The attached is one board, a MOSFET driver, from an old power amplifier in our lab that unfortunately suffered a mini-fire recently. (Prior to its untoward warming, it operated from 160V rails and had a max current output of ~75A DC, with a O(100 kHz) bandwidth). The board design is from about 1990.

What is the 4-lead component in the green box? It's unlabelled, and as far as I can tell there is a negligible resistance across all four pins. It looks like two thick, plated wires joined by something chunky and wrapped in black insulation.

Am I being dumb? Is it just a jumper?

enter image description here

  • \$\begingroup\$ What is the dark thing bridging the jumpers? \$\endgroup\$ – Sunnyskyguy EE75 Oct 2 '17 at 22:23
  • \$\begingroup\$ That was the question I was asking -- duskwuff/we think it's a low value, high-current resistor \$\endgroup\$ – Landak Oct 2 '17 at 22:25
  • \$\begingroup\$ It's hard to tell the material coating or contents. Copper , or axial capacitor? \$\endgroup\$ – Sunnyskyguy EE75 Oct 2 '17 at 22:29
  • \$\begingroup\$ There are a similar pair of resistors to the right without jumpers too and an SOT bias R added to a testpoint. But this thing looks like a plastic coating or something burnt \$\endgroup\$ – Sunnyskyguy EE75 Oct 2 '17 at 22:49

enter image description here

Figure 1. A 4-terminal current shunt.

The highlighted item is a 4-terminal current shunt. The two terminals circled in red are the current terminals. The two circled in blue are the voltage measurement terminals.

This arrangement is known as the Kelvin (4-wire) Resistance Measurement technique.

enter image description here

Figure 2. The Kelvin (4-wire) Resistance Measurement.

The principle of operation is that the voltmeter circuit draws such a little current that wire resistance does not induce any voltage drop. The voltage reading, therefore, is a true reading of the voltage across the shunt resistor.

On the circuit pictured note that the voltage readings are taken inside the current circuit soldered joints so they and the loops of wire to the actual shunt do not affect the reading.

  • \$\begingroup\$ Do you mean that the thing described as "something chunky and wrapped in black insulation" between two thick jumpers is the actual shunt? Why would this precision thingy be in a power amplifier? Some feedback? \$\endgroup\$ – Ale..chenski Oct 2 '17 at 23:18
  • \$\begingroup\$ OP doesn't specify what the thing was amplifying or driving. "160V rails and had a max current output of ~75A DC, with a O(100 kHz) bandwidth)" doesn't sound like audio. I'd say it's a programmable PSU or motor driver in which case the shunt would be monitoring output current. The PCB traces in the red circles look light for 75 A but maybe they're beefed up on the underside. I have no idea what the coating is. \$\endgroup\$ – Transistor Oct 2 '17 at 23:23
  • \$\begingroup\$ Maybe some plasma deposition or laser cutting equipment... \$\endgroup\$ – Ale..chenski Oct 3 '17 at 0:25
  • \$\begingroup\$ @Transistor -- Yep, you're right: it's a gradient amplifier for an MRI scanner. I also thought the traces looked a little light, but that entire board is mm away from a gigantic heatsink, bonded via its MOSFETs. There are larger traces on the underside, as you can see by comparing the top track by J19 with a 5V TTL at U103. A shunt is almost certainly correct -- the device has overcurrent protection! \$\endgroup\$ – Landak Oct 3 '17 at 8:08

I suspect that's a current shunt. Basically, a low-value resistor with extra terminals to make it easy to measure the voltage across it.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.