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Let's say I have this converter (ebay link):

It accepts 10-60 DC volts as input and outputs 12-80 DC volts.

What I'm asking is if it is possible for me to INPUT 60-70 Volts onto its OUTPUT pins and try and get 12 volts on its INPUT pins? (I have one but I'm afraid to fry its chips.)

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closed as off-topic by The Photon, Nick Alexeev Oct 13 '16 at 18:11

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Questions on the use of electronic devices are off-topic as this site is intended specifically for questions on electronics design." – Nick Alexeev
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    \$\begingroup\$ What?? No! Don't do it. Even for an expensive Japanese converter. \$\endgroup\$ – Eugene Sh. Oct 13 '16 at 17:41
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    \$\begingroup\$ I'm voting to close this question because the answer is only two letters long. \$\endgroup\$ – The Photon Oct 13 '16 at 17:44
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    \$\begingroup\$ Please, boiz, I'm new here. Why so much hate? I wouldn't treat you the same way if we've met at stackoverflow.com. \$\endgroup\$ – AgentFire Oct 13 '16 at 17:47
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    \$\begingroup\$ @AgentFire, Sure, but gcc can't. You can use a different tool to do a different job. You can use a (properly-chosen) buck converter to convert 70 V to 12 V. You can't just turn a boost converter around backwards, any more than you can turn gcc around backwards to turn it into a decompiler. \$\endgroup\$ – The Photon Oct 13 '16 at 18:15
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    \$\begingroup\$ I absolutely do not think this is a stupid question. It just has a simple answer. \$\endgroup\$ – pericynthion Oct 13 '16 at 18:44
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No, it will not. The circuitry of a boost converter is not set up to take in power, only output power.

For that you will want to search for a buck converter, instead of a boost converter.
https://www.amazon.com/s/ref=nb_sb_ss_c_1_8?url=search-alias%3Daps&field-keywords=buck+converter&sprefix=buck+con%2Caps%2C233

however, most don't accept voltages as high as 60V or 70V, they ususally cap off at the 30V to 40V range.

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  • \$\begingroup\$ Will I be able to connect two of the like in a sequence and be done with it? \$\endgroup\$ – AgentFire Oct 13 '16 at 17:45
  • \$\begingroup\$ If I understand you correctly most definitely not. These don't work in series the same way resistors do, they have a variable current draw, and thus a variable voltage drop. Once the input on one gets above the failure point, it will be toast. Then, depending on whether the first one to fail opened or failed short, the second will either not have both a power and a ground connection, or it will have the full 60 v and fail as well. \$\endgroup\$ – ambitiose_sed_ineptum Oct 13 '16 at 17:50
  • \$\begingroup\$ Thank you for your reply. What would be the best way to step down from 60-70 to 12V then, if the planned output power is no more than 25W? \$\endgroup\$ – AgentFire Oct 13 '16 at 17:52
  • \$\begingroup\$ You could try to find a high voltage input buck converter, but I don't know how successful you will be in that. I don't really know how else you could. What kind of power source are you using that provides 70Vdc? \$\endgroup\$ – ambitiose_sed_ineptum Oct 13 '16 at 17:57
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    \$\begingroup\$ 70v might almost be high enough to put into a 'universal input 85-265v' SMPS and get 12v out (yes, you can put DC into these) \$\endgroup\$ – Neil_UK Oct 13 '16 at 19:27
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While synchronous (two MOSFET) boost converters can indeed convert bidirectionally, your board isn't necessarily of the synchronous switching topology (it could use a diode instead).

In addition, that board certainly won't be regulating the input side so don't expect 12 V at the input. If it's set to produce e.g. 60 V and you backfeed 61 V it'll try to draw as much current as needed to reduce the output voltage back to 60 V, and all the energy will just end up at the input. If the input acts as a voltage source (like a battery) it will sink that current and maintain a fairly constant voltage, but if there is nothing to sink the current the input voltage will rise uncontrollably.

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Only the input side of this inverter has the pcm system. The output only has a feedback to the input side. These systems are not designed to work that way. If you try to reverse then say goodbye to the module.

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