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?)

  • \$\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 '20 at 17:49
  • \$\begingroup\$ @NatsuKage " ... 2 x boost ..." \$\endgroup\$ – Russell McMahon Jan 26 '20 at 17:59
  • 2
    \$\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 '20 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 '20 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 '20 at 18:25

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.


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


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.


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


  • \$\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 '20 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 '20 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 '20 at 23:03
  • \$\begingroup\$ @Geeky121 A voltage multiplier will work for just +100V also. \$\endgroup\$ – user4574 Jan 27 '20 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 '20 at 23:00

Let me try to answer you, I didnt look into your link but to fulfil your need all you have to do is build a Boost converter, but 200V boost converter (12V to 200V), Why 200 is, because +/-100V is 200V from one node to other. You will need 94% duty cycle to get there. 94% duty you can do using 555 time for experiment or microcontroller. Once you have 200V, (see image), you can center tap for +/-100V each. Note this is good enough for experiment and do not use this for any product requiring long term reliability without improving the design given here. i hope you will calculate capacitance and Mosfet , diode values. add two balancing resistor across output caps. my assumption is your load in not completely imbalanced load.
enter image description here


You could use a diode pump for your pos 100 VDC and use a second diode pump for your neg 100 VDC .This diode pump system could be driven by a boost convertor .The more stages in the diode pump system the lower the max volts on the boost convertor main switch .The pos and neg DC voltages will track reasonably so you can monitor the pos volts via a standard resistive divider to your error amplifier on the boost convertor .Using a chip will mean less peak switch volts and more diode pump stages .Going discrete simplifies the diode pump .Both approaches have worked for +/- 500VDC so it is really up to you how you do this .


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