Consider the following circuit:

Edited schematic with further information [edited schematic]

DC_gen1 and DC_gen2 are two +28V dc ideal generators. Fuse is real component and it has a finite and small resistance (0.1 Ohm for instance).

I need to design a Sense circuit that detects that the fuse is broken. I have already considered a few possibilities:

  • A differential amplifier could sense a voltage across the fuse. Anyway, since the two generators have the same voltage, no current is flowing through the fuse and therefore there is no voltage drop on it.
  • A galvanically isolated power supply could inject a current in A point to B point to create a voltage drop across the fuse. Anyway, the injected current would flow through DC_gen1, ground and DC_gen2, and therefore no voltage drop would be createdo on the fuse.

Can you please suggest how to sense the broken fuse?

EDIT: Following comments, I edited the circuit to add details:

  • The two generators are connected through a long power cable; the purpose of the fuse is to protect DC_gen2 from short circuits on the cable itself. DC_gen1 already implements its own protection so no fuse is required.

  • Due to physical constraints, the sense circuit must connect next to the fuse, at point A and B.

  • 2
    \$\begingroup\$ Can you tolerate a little inductance on each side of the fuse? If so you can inject an AC across the fuse. \$\endgroup\$
    – Transistor
    Commented Aug 7, 2019 at 15:43
  • 2
    \$\begingroup\$ The fuse is there to protect against imbalance between DC_gen1 and DC_gen2, so why not just monitor their output voltages directly? If one drops out, the fuse will blow, if one goes too high the fuse will blow. You're technically not sensing if the fuse has blown, but you'd accomplish the same thing. \$\endgroup\$
    – Stiddily
    Commented Aug 7, 2019 at 15:51
  • 7
    \$\begingroup\$ Then please explain the reason for the fuse. At the moment, it is unclear. \$\endgroup\$
    – Andy aka
    Commented Aug 7, 2019 at 16:08
  • 1
    \$\begingroup\$ Perhaps a shunt in series with the fuse then? There has to be an imbalance between the two at some point otherwise the fuse would never open. You'd be able to detect that there was an over current event as the fuse opened. Even instant blow fuses have a trip time of several milliseconds. \$\endgroup\$
    – Stiddily
    Commented Aug 7, 2019 at 16:52
  • 1
    \$\begingroup\$ Thevenin equivalent seen from your "sense circuit" is a plain short circuit, no matter the fuse is ok or not. This rules out any chance to detect anything. I am afraid you have to release some idealities and clarify what are you really trying to do \$\endgroup\$
    – carloc
    Commented Aug 7, 2019 at 21:51

6 Answers 6


Banish your anxiety and shed your stress. Use a miniature circuit breaker with auxiliary contact.

enter image description here

Figure 1. MCB with NO and NC auxiliary contacts.

Monitor the auxiliary contacts and enjoy the benefits of complete isolation of your monitoring circuit and the 28 V power. Fixing the fault is a matter of resetting the breaker. No fuse to stock.

Note that you'll need a DC-rated MCB. (The one in the photo is AC, I suspect.)


As suggested by this question, the proper way is the four wire method, connecting two cables for introducing a small controlled current and the other two for measuring the small observed voltage.

The value of the resistance will tell if the fuse is proper for its nominal capacity.

Another choice, also suggested in there, is using a Wheatstone Bridge with three known and fixed resistors close to the fuse resistance, and putting the DUT fuse as fourth arm of the bridge, apply a small known voltage at one side of the bridge, and measure the voltage in the other side.


Edit: I forgot to add a choke in series to the \$28V_{DC}\$ bus: this is used to prevent the problem which affects the galvanically isolated DC sensing circuit.

In safety railway electronics, the problem of univocally identify the presence of a conduction path occurs very often. A typical example is the problem of finding the state of the (usually remotely located) switch associated to every track branch of a lay-by: a closed switch means that the train will pass on the associated track branch, so being able to safely check this is of paramount importance.
A typical solution adopted in order to solve such kind of problems is to use a Meissner oscillator where the switch (or the conductive path to be checked) is part of the isolated \$LC\$ resonator: a schematics commonly adopted is the following one, which I simply adapted to the specific problem of broken fuse detection we are dealing with


simulate this circuit – Schematic created using CircuitLab

The working principle of the circuit is conceptually simple: it oscillates due to the phase inversion produced by the interstage coupling transformer \$\mathrm{TR_1}\$ (note the \$180°\$ phase difference between the primary winding \$\mathrm{W_p}\$ and th secondary wingding \$\mathrm{W_{s2}}\$) of at the frequency $$ f_{osc}\simeq\frac{1}{2\pi\sqrt{L_\mathrm{TR_1}C_1}}\sqrt{1+\frac{R_\mathrm{Fuse}+R_{L_\mathrm{TR_1}}}{r_o}} $$ where \$L_\mathrm{TR_1}\$ and \$R_{L_\mathrm{TR_1}}\$ are respectively the inductance and resistance of the secondary winding \$\mathrm{W_{s2}}\$, while \$r_o\$ is the collector output dynamical resistance, by which \$Q_1\$ drives the primary winding (unfortunately multi-winding transformers symbols are currently not available in Circuit Lab, so it may seem that we are using two transformers, but it is not so)

  • If the Fuse is not broken, then the oscillator oscillates at its characteristic frequency.
  • If the Fuse is broken, the the circuit does not oscillates.

Usually the value of the components in the circuit are chosen such that \$f_{osc}\$ ranges from few hundred hertz to few thousand hertz. The oscillating signal can be taken from the collector of \$Q_1\$ or from yet another auxiliary winding: properly rectified and loaded allows the detection of the state of the fuse.


  • Obviusly the choke inductance must be chosen in order to verify the following relation $$ |Z_{L_\mathrm{choke}}|_{\omega=\omega_{osc}}=2\pi f_{osc}L_\mathrm{choke}\gg R_\mathrm{Fuse}. $$ Doing so, the dynamical "shorting" of the fuse and consequently missing detection of its breaking is avoided.
  • The interstage transformer \$\mathrm{TR_1}\$ is the trickier component to design: the coupling between the winding \$\mathrm{W_p}\$ and \$\mathrm{W_{s2}}\$ should be the lowest possible (meaning that the mutual inductance \$M_\mathrm{ps2}\$ between these two winding should be very low), while the coupling between \$\mathrm{W_p}\$ and \$\mathrm{W_{s1}}\$ and between \$\mathrm{W_{s1}}\$ and \$\mathrm{W_{s2}}\$. However, it could be designed in a way such it adds a very high galvanic isolation: in railway applications, \$3\mathrm{kV}_{AC}\$ double isolation is a commonly reachable value.
  • There is no need to put the circuit near the fuse: it can be placed remotely, provided the proper \$f_{osc}\$ is chosen.
  1. Put a voltage divider across each side of the fuse. Select the resistor divider so that the total resistance is around 500K Ohms, and the output in the center of the divider is 5 V.

  2. Feed each output to the discrete inputs of a microcontroller

  3. Code the microcontroller to compare the two inputs to detect when the voltage at the load side of the fuse drops to zero volts when the fuse opens.

  4. If the input that monitors the load side of the fuse is zero volts and the input that monitors the line side of the fuse reads 5 v the fuse is open.

  5. If both are zero volts, the circuit is not powered.

  6. If both inputs read 5 V the fuse is ok.

  • \$\begingroup\$ I don't think that this will work (at least how you describe it). Both sides of the fuse are powered and there is also a load either side of the fuse. Even if you remove the fuse (or it blows) there will still be (more or less) 5V at both inputs of the µController. The "zero volt" situation will not happen. \$\endgroup\$ Commented Jul 3, 2023 at 18:24

\$I^2(t)_{[Amp²]}*R_{{[Ω]}_{(T)_{fuse}}}*t_{[s]}\$ roughly defines the fusing of a short thin metal fuse that has "low resistance" yet much higher than your source and load resistance to become the hottest link with a short thermal time constant but hold the rated loop fuse current and chosen to withstand the rated applied AC or DC current and eventually fuse open when exceeded. DC circuits must be for fusing are derated for voltage due to back EMF voltage and circuit inductance.

A fuse can never determine if the source voltage has dropped due to an open or short circuit in your feed cable thus your goal may be correct, but a fuse is the wrong solution.To accomplish protecting your Gen2 from a short circuit in the feed cable is "already" defined by you as protected by Gen1. However an intermittent cable-short can create \$V=L*dI/dt\$ high voltage from the sum of {cable & Gen1 inductance} =L, when the circuit becomes unshorted and reconnected. This requires an over-voltage clamp (TVS or MOV or gas tube) that can handle the energy,E absorbed from breaking open the shorted current \$E=½LI^2\$. Gas tubes are also negative resistance devices with some inert gas and thus must be fused before the gas tube in either AC or DC for over voltage thus MOV, TVS are more desirable with much faster than xx microsecond gas-tube ionization response times as well which can be useful to ignore sub-microsecond glitches.


A fuse cannot detect a short-circuit upstream in your cable. You want to fuse it upstream at Gen 1 before cable and consider OVP ( over-voltage protection) in case of intermittent shorts.

  • define the specs for your circuit protection with parameters like UVP, OVP, OCP and transient energy clamping
  • define your failure detection circuits as well.
    • a passive fuse detection can simply be a current limiting resistor with 60V neon bulb and or a 2V red LED.
    • active fuse open or cable short-cct detection requires an standby voltage to back-drive a low voltage that draws little load current thus the cable source impedance will either be detected as near 0V due to a short or near applied voltage due to an open circuit.

You can add LED with series resistor (parallel to fuse) so that when fuse breaks current will flow through LED and it lights up. You need two of them in opposite direction so that imbalance in any source will get detected.

  • \$\begingroup\$ "Ideal generators." No potential difference at all, never mind enough to light an LED. Welcome to EE.SE. \$\endgroup\$
    – Transistor
    Commented Aug 7, 2019 at 16:00

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