-1
\$\begingroup\$

I'm doing an industrial project, i need to step down the high voltage of 2kV to 5V, is it possible to step down to around 5V without transformer, is there any other alternatives? please suggest me :)

\$\endgroup\$

closed as unclear what you're asking by Olin Lathrop, laptop2d, DoxyLover, Bimpelrekkie, jonk Nov 1 '17 at 20:50

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    \$\begingroup\$ Resistor divider + bridge rectifier. You didn't say anything about how much power you needed, or isolation ... \$\endgroup\$ – Bryan Boettcher Oct 30 '17 at 14:43
  • 3
    \$\begingroup\$ Why would you not use a transformer? \$\endgroup\$ – Transistor Oct 30 '17 at 14:44
  • 6
    \$\begingroup\$ Better tell us what you are trying to do. \$\endgroup\$ – Eugene Sh. Oct 30 '17 at 14:57
  • 5
    \$\begingroup\$ "My device would be bulky so i need to design without transformer." Safety first. Size later. Put all the information in the question not scattered through the comments. \$\endgroup\$ – Transistor Oct 30 '17 at 15:25
  • 1
    \$\begingroup\$ There is way too little information here. How much power do you need at 5 V? What's wrong with the obvious answer of using a transformer? What frequency is this AC? Is the 2 kV ground-referenced or isolated? What about the 5 V? What will be connected to the 5 V? This smells like a X-Y problem. Pop up two levels and explain what you are really trying to accomplish. Starting the close as unclear process, so answer soon. Also, answer all the questions, not just what you think matters. \$\endgroup\$ – Olin Lathrop Oct 30 '17 at 16:43
1
\$\begingroup\$

If all you need is to get a sample the voltage with an ADC (I saw your comment in Peter Green answer), just make a voltage divider.

Hook some 1/2W high ohm resistors in series between 2kV and ground, trying to make each one see at most 500V, and a last resistor proportional to the 5V you want. Something like this:

Vcc --- 20MΩ --- 20MΩ --- 20MΩ --- 20MΩ --x-- 200kΩ --x-- GND

The first four resistors are of high value because too low resistors will load your 2kV source, drawing a big current which may cause voltage drop and overheating on the resistors. And you want to use several of them (four), instead of just a big one, because each resistor has a voltage limit. Above that, voltage may arc from a lead to another, creating unwanted fireworks.

This network will draw a low current: about 25µA. The last resistor is where you will hook your ADC (note the x), and it´s calculated this way: 5 * 80000000 / 2000 = 200kΩ. This resistor will have about 5V of the total 2kV. The remaining 1995V will be distributed between the 20MΩ resistors. If you want to be on the safe side, decrease the 200kΩ a little, to have less voltage on it, preventing accidents by overvoltages on your ADC.

\$\endgroup\$
  • 1
    \$\begingroup\$ All this assumes that the other phase is grounded. Even if it is a direct connection between a 20 kV circuit and a 5 V micro is crazy. \$\endgroup\$ – Transistor Oct 30 '17 at 17:26
  • \$\begingroup\$ Some MCUs have a separate analog GND pin, that may help when the 2kV is not referenced to the ground, but I'm not sure. About direct connection, I don't see a way to isolate it and still have a compact design as the OP asked. Personally, I don't think it's crazy. People tend to be afraid and over concerned about high voltage, but the danger is not so high, if proper isolation is provided. And finally, it's 2kV, not 20kV. \$\endgroup\$ – Marcovecchio Oct 30 '17 at 17:42
0
\$\begingroup\$

You need to answer a few questions.

  1. Do you need isolation?
  2. How much power do you need?
  3. What efficiency and/or power factor do you need?

If you need isolation that pretty much pushes you into using a transformer somewhere in the design though it may make more sense for it to be a high-frequency transformer in some kind of switched-mode converter.

If you don't need isolation then other approaches become possible. For example a buck converter or for small ammounts of power a capacitive dropper.

\$\endgroup\$
  • 1
    \$\begingroup\$ With 2kV then isolation should be standard... \$\endgroup\$ – Solar Mike Oct 30 '17 at 16:08
  • \$\begingroup\$ I'm making a device so that it should measure voltage, current and power factor... for Voltage of around 2kV i need to step down to 5v and then to adc and micro controller. But using transformer my product would be bulky, i thought of voltage divider first which is kinda very dangerous. So i thought of any other methods i can get \$\endgroup\$ – Dexter Oct 30 '17 at 16:19
  • \$\begingroup\$ My device should be a size of a mobile charger, is it possible to do it for such a high voltage? \$\endgroup\$ – Dexter Oct 30 '17 at 16:20
0
\$\begingroup\$

2kV components are bulky. That's because of the insulation needed, but more important, because no one bothers with 2kV when he doesn't need 2MW. Even 200kW stuff would be done at 690V, to allow 1kV insulation and a wider and cheaper selection of components.

So … if you want something small at 2kV, you have to build most of the components yourself. Sometimes a series connection of low-voltage components will help (e.g. 1N4007 goes up to 1kV) but most times, you are out of luck.

So winding a 2kV->690V transformer yourself will solve a lot of headaches. You will still hardly come down to the size of a cell phone charger. These are incredibly tiny.

\$\endgroup\$
  • \$\begingroup\$ You can't possibly say how small the result may be because we have no indication of the power requirement. \$\endgroup\$ – Olin Lathrop Oct 30 '17 at 16:46
  • \$\begingroup\$ That's what I tried to explain. The power requirement of the piece in question doesn't count. You simply cannot buy low-power 2kV components. Because it makes no sense in most cases. \$\endgroup\$ – Janka Oct 30 '17 at 16:50
  • \$\begingroup\$ But he could want 500 W, in which case it would certainly be larger than a cell phone charger, whatever that actually means. \$\endgroup\$ – Olin Lathrop Oct 30 '17 at 16:55
  • \$\begingroup\$ I've asked the industry guy for the power range, if the requirement for high power then can it be done? \$\endgroup\$ – Dexter Oct 30 '17 at 17:05
  • \$\begingroup\$ High power means big size. So, no. \$\endgroup\$ – Janka Oct 30 '17 at 17:16

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