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I have an objective to make a pulser for ultrasonic 1Mhz and 120V, for step up DC i used a boost converter with switching frequency 10KHz, 0.9 Duty Cycle and 12V supply, the boost converter work perfectly for static load like 1Mohm, but when i attach the converter for supplying my switching ultrasonic, the voltage from the converter was continuously decreased. below is the simulation and the circuit

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the green line is the supply from converter right before the zener diode, and the red line is voltage in R3, note that it is 1 MHz switching so it seems just like a solid shape. the green voltage dropped down when the switching circuit does the work, after a while the voltage will be below zener diode and the diode cannot maintain the output voltage. so how to maintain the voltage for supplying the switching load circuit

  1. do i need to increase the converter frequency to 1Mhz, same as the switching load? but when i do this, the voltage is slightly less than 10KHz converter.
  2. i have tried to increase C1 but the voltage still drop down but just with less step, so will huge C1 solve the problem?

thank you so much for the help

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You could use a boost controller IC that has feedback divider at C1 and changes its duty cycle to maintain the set voltage. This would eliminate the need for Zeners and increase duty cycle of the converter automatically once 1Mhz pulse load starts.

Texas Instruments over 100V output

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  • \$\begingroup\$ Thank you for your reply, is there any simplified boost controller circuit? i have searched but no good result, thanks \$\endgroup\$ – Zahi Azmi Apr 17 '18 at 7:57
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You have a 2.2 ohm resistor in series with your 12 volt supply: -

enter image description here

This is going to reduce your perceived input voltage as you draw more load at the output. Also, you need to remember that a boost converter blindly pushes energy per cycle to the output and it doesn't understand about regulating voltage. If the energy you push each cycle corresponds with the energy taken by the load each cycle then it regulates but, as you draw more load current, the output drops unless you put more energy in and that means raising the duty cycle above 90%.

Given that 90% doesn't leave much headroom I'd be tempted to increase the switching frequency and lower the inductance. A higher switching frequency means that you have a potential for delivering energy at a faster rate.

A lower inductance allows the inductor current to ramp to a higher level in a shorter time hence this also has the potential to deliver more energy per cycle.

Then, put it all together into a closed loop that regulates the output voltage by varying the duty cycle. That is what commercial chips do.

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  • \$\begingroup\$ Thank you for your reply, yeah, that's resistor is used for limiting current flow through inductor because i cannot find power inductor, thanks \$\endgroup\$ – Zahi Azmi Apr 17 '18 at 8:00
  • \$\begingroup\$ @ZahiAzmi You may not know this but it's common practice on stack exchange to upvote instead of thanking. Did you follow what I was saying about varying the duty cycle to regulate output voltage? \$\endgroup\$ – Andy aka Apr 17 '18 at 9:34
  • \$\begingroup\$ thank you, i have done that but the voltage will be lower than 120V, i have tried to find duty cycle control method but no result \$\endgroup\$ – Zahi Azmi Apr 17 '18 at 10:00
  • \$\begingroup\$ You have to vary the duty cycle in discontinuous mode (the mode you are likely using) because the output voltage depends on how you use the energy each switching cycle and this is dependent on load current. \$\endgroup\$ – Andy aka Apr 17 '18 at 13:41
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M2, when conducting, shorts (well, through R1) your output to ground. This consume much more power than your static 1Mohm load (actually, ~7.5W at 50% duty cycle), and the converter can't keep up.

Put this transistor in series with the load rather than in parallel, so there is no power wasted when the load is unpowered. If the load doesn't allow this simple change (too resistive and some capacitance), use a totem pole. Something like the circuit below could work, provided that you choose small mosfets with low gate charge (still withstanding 120V):

enter image description here

The simulation of the above circuit shows the supply doesn't need to provide more than 1W, and the control pulse current is reasonable (e.g. a NE555 will be able to provide it).

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  • \$\begingroup\$ Thank you for your reply, my load is high resistive, so if attach it to the mosfet drain directly, the transient time of the switching wont be good \$\endgroup\$ – Zahi Azmi Apr 17 '18 at 7:59
  • \$\begingroup\$ Yeah, but currently, you're wasting ~7.5W (at 50% duty cycle) through R1. Do you realize the size the resistor has to be? You could probably use a totem pole, this would solve both the size of the resistor and your problem. \$\endgroup\$ – dim Apr 17 '18 at 8:06
  • \$\begingroup\$ I've tried to use a simple circuit totem pole and try that in ltspice, the output in load not same as vcc, just around 0.5% \$\endgroup\$ – Zahi Azmi Apr 17 '18 at 8:09
  • \$\begingroup\$ Probably because you had a timing problem and both transistors were on at the same time. It is true that designing a 120V totem pole running at 1MHz is probably a bit challenging. But certainly still feasible, especially since in your case, the load is very low. \$\endgroup\$ – dim Apr 17 '18 at 8:15
  • \$\begingroup\$ So how could i fix the timing problem? Any circuit name for that? Big Thanks \$\endgroup\$ – Zahi Azmi Apr 17 '18 at 8:25

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