I need help to design a circuit that turns 12 VDC input to 60 VDC output while supplying 2mA of current. I tried a boost converter but i keep getting microamps of current and I'm having difficulty lowering the voltage. Also, the voltage and current keep fluctuating, it needs to be stable. It is currently at 125V and 5.96uA. I'm thinking that I may need another transistor or change one of the components values. I have increased and decreased every component's value but it causes other problems such as the voltage going down the 11V, the current being too high, etc. enter image description here

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    \$\begingroup\$ The boost converter is the way to go. The fact you didn't get what you want is hinting that you have some fundamental mistake in either your input data or your expectations. So you better tell us what you are trying to do and what exactly you have. \$\endgroup\$ – Eugene Sh. Jul 26 '18 at 20:22
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    \$\begingroup\$ PCB layout is critical for switch-mode regulators. Have you looked into using a TI Simple Switcher design? \$\endgroup\$ – MarkU Jul 26 '18 at 20:26
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    \$\begingroup\$ You have a circuit that can boost the voltage, but you need to consider that there is no regulation. How would you prevent the output rising above your required voltage? How much capacitance is needed to supply current when your switch is on ….and what ripple do you think you'd need? \$\endgroup\$ – Jack Creasey Jul 26 '18 at 20:34
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    \$\begingroup\$ You have nothing in your circuit (like a resistor) that will draw 2 mA from the output. \$\endgroup\$ – W5VO Jul 26 '18 at 20:59
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    \$\begingroup\$ 100 Hz is ridiculously low for a boost converter. 9 Amps into base of Q1 is extremely high. A transistor with Max V=15 V for a 60-V circuit won'f fly. \$\endgroup\$ – Ale..chenski Jul 27 '18 at 1:13

This is a basic boost converter.

You could have made it work unregulated with a bunch of changes like;
V1=100k~1MHz, C1=50nF, R1= 1k

But here is a simple conceptual design with a dynamic load of 6mA to see the regulation error of -1.7V or 58.3V.

A simple boost IC is simpler and both work.

enter image description here

  • \$\begingroup\$ in the circuit you posted, what is that amplifier looking part called? also what is the name of the transistor you used in the circuit? What voltage does that 10KHz generator have? \$\endgroup\$ – Fried Jul 28 '18 at 0:54
  • \$\begingroup\$ 1) CMOS Schmitt Inverter In a "Relaxation Oscillator 2) any NPN (PN2222A) 3) 10kHz is just a square wave load thru a 10k or -6mA looking at load regulation. Box on left is a simple DC-DC 60V Boost Regulator \$\endgroup\$ – Sunnyskyguy EE75 Jul 28 '18 at 2:41

I read this to mean you want to produce 60VDC from 12VDC at a minimum of 2ma. Because of the extremely low output current, you could build a voltage multiplier, which has an intrinsically limited peak output voltage, so it will give you a steady output voltage so long as the load is adequately small compared to it's current carrying ability and the input voltage remains steady. When I was first getting into this stuff I found the concepts much easier to understand than boost converters. A boost converter is likely more compact, depending on frequency and inductor size. All of that said, I'm going in depth into buck converters before I consider myself "ready for" boost converters, but I would investigate the following:

1.) Is the inductor being oversaturated? Increasing switching frequency will alleviate this. Voltage and ON pulse length will determine whether this occurs, meaning that the problem may not show up at low duty cycles, even if the frequency is too low.

2.) The duty cycle (time on/(time on+time off) will determine energy delivery to the inductor, and therefore output voltage at a given output current level. You do not have an active feedback mechanism in your circuit, so if you want to change the output voltage, you have to manually change the duty cycle. If the load is always the same, you can make a device that operates permanently at the suitable duty cycle for the voltage or current you want to run.

3.) If your load is resistive (not capacitive or inductive or switching electronic), your output current will be determined by the voltage you apply, and as a result, if you are getting 5.96 uA at 125v with such a load, you will get 2.86 uA at 60v.
4.) As mentioned above, with a boost regulator, you need active feedback if your supply or load is expected to vary. That means if you're using batteries, you need to compensate for the fact that their voltage will drop during current spikes and also over time as the battery is depleted. This is the part of a boost regulator I'm not ready for yet, and the reason I'm not building them yet is that active feedback is great! If I really need a boost regulator and can't put it off, I'll use someone else's design and be sure it includes active feedback.

5.) You may wish to figure out what the switching time of your transistor and diode are, figure out what the highest frequency you can switch the slower at without your switching losses going above say 1%, and operate at that frequency. 50uH is a plump little inductor, so I'm guessing in the 1-5khz range you may see an improvement. The higher the frequency you can switch at, the less of an inductor you need, which will decrease output lag and make your active feedback mechanism perform better if you add one. It also means that you can use an inductor with fatter wires and therefore lower resistance(keep resistance as low as you can for voltage converters)

5.) I prefer mosfets to transistors, partly because of the extra resistors, but you appear to be running 10v over 1 ohm to the gate of that transistor, which would give you a current of 10A, which seems like 250 to 1000 times too much current there.

6.) If that is a circuit simulation and not a record of real life measurements, it is at least somewhat working. The 5.96uA would be the leakage current through your capacitor, and simply add to the load current when you actually attach a load.

7.) Be aware that there is a wide variety of premade boost converters on the market, and you can also get boost regulator controller ICs, follow the datasheet to select components, and build your own with as many bells and whistles as you like, particularly active feedback.

All of this is based on the assumption that you want to regulate your voltage to 60V, with a max current of 2mA. If you actually want to regulate current to 2mA, that's a bit different, and requires a different form of active feedback.

Any of these help or interest you? If you can figure out the questions to ask, there are people here who can go deeper than I can on all of this material.


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