Choosing a high-voltage tester starts with AC or DC

When selecting a high-voltage tester, people often start with the specifications.
How many kV can the system deliver? Which options are included? Does it fit the budget?

But starting there involves a major risk.

The most important choices are made before you even look at voltage.
In fact, if those initial choices are wrong, even the best tester will be unsuitable in practice.

The correct order is always:

1. What does the standard require
2. AC or DC
3. How much test current do you need
4. Only then: maximum voltage, configuration, and automation

In this blog, we explain why this sequence is crucial, how AC and DC behave technically, and how simple calculation examples help you avoid selecting a tester that looks right on paper, but fails in real-world use.

A high-voltage test is not a “measurement” like a multimeter performs.
It is a stress test: you subject the insulation system to stress to see whether it fails.

And the way you apply that stress, with alternating voltage or direct voltage, makes a fundamental difference.

AC high-voltage test: dynamic loading of insulation

During an AC test, the insulation system is continuously charged and discharged, typically at 50 Hz.
That means: the electric field goes through 50 full cycles per second, so it reverses direction 100 times per second.

Electrically, almost every product behaves like a capacitor.

What does that mean in practice?

Even if the insulation is perfect, current flows.
That current consists of:

• capacitive current
• dielectric losses
• any real leakage or breakdown currents

A basic AC tester cannot separate these components and sees one total current. Only testers that measure the resistive (real) current separately show that part on its own.

Calculation example 1 – why AC suddenly “requires a lot of current”

The capacitive current with AC can be calculated using:

I = 2 · π · f · C · V

Assume:

• frequency = 50 Hz
• test voltage = 2 kV AC
• product capacitance = 100 nF

Step by step:

• 2 · π · 50 ≈ 314
• 314 · 100 nF = 314 · 100·10⁻⁹ = 3.14·10⁻⁵
• 3.14·10⁻⁵ · 2000 V ≈ 0.063 A = 63 mA

Practical:

• A product that “only” requires 2 kV already draws more than 60 mA.
• Many compact AC testers already reach their limits here.

What happens when the voltage increases?

Same product (100 nF), but now at 10 kV AC:

• 3.14·10⁻⁵ · 10,000 V = 0.314 A = 314 mA

And that is exactly why:

• “maximum kV” says very little
• maximum current at that kV is decisive

When do you choose AC?

AC is usually the right choice when:

• standards prescribe it (e.g. IEC 60335, 60601, 61010)
• you want to perform dielectric strength (withstand) tests

AC tests are normatively strong, but technically demanding.

During a DC test, the product is charged once up to the test voltage.
After that, the voltage remains constant.

After the charging phase, you see almost exclusively:

• real leakage current through the insulation, plus a slowly decaying absorption current

This makes DC testing easier to interpret, but not automatically simpler.

Calculation example 2 – DC charging current

The charging current with DC follows:

I = C · dV/dt

Assume:

• C = 100 nF
• you want to reach 10 kV

If you ramp up in 1 second:

• dV/dt = 10,000 V/s
• I = 100·10⁻⁹ · 10,000 = 1 mA

If you ramp up in 0.1 seconds:

• dV/dt = 100,000 V/s
• I = 10 mA

Practical:

• DC requires less continuous current than AC
• but fast ramps do require peak power

In production, testing is often done quickly, and the DC source must be designed for that.

Energy: the underestimated difference between AC and DC

With DC, energy is stored in the product.
That energy must be safely dissipated after the test.

The stored energy is:

E = ½ · C · V²

Calculation example 3 – stored energy

C = 100 nF, V = 10 kV:

• E = ½ · 100·10⁻⁹ · (10,000)²
• E ≈ 5 joules

C = 1 µF, V = 10 kV:

• E ≈ 50 joules

Practical:

• 5 J is already dangerous
• 50 J is serious energy
• discharge is not a side issue, but a core function of the tester

Discharge: why DC sometimes feels “slow”

Discharge is often performed through a resistor.
The time constant is:

τ = R · C

After approximately 5τ, the voltage is essentially gone.

Calculation example 4 – discharge time

R = 100 MΩ, C = 100 nF:

• τ = 10 s
• 5τ = 50 s

Practical:

• With larger capacitances, discharge determines the cycle time
• Professional DC systems therefore use active discharge circuits and monitoring

Step 3 – The real decision question: how much current do you need?

Now it becomes clear why this order is so important.

With AC

• The required current is determined by C and V
• You can calculate this in advance
• Too little current → unstable test, false rejects

With DC

• Continuous current is lower
• But ramp speed, energy, and discharge define the requirements
• “Lightweight” DC systems make testing slow or unsafe

Imagine:

• AC tester: max 100 mA at 5 kV

What is the maximum capacitance you can test?

C = I / (2 · π · f · V)

• C ≈ 0.1 / (314 · 5000)
• C ≈ 64 nF

If your product exceeds this, you will hit the current limit, even if the tester can reach a higher voltage on paper.

Conclusion: think first, then specify

You don’t select a high-voltage tester based on kilovolts alone.
You select it based on electrical behavior.

The correct order is always:

1. What does the standard require
2. AC or DC
3. Required current and energy
4. Only then voltage level and configuration

Those who follow this sequence:

• avoid wrong investments
• achieve stable and repeatable tests
• and build test setups that scale with future needs

At IONIO, we therefore never start with the instrument. We start with the product and calculate it first.

IONIO can help you work out your test requirements based on your product, standard and capacitance. View the mobile DC high voltage tester HP80KV or contact IONIO for a no-obligation consultation.