Building a Capacitive Load for a High-Voltage DC Network Test Station

Test stations for high-voltage DC networks are complex undertakings, especially when the voltage exceeds the European three-phase voltage of 400V and a capacitance greater than 1mF is required. This was the primary challenge in building this test station. Additionally, a continuous current had to flow through load resistors, which added a computational challenge in charging, discharging, and balancing the capacitors.

Project Requirements:

DC Network Voltage: The load voltage should increase from 0 to 830V within 2 seconds.
Capacitance: The system should support capacitances of 1mF, 2mF, and 10mF.
Parallel Resistive Load: Each capacitance should be paired with a parallel resistive load of 8kΩ, 4kΩ, and 800Ω, respectively.
Current Measurement: Implement current measurement for the charging current.
Temperature Monitoring: Basic temperature measurement for capacitors and resistors within their operational limits.
Integrated Emergency Shutdown: Include an integrated emergency shutdown for all loads.

In this article, I will share my theoretical approaches to designing the load, the mistakes I made during development, and my best-effort solutions to resolve them. As is the practice in the company I work for: mistakes are to be experienced and learned from, and there's nothing that can't be fixed. So, feel free to share your suggestions on how I could have implemented this project better, but please, be fair!

Starting with a Block Diagram

At the beginning, there is a block diagram.

To protect company-sensitive information, I will only be publishing half of the block diagram here.

In the block diagram, we aimed to achieve the highest capacitance and the lowest resistance through simple parallel connections of the components. For this, we selected special high-voltage relays that offer adequate insulation and can interrupt DC currents up to 30A without sticking. Unlike AC current, which crosses zero amperes and allows switching without generating welding sparks caused by line-bound inductances, DC current does not have this characteristic.

Additionally, the block diagram includes a Hall sensor for current measurement, temperature sensors, and the emergency shutdown mechanism. I will briefly explain these circuit components. However, to avoid going off-topic, I will focus on the actual load and the challenges associated with it.

Let's Start with the Most Challenging Part: Capacitor Balancing

Why do capacitors need to be balanced, and where does this occur?

Here's the situation:

Affordable capacitors with capacitances in the range of several millifarads typically have a maximum voltage rating of 450V ± 15%, if you're lucky (calculation follows). Since 850V was required, this isn't sufficient.

So, what can be done? Capacitors can be connected in series to achieve a higher voltage rating.

How does this work? The electric field strength $E$ within a capacitor, which is responsible for a potential breakdown, is simplified as $E=U/dE = U/d$ , where $UU$ is the applied voltage and $dd$ is the distance between two equivalent plates, between which a uniform electric field is formed.

If two such plates with the same spacing are placed in series and connected by a short wire, the formula becomes $Eges=Uges/(2×d)E = U/(2 \times d)$ . The resulting field strength is reduced by half, allowing the voltage across the entire series connection to be double the original voltage while maintaining the breakdown voltage of an individual capacitor.

By the Way: I know there are some missing vector arrows in the mathematical expressions, but if you're looking for mathematically perfect formulas, then you're in the wrong place and should probably check out a physics blog. I'm an engineer, and for us, it's all about practical understanding and application.

Of Course, It’s Not That Simple…

When ideal capacitors are charged, the voltage across and the current through the capacitor behave as follows:

This is a typical e-function and after 5*T the capacitor is charged to the applied voltage. Butttt...In a real capacitor, there is an insulating material between the plates that, together with the flux density

$D$ in the capacitor, attempts to further reduce the prevailing field strength $E$ , according to the formula:

The advantage, as mentioned, is that the voltage withstand capability increases due to the material constant $ϵr$ . The downside, however, is the material itself.

When a capacitor with air as the insulating material is charged, charges Q accumulate on the positive plate, originating from the negative plate. Once the area $A$ of the positively charged capacitor plate is saturated, which is described by the flux density $D$ , the charging process is complete. This separation of charges, maintained by the air as the insulator, creates the electric field $E$ toward the negatively charged plate.

Since every insulator also has a conductivity $κ$ , partial discharges occur within any material, leading to a leakage current in the capacitor.

This fact highlights the first problem with series-connected capacitors. Each capacitor discharges by a voltage $Ux$ , but since the total voltage is applied and the loop equation

must be satisfied, one capacitor must have more or less voltage across it.

If a series connection of two capacitors, each rated for a maximum of 450V, is chosen to achieve a total voltage of 900V, it's possible that one capacitor could have 400V across it while the other has 500V. This would lead to the destruction of one of the capacitors.

Solution: Active Balancing Circuit

To solve this problem, the same voltage must be present across each capacitor at all times. According to Ohm's Law, this requires more or less current to flow through the resistor formed by the conductivity of the material, to compensate for the voltage asymmetry.

This is where balancing circuits, which can be either active or passive, come into play. An active solution might look something like in the picture on the right side.

The circuit discharges the capacitor with the higher voltage so that both voltages equalize. This is achieved by activating either $V1$ or $V2$ through the voltage divider consisting of $R11$ to $R16$ . The diode $D1$ ensures a stable base-emitter voltage.

This Active Circuit is Relatively Easy to Design for Smaller Capacities up to 100µF.

For larger capacitances, different MOSFETs must be used—ones that are not only fast but also have a suitable Safe Operating Area (SOA). At this point, it also becomes necessary to use more powerful resistors, whose form factor might no longer fit onto a PCB, as will be shown later.

Solution: Passive Balancing Circuit

In a passive balancing setup, only power resistors are used to discharge the compensating currents of capacitors that are either overcharged or undercharged. The circuit for this is relatively simple and looks as the circuit on the left side.

The resistors $R L_{1}$ and $R L_{2}$ are used to quickly represent the leakage current, while $R B_{1}$ and $R B_{2}$ are meant to represent the actual balancing resistors. The connection point between the resistors is used to create the balancing effect.

Source: Reese, V. (n.d.). Electrolytics. Massachusetts General Hospital. Retrieved from https://www.nmr.mgh.harvard.edu/~reese/electrolytics/

Circuit Explanation and Calculation:

Using a simple voltage divider, formed by RB1 and RB2, the total voltage of the network is divided in half, assuming the resistors have low tolerances.

This ensures that, in order for the loop equation to be satisfied, the voltages across the capacitors behave in the same way. This forces the balancing currents of the capacitors to flow through the resistors, as the voltage division across the resistors "forces" them to do so. This gives us the simplest balancing circuit for capacitive loads.

And here’s how to calculate it.

Tim Reese mentioned on his website that a few standard values are typically used for effective balancing. For example, the leakage current of a capacitor can be calculated based on its maximum voltage and capacitance as follows:

The solution of the leackage current is in when the capacity is in .

For the 10 mF capacitor specified in the requirements, a leakage current of 1060 is observed.

To ensure that any asymmetry is compensated by the balancing resistors the current through the resistors should be chosen to be 10 times greater than the leakage current.

Now we have the solution, that the resistors should be:

Tim Reese also points out that this design is too tight because the tolerance of such capacitors must be taken into account for the worst case. The electrolytic capacitors used in this case had a tolerance of 20%. This tolerance must be factored into the maximum voltage that can be applied to a single capacitor accordingly. With and the voltage overshoot is calculated as follows:

With N representing the number of capacitors connected in series.

This means that, in fact, 48 V less should be applied to these two capacitors connected in series.

Due to space constraints in the given enclosure, as will be seen later, it was nevertheless decided, after consulting with the experienced senior engineers, to connect only two capacitors in series. This effect was mitigated by carefully selecting the capacitors, adding high-power TVS diodes in parallel, and increasing the balancing current to reduce the impact. Otherwise, more capacitors would need to be connected in series and parallel to achieve the capacitances specified in the requirements.

Intelligent Parallel and Series Connection of Capacitors

To achieve the specified capacitance and resistance values, it was necessary to connect them as efficiently and cost-effectively as possible. In order to obtain total capacitances of 1mF, 2mF, and 10mF, it was already clear that these values, as shown in the block diagram, would be achieved through parallel connections.

So...

So, how do we maintain the necessary balancing circuit and generate the specified capacitance values as accurately as possible? Especially given the wide range of 450V capacitors and the maximum dimensions allowed by the enclosure?

Our designer quickly produced a drawing to determine what could be maximally achieved with the already selected components (the configuration of the resistive load was done in advance). The maximum parameters we were given for a single capacitor were a diameter of 30mm and a height of 60mm. In total, this amounted to 40 capacitors, along with the relays responsible for switching them in parallel.

The remaining challenge was how to arrange the connections. I thought to myself, why spend so much time browsing well-known distributors and search engines like Octopart, Mouser, etc., and trying to figure out how to connect each potential capacitor most efficiently? There are better ways nowadays that save both time and money.

In the age of ChatGPT, which our company has acquired in the full version, it is the task of an engineer to think creatively about what the final result should look like. You define the input parameters and the goal (the specification) and provide this information as accurately as possible to the AI. This way, engineers can spend more time thinking about essential aspects of the project. I believe this approach can bring us significant progress, helping us reduce overtime in the engineering profession and, as a result, make our families happier.

With a program I developed for the automated simulation of an integrated LLC transformer, I applied this approach and plan to continue doing so. But more on that in a future blog post...

So, how did it go with ChatGPT? I provided it with the mechanical and electrical specifications we had come up with, gave it the balancing circuit diagram as an image to follow, and instructed it to search Octopart for the most cost-effective configuration. After some discussions with it, which cost me an hour of work, we arrived at the following result:

1mF Load:

Composed of 4 capacitors
Each had a value of 1mF
Two 1mF capacitors in series result in 500µF
If you connect the same configuration in parallel, you achieve a total capacitance of 1mF

2mF Load: