3. The problem of the Open-Loop Startup of PCs

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Tags/Project: Operational-Notes-wiki

Metadata Field	Document Information
Status	Draft
Created	2026-09-24
Last Updated	2026-09-29
Author	Davide Cuneo
Rev. #	0.4
Reviewers	TBD

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Introduction and Context

This note describes a behaviour of current-controlled power converters that must be kept in mind whenever a PC is operated on a load with very low resistance and inductance, for example when its output cables are short-circuited to test the converter without the magnet. At startup, for a very short time, the converter is not yet regulating its current, and a small voltage applied to an almost-zero resistance can produce a large current peak. The note explains why this happens and how to avoid it with a series resistor.

How a Current-Controlled Power Converter Works

A current-controlled power converter does not impose a voltage: it imposes a current. To do so, it continuously measures the current flowing in the load and adjusts its output voltage to bring the measured current to the requested value (the setpoint). This is done by a feedback loop, typically a PID controller:

1. the converter measures the output current;
2. it compares it with the setpoint;
3. it corrects the output voltage accordingly;
4. the cycle repeats continuously.

As long as the loop is running, the current is under control. The key point is that this loop is not instantaneous: it needs some time to measure the current and compute the first correction.

The Open-Loop Startup Problem

At the exact moment of startup, before the feedback loop has read the actual current and computed a correction, the converter operates effectively in open loop. It applies a small initial base (or probing) voltage $V_0$, typically between 0.5 V and 1 V.

If the output cables are short-circuited, the only resistance in the circuit is the tiny parasitic resistance of the cables and of the internal stages of the converter. Applying Ohm’s law with a total resistance of, for example, 20 mΩ:

$I_{\text{peak}}=\frac{V_0}{R_{\text{int}}+R_{\text{cables}}}=\frac{1\text{V}}{0.02\Omega }=50\text{A}$

Even a probing voltage of only 1 V therefore causes an uncontrolled, immediate current surge of tens of amperes before the digital control has had time to react.

Why this is not a problem with the magnet

In a real circuit the current cannot jump instantaneously: the inductance of the load limits how fast it can grow. A superconducting magnet has a large inductance, so at startup the current rises slowly and the feedback loop has all the time it needs to take control. The problem appears with loads that have both very low resistance and very low inductance, such as a short circuit made with the output cables.

Solution: A Series Resistor

The problem is solved by inserting a resistor in series with the circuit. With a known resistance in the loop, the initial peak is physically bounded by Ohm’s law, whatever the converter does before the regulation starts.

Consider a load of $R=0.1\Omega$, a probing voltage $V_0=1\text{V}$ and a requested setpoint $I_{\text{set}}=2\text{A}$:

• Startup step ($t=0$). The converter is still in open loop and applies $V_0$. The current is limited to:
  $I_{\text{initial}}=\frac{V_0}{R}=\frac{1\text{V}}{0.1\Omega }=10\text{A}$
• Regulation step ($t>0$). A few milliseconds later, the controller reads 10 A, which is above the setpoint of 2 A. The algorithm then reduces the output voltage to:
  $V_{\text{regulated}}=I_{\text{set}}\cdot R=2\text{A}\times 0.1\Omega =0.2\text{V}$
  and the current settles at the requested value.

The peak drops from about 50 A (short circuit) to 10 A, and it can be reduced further by choosing a larger resistor.

Table 1: Initial current peak for $V_0=1\text{V}$ and different series resistances.

Total resistance	Initial peak $I=V_0/R$
0.02 Ω (short circuit, cables only)	50 A
0.1 Ω	10 A
1 Ω	1 A

Sizing the Resistor

Three conditions should be checked when choosing the resistor.

1. Limit the startup peak. The resistance must be large enough to keep the initial peak at an acceptable value:

$R\ge \frac{V_0}{I_{\text{peak, max}}}$

2. Stay within the voltage capability of the converter. At the highest test current, the converter must be able to provide the voltage needed across the resistor:

$V=I_{\text{max}}\cdot R\le V_{\text{max, PC}}$

3. Withstand the dissipated power. The resistor turns the electrical power into heat and must be rated for it:

$P=R\cdot I^2\text{or equivalently}P=\frac{V^2}{R}$

For example, running a continuous test at 3 A through a 1 Ω resistor requires:

$P=1\Omega \times (3\text{A}{)}^2=9\text{W}$

A resistor rated for at least 15–25 W should then be used, such as a chassis-mounted power resistor, to keep a safety margin.

Power rating of chassis-mounted resistors

The nominal power of chassis-mounted resistors is usually given for the resistor mounted on a suitable heat sink. Without it, the power it can actually dissipate is much lower: check the datasheet and keep a generous margin.

Feedback

Found an error or something outdated in this note? Please contact Davide Cuneo (davide.cuneo@physics.uu.se).