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# High Voltage Transformer Design

When it comes to high voltage transformer design, it can be a bit tricky. When you’re trying to target a particular voltage one needs to account for things such as DCR losses in the windings. This is important when regulation is important.

## Design Procedure:

The following high voltage transformer design procedure will get you close to the ballpark for your needs. I will account for the DCR losses but I will ignore things such as core loss.

For this analysis I will consider a step-up transformer that takes in 120V @ 60Hz and outputs 6,000VAC @ 10mA; so a 60VA transformer.

Given the frequency is 60Hz, silicon steel is the core of choice here. It has a MAX flux density of perhaps 1.8T. We’re going to use half of this so that there is some margin inserted. Plus lower flux density in the core means lower core loss which further justifies ignoring the core loss.

We need to first determine the PRI turns to meet this 0.9T of flux density. This will be a function of the applied PRI voltage, the frequency, and the core area. Let’s call the core area Ac.

Next, let’s find the MIN number of turns needed to meet the output voltage requirement. I say MIN because the SEC turns will end up being more after we consider the loss due to DCR.

Next, let’s find a wire diameter for both the PRI and SEC windings such that the wire doesn’t get hot. We will then find the total copper area used by both windings which will help us determine the core area we want to choose.

For the PRI we will use 4A/mm^2 for a current density and for the SEC we will use 2A/mm^2. The SEC will have many more turns and a much higher DCR so we want to keep that current density low.

The power is 60VA, the SEC current is 10mA and therefore the PRI current is 0.5A (ratio of power to PRI voltage).

For the PRI we have:

So the total copper area used is:

We can use a “rule of thumb” and set 35% of the total window winding area of the core to this value. In this way we leave 65% of the window open for wire insulation and in most cases because a bobbin is used which also takes up some of that area.

Let’s also assume we are using an EI core. So 35% of the window area for some EI core is this total copper area.

Now lets’ set the product of the winding window and the core area equal to all the other stuff… Also, we need to remember that Ac was in m^2 while the copper area we found is in mm^2, so we will convert the Ac to be in mm^2 so all the units match.

Now we choose an EI core whose WWA and Ac have the product that best matches this value.

I found that an EI76*30 works just fine. The WWA and Ac product work out to 368,709mm^4, which is a bit more than required. This is good because the SEC turns are going to increase after we account for DCR plus I would suggest if a bobbin is to be used, that it be a dual section bobbin.

Now let’s get the particulars solved since we have a core area for the EI76*30, which is 762mm^2 (0.000762m^2) and a WWA which is 483.87mm^2

Now, let’s just check and be sure that our total copper area is about 35% and that our flux density is 0.9T.

For the high voltage transformer design, this is all good so far… Our unloaded output voltage will be about 6,000VAC, a bit less due to core losses we are not accounting for, but as soon as we draw a current when a load is attached that voltage is certainly going to drop due to DCR losses. Let’s figure out those losses and compensate by increasing SEC turns appropriately.

The average length per turn (Lpt) for an EI76*30 core is 0.1616m. This can be found by looking at the dimensions for this core.

The PRI voltage that “gets into” the transformer is the 120V minus that lost to PRI DCR….let’s find this value…

We’re already losing a fair amount in PRI voltage due to the PRI DCR… Next, we’ll find out how much the SEC voltage is due to this smaller applied PRI voltage and due to its loss because of its DCR:

Ok so as one can see, losses due to DCR can cause the SEC voltage to drop considerably from that desired. Now let’s find the required SEC turns to target 6000V when the SEC is under load…

We do this by setting the Vs in the equation below to 6000V and then re-arrange this to solve for Ns:

With this many SEC turns the SEC output voltage will be 6000V @ 10mA (again ignoring any decreases due to core loss). This also means the unloaded voltage will be more than 6000V.

The unloaded voltage will be:

**Just as a check, let’s check the loaded voltage with these increased SEC turns…**

So there we have it; this transformer has the following specification:

Now given the high output voltage construction this would require additional steps to avoid voltage breakdown.

- CONSIDER USING AT LEAST A DUAL INSULATED, PERHAPS A TRIPLE INSULATED WIRE.
- CONSIDER COVERING ALL SOLDER JOINTS WITH TEFLON TUBING.
- CONSIDER A DUAL SECTION BOBBIN.
- CONSIDER LAYING KAPTON TAPE AFTER AND BETWEEN SOME NUMBER OF SEC LAYERS.
- CONSIDER WINDING THE SEC IN ONLY ONE DIRECTION TO REDUCE THE LAYER TO LAYER VOLTAGE BY HALF. THAT IS, START EACH LAYER AT THE SAME PLACE.
- CONSIDER ENVELOPING THE PRI AND SEC IN KAPTON TAPE WITHIN THEIR SEPARATE BOBBIN SECTIONS, AND COVERING WITH MULTIPLE TAPE LAYERS.
- CONSIDER POTTING THE ENTIRE TRANSFORMER IN A THERMALLY CONDUCTIVE, HIGH DIELECTRIC POTTING EPOXY.
- REMEMBER FLYING LEADS CONNECTED TO THE SEC SHOULD BE RATED FOR A HIGH ENOUGH VOLTAGE, WELL BEYOND THE SEC VOLTAGE VALUE.

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