0
\$\begingroup\$

I see spellman/bertan hv power supplies achieve this but are expensive. Is there a circuit that can regulate cheap flyback transformer's voltage ripple just as well? I've seen some circuits referred to as smoothing circuits and filter circuits but it's hard to find quality search results that actually mention design related to voltage ripple control. I would only need around 30 uA to 400 uA.

\$\endgroup\$
  • 1
    \$\begingroup\$ Given Q = C * V, and the time derivative produces I = C * dV/dT, then you have a tradeoff in the size of the capacitor, the size of the load current and the ripple, versus the frequency of recharging. (The formula does assume constant C). \$\endgroup\$ – analogsystemsrf Apr 13 at 18:51
  • \$\begingroup\$ do you need 200mV regulation too? \$\endgroup\$ – Jasen Apr 14 at 4:56
0
\$\begingroup\$

With those modest current requirements (30-400uA), why not just use an LDO on the output as post regulator? You don't mention the voltages, so that may not be an option. If the voltage is higher than 36V, then you'll have to start looking at RC or LC low-pass filters. For the filters you'll have to know the highest frequency of the ripple and just filter it.

\$\endgroup\$
  • \$\begingroup\$ The OP doesn't mention the switching frequency, but it would be important to select an LDO that has good PSRR at the switching frequency. Also, it's important to be careful with LC filters as a pure LC (outside the control loop) could ring if there are transients with repetition rate or frequency content at the resonant frequency. Inside the loop the filter can cause excess phase shift and instability. All these can be good options if done correctly. \$\endgroup\$ – John D Apr 13 at 19:10
  • 1
    \$\begingroup\$ If the L+C filter is used, then add a resistor in parallel with the inductor, of value sqrt( L / C ). Thus for 100uH and 1uF, use 10 ohms in parallel, for excellent dampening. \$\endgroup\$ – analogsystemsrf Apr 13 at 19:32
  • \$\begingroup\$ I thought it was well known that flyback transformers output around 15,000 to 30,000 volts. Of course they can be driven higher. I measured 27kv on one the other day. \$\endgroup\$ – agileOne Apr 14 at 3:42
  • 2
    \$\begingroup\$ flyback is a topology, it can be used at lower voltages, even for step-down! most phone chargers are flyback topology. \$\endgroup\$ – Jasen Apr 14 at 4:54
0
\$\begingroup\$

pretty much you need to put active regulation after the rectifier of the flyback. a capacitance multiplier is probably good start.

\$\endgroup\$
0
\$\begingroup\$

enter image description here

Disclaimer: I am not responsible for the actions of others, this post is merely educational. Do not make this circuit.

The best way I could find to have high voltage ripple brought to negligible amounts is to drive the flyback transformer(or other high freq. AC source) at the highest frequency possible within the range of 20khz to 100khz, take its AC (remove diode) and use this as input to a full wave Greinacher multiplier, with a parallel filter network. It may be more practical to include a voltage divider after the flyback, as I intend to do.

This spreadsheet calculator helps to calculate capacitor and diode values for low ripple, according to your own desired current and voltage.

I used google translate on this pdf with OCR to learn of this design as Fig. 7. The final ripple is shown in Fig. 9.

Fig. 10 implies that inductors stray capacitors are added; the roughly translated text claims the following:

"Stray capacitors equivalently between two condenser columns Must compensate for the lead current flowing through the As shown in Figure 10, cw circuit capacitors Insert an inductor in parallel between the column and the boost transformer true Sky The gain of the step-down transformer at this time was increased."

I believe google translate mistakenly translated "step-up" as "step-down".

I recommend making a negative voltage version of this circuit to accomplish the ranges of voltages closer to earth ground and to reduce the number of stages and their consequential voltage drop.

Example component values are provided for Fig. 7, where any capacitor labeled "Cf" is equal to Co, and N is the number of stages in the Greinacher/Cockroft-Walton multiplier:

f=100KHz, I=75μA, I2=10μA

I3=10μA, Cp=36pF, Co=3900pF

R1=250KΩ, R2=125KΩ, R3=62.5KΩ

Keep in mind any load resistances affect the low pass filter cutoff frequency. I could not find much of example values for the "RL1" and "RL2" load resistances, seemingly they are just there to show how to connect the circuit to its application.

Due to low capacitance values and high voltages, it is cheaper to make your own capacitors according to this formula so long as large size is not a problem. Caps and diodes minimum ratings should be twice the input voltage and current.

It is best to avoid parallel diodes to achieve current rating because of their tendency to heat up and provide resistance, taking current disproportionately. Also, apply soldering heat conservatively to avoid breaking them.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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