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I am looking to power a phone ringer circuit that requires +100 V DC and a -100 V DC. I have found a circuit here but this converts it to 120 V DC with a transformer. Is it possible to get +/-100 V DC from 12 V DC without a transformer? If a transformer is the best option then I will use the circuit here and step the voltage down to around 100 V DC (If this circuit is the best way of course!).

UPDATE: I did find this instructable, Would this work? (I could use 2 for a +100 and -100 V DC correct?)

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  • \$\begingroup\$ Hello there. Are you using a fixed DC source? This will require a more complex circuit. If you have a mains AC line available, it is easier to create. But two buck converters like your instructable might do the trick. \$\endgroup\$ – Natsu Kage Jan 26 at 17:49
  • \$\begingroup\$ @NatsuKage " ... 2 x boost ..." \$\endgroup\$ – Russell McMahon Jan 26 at 17:59
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    \$\begingroup\$ Yes - two circuits as per instructable with correct polarity outputs would work. Need an NPN and PNP version. WARNING THESE VOLTAGES COULD EASILY KILL YOU - OR SOMEONE ELSE. \$\endgroup\$ – Russell McMahon Jan 26 at 18:00
  • \$\begingroup\$ I suggest isolating the 12VDC source of each boost converter from each other using an Isolated DC/DC Converter. \$\endgroup\$ – Natsu Kage Jan 26 at 18:10
  • \$\begingroup\$ If you used the circuit I suggested in your earlier question you wouldn’t have to convert to DC because a regular 230 volts transformer having a secondary of 12 volts can be employed to produce 90 volts at 25 Hz directly. \$\endgroup\$ – Andy aka Jan 26 at 18:25
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The "instructable" in update section really generates the needed voltage, but it's not practical. It's a teacher's demo for the basic DC voltage boosting with an inductor and a switch. Every operating cycle pushes a portion of new charge to the output capacitor and the output voltage grows. The growth stops when one of the following happens:

  • the load takes as much current as the inductor pushes as an average
  • the diode starts to conduct reversely due its limited voltage range
  • the output capacitor or the load break down

What makes it impractical:

  • there's no regulation which limit the output voltage to a safe value by stopping the switching as long as it's not needed
  • there's no security measures which prevent the switch current to grow dangerously in case the switch is ON too long time. In inductor the current grows gradually from zero to the limit caused by Ohm's law at rate U/L.
  • you need 2 circuits; one for plus and one for minus.

Use at least properly controlled switching regulators. They have been around so long time that they can be considered to be mature.

The next is effectively a suggestion to redesign the ringer circuit.

One DC voltage is enough if you make pulses of it and remove the DC component with a capacitor. Landline phones can have that capacitor, so switching the voltage between 0 and 100V where 0 means active pull down can be an acceptable ring voltage. But that's not quaranteed. Old phone specs had AC ring voltage added to the DC.

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Is it possible to get +/-100 V DC from 12 V DC without a transformer?

Yes.

One way would be to use the 12V to power a 12V square wave generator. The wquare wave would feed into two voltage multiplier circuits made out of capacitors and diodes. You would probably need about 8 ~ 9 stages to go from 12V to 100V.

https://en.wikipedia.org/wiki/Voltage_multiplier

The square wave generator could be a 555 timer if you want. If you need any significant power you would buffer the timer output with an H-bridge driver or some transistors.

I am not sure how many of these you are making. But if you are just making one then consider that your time is worth something too. If you are willing to spend a little money there are lots of off the shelf modules that will convert 12V to +/- 100V.

Here is one example.

XP Power F02CT

https://www.digikey.com/product-detail/en/xp-power/F02CT/1470-3998-ND/6802048

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  • \$\begingroup\$ Not very cost effective, but at ~7x4x2cm, it's quite compact and powerful enough for the circuit needed by Geeky121. \$\endgroup\$ – Natsu Kage Jan 26 at 20:50
  • \$\begingroup\$ @NatsuKage Not cheap, but looking at the decision of make-vs-buy one has to consider the cost of parts, labor, and engineering time. If this is a one-off then buying it is probably cheaper than making it all things considered. \$\endgroup\$ – user4574 Jan 26 at 22:57
  • \$\begingroup\$ I will be making several of these circuits So i believe making the circuit myself will be worthwhile. @user4574 do you think that the voltage multiplier circuit would be worth while if I only did +100 v DC like user287001 said? \$\endgroup\$ – Geeky121 Jan 26 at 23:03
  • \$\begingroup\$ @Geeky121 A voltage multiplier will work for just +100V also. \$\endgroup\$ – user4574 Jan 27 at 0:05
  • \$\begingroup\$ @user4574 I forgot I had a motor shield for arduino with an H bridge! I was able to test the theory with several diodes and 10uF capacitors and I was able to get ~92v DC from 6 stages with 12v Square wave at about 1k Hz. I was looking at other multiplier circuits and do you think I am using an appropriate sized capacitors? Should I use smaller caps? Thanks! \$\endgroup\$ – Geeky121 Jan 27 at 23:00

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