Bridge rectifier circuit with a step-up transformer

I have been experimenting with a full-wave bridge rectifier in LTspice.

I am trying to step up 120 V AC/60 Hz to 1 kV DC for a load of around 2 kΩ. The step-up transformer has a ratio of 25:9. The smoothing capacitor was calculated in order to have a ripple voltage of around 10 V.

Since I am new to this, can someone help with the following;

• If I were to have to design this on a PCB, what are some tips that I need to follow?
• Are there some obvious problems with the circuit, and if so, what are they, and how do I mitigate them?
• The fundamental harmonic of the current across the capacitor is 140 A which is a lot, should I be worried?
• How can I increase the current across the load?

The circuit I have is this:

The voltage waveform is the following:

Here is the waveform zoomed in:

Here is the waveform of the current across the smoothing capacitor:

Here is the waveform of the current across the load:

• "current across the capacitor is 140 A" Yes. Something will break or not survive for very long. You need some inrush current protection, for example an NTC resistor. "How can I increase the current across the load?" Decrease R_LOAD. Commented Apr 14, 2023 at 15:17
• The heat of just how many kW are you ready to cope with? And what do you need smoothing for? Commented Apr 14, 2023 at 20:34

If I were to have to design this on a PCB, what are some tips that I need to follow?

Creepage distances - keep them conservative. 8mm spacing for 1000V would be a good bet I think.

Are there some obvious problems with the circuit, if so what are they, how do I mitigate them?

Using smoothing capacitors to stabilize current will require impractical capacitances and currents. You do want a linear regulator there.

The fundamental harmonic of the current across the capacitor is 140A which is a lot, should I be worried?

No way - not for real capacitors, not for real transformers. Never mind that 400μF at over 1000VDC is not something you'll buy in any single capacitor. You'll need four 450VDC capacitors in series, with balancing resistors. And you won't want 4x1.6mF in series either. 4x100μF/450V in series would be most practical, anything over that may become huge, unwieldy, and an arc flash hazard.

How can I increase the current across the load?

The current will be determined by the power rating of the transformer - everything else in the circuit is then sized to accommodate a somewhat derated rating of the transformer.

Now let's look at this design step-by-step.

First of all, this is a 500W supply: $$\P=(1000{\,\rm V})^2/2{\,\rm k \Omega}=0.5\cdot 10^{6-3}{\,\rm W}=0.5{\,\rm kW}.\$$, and with derating you're looking at a 750W design. The transformer should be rated 600VA or higher if you want to run it continuously and have some margins left, perhaps 750VA would be better. Giving transformer's insulation an easier job is always beneficial, so running that transformer not very hot is the idea.

It doesn't have to be said that at such voltages and power levels, not only is electrocution is a hazard, but also arc flash. With an inadvertent low-impedance short (say with a tool), chunks of traces will readily vaporize. Having copper plasma thrown at you is not advisable to say the least. In other words, this project will very happily kill you and/or set the room on fire. Do not work on this thing alone.

Once the power level is given, we need to define the input voltage range specification. Conservatively you'd want it to work within ±15% of nominal input voltage.

Then we need to decide on a reasonable power loss in the regulator circuit - this is driven by the ripple voltage, and the requirement on the output capacitance. The output current at 500W is 0.5A, so let's go with that - we can always "upsize". Suppose we want the regulator to dissipate 10% on top of the load power, so 50W.

To reduce dissipation, we'll use a two-stage regulator. The first stage will have a fairly small conduction angle and will pre-regulate the bulk capacitor voltage. That is, we have one series regulator directly between rectifier output and the bulk capacitor. The bulk capacitor is charged in pulses up to the preregulator voltage VR. This voltage needs to be at least about 15V higher than the output voltage, to maintain regulation across the 2nd regulator stage.

The voltage on the bulk capacitor will be approximately a sawtooth. Let's say we set 100Vp-p ripple. That's $$V_{rms\,saw}=\frac{V_{pp}}{2\sqrt 3}=\frac{100{\,\rm V}}{2\sqrt3} \approx 30V_{rms}.$$

At 0.5A, that's about 15W dissipated on the 2nd regulator stage. The 1st stage should be dissipating about 35W - we'll check that.

Given the 100Vp-p ripple, we can size the bulk capacitor: $$C=Q/V=\frac{0.5{\,\rm A}\cdot 1/120{\,\rm s}}{100{\,\rm V}}\approx 42\mu F.$$

For that capacitance, it is feasible to either use a parallel bank of 2kV rated foil capacitors of about 0.5μF each, or - assuming -50% tolerance - four 470μF/450V in series.

Thus far, the circuit looks as follows:

simulate this circuit – Schematic created using CircuitLab

There are two regulator loops:

1. OA1 maintains the output at the center point of R19 and R20. After C9 has charged up, and OUT is at 1.0kV, there's exactly 12V across R19 and R20. R9-R14 scale the output voltage for feedback.

2. Q6 maintains the pre-regulator gate voltage VG1 at such a level that the lower envelope of VB - at VBLOW - is at about 25V. The VBLOW envelope detector consists of R17,C6,D7.

M2 dissipates about 20..30W, depending on line level, thanks to the VBLOW envelope regulation low.

M1 dissipates another 20..50W, in fairly large peaks, and would probably be a SiC or GaN device rated for such use. It needs to withstand full rectified voltage during capacitor ramp-up.

Both M2 and M1 need suitable heatsinks and fans. Probably the cheapest option is CPU+fan heatsinks for PCs - for CPUs that dissipate 100W at least.

A few more protection features would be desirable, such as M1 current limit proportional to OUT voltage, i.e. a limit set for a certain constant load resistance.

OA1 floats on the OUT potential - that's its VSS rail, and the VCC rail is 12V above OUT, generated by U1.

The capacitor bank is shown at the -50% tolerance. Nominally those would be 450μF capacitors, or about 125μF series equivalent. The ripple voltage is regulated by the VBLOW control loop, i.e. as the capacitors lose capacitance, the ripple voltage increases, its minimal voltage always maintaining regulation across M1.

Diodes D1-D4 should be avalanche rated, 2kV reverse voltage (or more).

Another approach uses a pre-regulator on the mains (input) side of the transformer, with an optocoupler to provide feedback. It's easier to implement, since we don't need a 1700V-rated pass element M1. A 500V transistor or mosfet will work fine on the primary side, for all line voltages up to 240Vrms nominal. For use with 120Vrms nominal. a 400V device is perfectly fine.

I won't be putting a schematic together here, but for some inspiration:

A schematic of an implemented pre-regulator mains-side driver from the Dabbledoo blog above:

And a schematic of the regulator that drives the optocoupler above from the secondary side - that's an excerpt from Valhalla 2701C service manual schematics:

• Thank you so much for your help !
– DRF
Commented Apr 14, 2023 at 19:59
• Capacitors substantially larger than 400 μF at 1 kV do exist; I've used them before. They're expensive, but sometimes you need the high pulsed current this sort of large film capacitor can provide. Commented Apr 15, 2023 at 3:52
• I think you may have meant to use a parallel bank of eight 5 uF 2 kV capacitors for 40 uF total. Commented Apr 15, 2023 at 20:19
• @Hearth Yes, they do exist, and when you need them you need them. I’m not sure whether the question is about a future product or one-off. For a one-off, the little bit of dissipation in the balancing resistors is fine, especially that this is not a precision voltage source I don’t think. Commented Apr 16, 2023 at 13:17
• @DimitarZhekov Then try to look for 2kV rated foil or electrolytic capacitors, and forgo the balancing resistors since there won’t be capacitors in series. It’ll reduce the dissipation a tiny bit, and you’ll have fewer parts to buy. The Valhalla 2701C pre-regulator works well and is another way of implementing this with a fixed dissipation that is not dependent on the absolute mains voltage. Commented Apr 19, 2023 at 12:26

Your supply is 60 Hz and, your primary inductance is 5 μH <-- that is a massive problem on a supply voltage that peaks at 120 volts: -

5 μH at 60 Hz has an impedance of 1.88 milli ohms and, will take a peak magnetization current of nearly 64,000 amps. OK, it will be a bit less due to the 1 milli ohm resistance in the source.

Of course this cannot happen so, we build transformers that operate at 60 Hz with a primary inductance of several to tens of henries inductance.

The step-up transformer has a ratio of 25:9.

That is untrue according to your diagram (see snippet above); the turns ratio is the square root of the inductance ratio and, given your inductance ratio is 69.44, your turns ratio will be 8.333; 25:9 is 2.777 and not 8.333.

• I understand, that is a great point! Are there some other obvious issues with the circuit?
– DRF
Commented Apr 14, 2023 at 15:27
• You might want to check that the reverse voltage rating of the diodes is good for 1500 volts and, your electrolytic capacitor voltage rating ought to be around 1500 volts too. Commented Apr 14, 2023 at 15:48
• great, thanks for the heads up
– DRF
Commented Apr 14, 2023 at 16:16
• sorry i meant the turn ratio is 25/3
– DRF
Commented Apr 15, 2023 at 20:06

A 500 watt 60 Hz linear power supply is not really something suitable for a PC board. The components (except maybe the rectifiers) are too large.

It might be worthwhile to consider a voltage doubler, which can be constructed with a 120 to 360 volt transformer (with 100 mOhms input resistance and 1 ohm output), two capacitors, and two rectifiers. Note that peak capacitor current is only about 6 amps. This design is inherently current limited, but OTOH voltage regulation is very poor:

If you really want to use a full wave bridge design, a series inductor can help to reduce the capacitor peak current. I'll add that circuit in a little while.

Here is a full wave bridge simulation, including an inductor, which reduces output ripple and peak capacitor current to about 2 A. But even without the inductor, capacitor current is only about 5 A. The transformer has a 750 V secondary with 1 ohm output series resistance.

• The inductor idea looks very interesting. thank you
– DRF
Commented Apr 15, 2023 at 22:35