.4 % or 2 V. \$\endgroup\$
    – Uwe
    Commented Jun 1, 2022 at 15:22
  • \$\begingroup\$ @Uwe the input voltage needs to be 10s of millivolts accurate, not the output voltage, though they are proportional. \$\endgroup\$
    – earl
    Commented Jun 1, 2022 at 15:27

2 Answers 2

Reset to default
3
\$\begingroup\$

You should consider using this converter inside of a closed-loop controller. The output voltage could be precision divided down to produce (say) 1 volt out per kV in and you use an op-amp controller to compare that potted down voltage with a precision 1.5 volt reference. This will ensure that if anything drifts inside the XP device, you have a closed-loop mechanism that keeps the output voltage stable (even under varying load conditions).

From the data sheet: -

A separate high impedance control pin is standard and is designed for external error amplifier and/or DAC control in closed or open loop systems

Control Voltage Input Analog Control Voltage adjusts output from 0 to 100%

enter image description here

Thanks to @soup for finding a decent article on controlling the device: -

enter image description here

Make sure you choose an output resistor divider that doesn't get too warm. Maybe 10 MΩ for the upper resistor (rated at the voltage) and the appropriate low kohm resistor for the lower part of the divider. This network will also discharge the output to safe levels when turned off rather than leaving a high voltage present that could be a nasty surprise for someone.

Be aware that the device in your question can only produce 100 volts on the HV output: -

enter image description here

Maybe you meant the AG10P-5?

\$\endgroup\$
8
  • \$\begingroup\$ I am not familiar with closed-loop control circuits. Could you point me to reading materials useful for my case? \$\endgroup\$
    – earl
    Commented Jun 1, 2022 at 15:18
  • 1
    \$\begingroup\$ What Is a Control System and How To Design a Control Loop for a DC-to-DC Converter might be a useful read. \$\endgroup\$
    – Andy aka
    Commented Jun 1, 2022 at 15:23
  • \$\begingroup\$ Yes, the device I have is actually the AG15P-5. I'll edit my question. \$\endgroup\$
    – earl
    Commented Jun 1, 2022 at 17:14
  • \$\begingroup\$ I went ahead and selected this as the best answer, given how fleshed out you made it. Thanks for that. Could you recommend an opamp to use in this case? I have a plethora of TL072s and some other more high end ones. Additionally, how should I determine the values of the resistor/capacitor on the opamp output? \$\endgroup\$
    – earl
    Commented Jun 1, 2022 at 17:16
  • 1
    \$\begingroup\$ @earl you probably want an op-amp that works on a single 5 volt supply rail and that just about covers the TL072 but, the trouble with that part is that its inputs don't work when the input is close to the negative supply (which is 0 volts unless you have another supply). Realistically, if you have a single rail available (5 volts) you should look for a rail-to-rail op-amp (inputs and outputs). The RC feedback on the op-amp might be a bit of a guess but 100 nF and 100 ohms should be a good start with an aim to lower the capacitor to around 10 nF (better control bandwidth). \$\endgroup\$
    – Andy aka
    Commented Jun 1, 2022 at 17:23
4
\$\begingroup\$

As said by Andy aka, using a version with a control pin is the easiest way to accomplish your goal. However if you already have the EMCO converter and can't buy a new one (they are fairly expensive) EMCO's application note 9 has a recommended design that has worked well for me.

enter image description here

Instead of using a control pin, the output voltage is monitored through a voltage divider by an error amplifier (op-amp + transistor) which allows the input voltage to be controlled. Since the EMCO device's output voltage varies linearly with input voltage this allows fine control of the output voltage. With this you can provide the 1.14V to the non-inverting side of the op-amp to produce the desired output voltage.

The 1.14V can be generated with a high bit DAC, a precision voltage source, or if you know the voltage will never need to change a voltage divider between 5V and GND could be used instead (note that the ripple of the 5V supply will affect the EMCO output here).

\$\endgroup\$
2
  • 1
    \$\begingroup\$ Good find +1 for that. \$\endgroup\$
    – Andy aka
    Commented Jun 1, 2022 at 15:39
  • \$\begingroup\$ This is great. I actually have the version with the control pin, I'll give this a read. \$\endgroup\$
    – earl
    Commented Jun 1, 2022 at 15:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.