Oscilloscope Current Measurement - Page 1
Forum section: electronics / beginners

author: wardenclyffe
author_rank: Contributor
timestamp: June 28, 2011, 07:48:36 pm
content: I was trying to see the current going through some LEDs with my oscilloscope the other day. I realized that the scope only measures voltage so I decided to insert a 1-ohm shunt in series with the LEDs and measure the voltage. The circuit uses an LC oscillator to drive a transistor and pulse current through the LEDs at about 700kHz so there is no current limiting resistor already present in the LED string.My question is why adding a shunt resistor is a common practice for measuring current with an oscilloscope. Why wouldn't it alter the current drawn? Or, if it does, is there a more accurate way to measure current with an oscilloscope (transformer, current probe ($$), etc.).

author: ejeffrey
author_rank: Super Contributor
timestamp: June 28, 2011, 09:06:43 pm
content: Current shunts are used because they are cheap, easy, and provide very predictable loading at all frequencies. You could use a transimpedance circuit to eliminate the voltage drop across the shunt, but then you would be limited by the amplifiers bandwidth. Above some frequency the virtual ground would start to look high impedance. Even a hall effect probe loads the circuit a tiny bit. You just can't measure without diverting some of your signal power to the scope. This is equally true of voltage measurements -- the 1 MOhm / 10 meg impedance of a scope / 10x probe is not infinite, and due to the shunt capacitance it can load your circuit quite badly at high frequency.

author: Unknown
author_rank: Unknown
timestamp: June 28, 2011, 10:27:35 pm
content: One issue with resistors is that you need a differential probe unless the circuit is floating or the shunt is on the low side (often inconvenient due to multiple paths to ground).More advanced than shunts are current transformers (used by AC-only current probes). They add some inductive loading, but can have bandwidths up to a few GHz. These tend to be relatively inexpensive (of course still much more than a resistor). For DC to a few hundred kHz, Hall effect sensors are used. For DC to something like 100MHz, there are probes that combine a Hall effect sensor with a current transformer. This is extremely complex to manufacture, since any gaps in the magnetic core will screw up the linearity. I believe Tektronix and Hioki make them, and Lecroy licenses them from Hioki(?). Especially the latter ones are quite expensive.

author: vk6zgo
author_rank: Super Contributor
timestamp: June 29, 2011, 02:47:47 am
content: If itisin series,by definition,it is a series resistor,not a "shunt".Of course it is "in shunt" with the Oscilloscope probe/'scope input!VK6ZGO

author: w2aew
author_rank: Super Contributor
timestamp: June 29, 2011, 02:54:49 am
content: ...and of course if your current "sense" resistor can't have one leg connected to ground in your circuit, and you don't have a differential probe, you can use two probes - one connected to either end of the resistor, then use the ADD and INVERT vertical functions to make a differential measurement.

author: Mechatrommer
author_rank: Super Contributor
timestamp: June 29, 2011, 07:03:13 am
content: Quote from: w2aew on June 29, 2011, 02:54:49 amthis is where floating/handheld dso excel, imo.

author: Unknown
author_rank: Unknown
timestamp: June 29, 2011, 10:25:04 pm
content: Quote from: w2aew on June 29, 2011, 02:54:49 amThis works just fine as a differential probe, the only issue is that common mode rejection ratio sucks at DC, and sucks even more at high(ish) frequencies.Quote from: Mechatrommer on June 29, 2011, 07:03:13 amExcept that the capacitance to ground tends to screw things up for anything resembling high frequency. Isolated channels also tend to increase noise, someone complained that the Tek TPS2000 series is basically unusable for measuring voltage across current sense resistors for this reason. The advantage of a real differential input is that the impedance of both inputs is symmetrical, both are high-impedance inputs.