uCurrent Gold current probe? - Page 1
Forum section: electronics / beginners

author: alin_im
author_rank: Regular Contributor
timestamp: January 11, 2016, 11:32:31 am
content: Hey guys,I want to buy a uCurrent Gold and I have a question. Can I use it to display the current consumption (2 - 70 mA) of a micro controller on an oscilloscope (rigol ds1054z) ?Regards,Alin I.

author: bitshift
author_rank: Regular Contributor
timestamp: January 11, 2016, 12:18:59 pm
content: According to Ohms law, if you put a current through a resistor there will be a voltage dropped across the resistor. To measure current, an ammeter typically puts a resistor in series with the load and then measures the voltage drop across this resistor. If you know the voltage drop and the resistance you can calculate current. The resistor in an ammeter is called a shunt resistor.Lets take the scenario where you want to calculate the current a widget you are designing consumes. Lets pretend the input voltage to your widget is 5V and it behaves like a resistor with a value of 5 ohms. Using ohms law you calculate that the current through your widget should be 1A.To validate this calculation, you attach an ammeter in between the input voltage of your power supply and the input terminal of your widget. For illustration lets assume the value of shunt resistor in the ammeter is 1ohm. Now there is 5V across 6 ohms (5 ohms of the widget and 1 ohm of the ammeter) instead of the intended 5ohms meaning that the current draw is 0.83A instead of the expected 1A. As you can see this resistor in the ammeter can greatly skew your measurements. The technical term for the voltage dropped across the shunt resistor is called burden voltage.The more current you put through the shunt resistor of the ammeter the greater the burden voltage will become. For example, if you are trying to measure 190mA on the 200mA range of your multimeter, the large voltage drop may make the measurement very imprecise. When trying to measure smaller and smaller currents, the effect of the burden voltage becomes greater because of the change in voltage to the load that it creates.The uCurrent was designed to have a low burden voltage which allows you to make very accurate measurements of small currents. In this case when measuring 2-70mA, a standard multimeter like a Fluke 87 can be used to measure the current fairly accurately. A low burden voltage measurement is not required.If you don't have a standard multimeter, I would strongly recommend purchasing one instead of a uCurrent since it will allow you to much more than just measure current.

author: rs20
author_rank: Super Contributor
timestamp: January 11, 2016, 12:39:38 pm
content: alin_im: Yes, that's right, you can see a graph of time vs current that way. But be aware that the bandwidth of the uCurrent gold is limited to 300 kHz or so.bitshift: Burden voltage does not reduce accuracy; in the sense that the displayed reading is not erroneous due to burden voltage. The issue is that when you measure current in a circuit, you have to break a wire link and replace it with the ammeter (with its burden voltage) -- and this affects therest of the circuit(changing the actual current, and the reading along with it). It's not the accuracy of the meter that's at issue, it's the sensitivity of the circuit to supply voltage. And without knowing the OP's usecase or intended accuracy, you can't presume to claim that the burden voltage at 70mV will be acceptable? 87V's shunt resistance of 1.8 ohms will cause a 120mV shift in supply voltage; if that supply voltage is 1.8V, you're talking an error of nearly 10%, assuming the device even works anymore!

author: Brumby
author_rank: Unknown
timestamp: January 11, 2016, 12:43:56 pm
content: I don't think there was ever any inference that there wasinaccuracy in thereading(the value displayed on the meter), just that there would beinaccuracy in themeasurement(the current flow in this arm of the circuit).

author: alin_im
author_rank: Regular Contributor
timestamp: January 11, 2016, 01:09:59 pm
content: Quote from: rs20 on January 11, 2016, 12:39:38 pmThank you for your response, I do not think I will have trouble with the bandwidth, because I will measure the current consumption of the MCU, when it is awake consumes 70 mA for 2 -3 seconds and then goes to sleep for 58 seconds and consumes 2 mA (I am not so picky about the 100% power consumption wave reproduction, that has frequencies over the 300 kHz range (like noise or things like that)). Plus I wanted to buy this from a long time, to measure the MCUs and transistors currents.At the moment the maximum frequency that I use is 100 kHz if I want to modulate a signal of 1 kHz, so I again the frequency range will not be a problem.I do not afford to buy a better multimeter (like a 87, I am still a student), so this is the best solution that I could think and also it is relatively affordable for my budget.Also I wanted to put some nice screenshots of the power consumption of my project from an oscilloscope, instead of making them in excel or other graph program generator.Do you think I do the right thing ?

author: bitshift
author_rank: Regular Contributor
timestamp: January 11, 2016, 05:55:37 pm
content: Quote from: rs20 on January 11, 2016, 12:39:38 pmThat is precisely what I was trying to convey. I will make an effort to make my posts more clear and concise.Quote from: rs20 on January 11, 2016, 12:39:38 pmVery true. In that case the uCurrent would come in handy. Thank you for pointing out the error. Helping others helps to solidify ideas in my head and this takes it one step further

author: bitshift
author_rank: Regular Contributor
timestamp: January 11, 2016, 06:08:14 pm
content: Quote from: alin_im on January 11, 2016, 01:09:59 pmA very cheap option would be to get a 1% 1ohm resistor and put that in front of the MCU. Connect one probe of the oscilloscope to one side of the resistor and another probe to the other side of the resistor. Then if you turn on the math function A-B, the resulting wave will be the current where 1V = 1A.

author: MLXXXp
author_rank: Frequent Contributor
timestamp: January 11, 2016, 06:10:54 pm
content: If the power supply of the circuit that you're measuring isn't isolated/floating, or if you're measuring voltage or something else on another channel on the scope, be careful of how you hook up the uCurrent. The uCurrent's input and output minus terminals are tied together.https://youtu.be/xaELqAo4kkQ

author: alin_im
author_rank: Regular Contributor
timestamp: January 12, 2016, 12:07:13 am
content: Quote from: bitshift on January 11, 2016, 06:08:14 pmThank you for idea !!! If I want 1 V = 1 mA, I need to use a 1 k resistor and If I want 1 V = 100 uA, I need to use a 10 k resistor right

author: rs20
author_rank: Super Contributor
timestamp: January 12, 2016, 03:20:30 am
content: Quote from: alin_im on January 12, 2016, 12:07:13 amIf you want 1 V = 1mA, then you need to use a 1k resistor, correct. However, keep in mind the issue of burden voltage: if you have 3mA flowing through your 1k resistor, then you measure 3 volts, but that 3 volts is deprived from the circuit under test. For example, if your microcontroller has a nominal 5V supply which is instrumented with a 1k resistor, and it draws 3mA, there will be a 3V drop across the resistor and the microcontroller will only see 2V (5V - 3V), which may prevent the MCU from working at all, let alone get an accurate reading.The whole idea if the uCurrent is that it uses a 100x amplifier inside, which allows a 100x smaller resistor and a 100x lower burden voltage while giving the same V/A ratio. For example, with the same example as above, if you used the uCurrent with its internal 10ohm resistor, that gives 1V = 1mA after the 100x amplifier, but the burden voltage is only 0.03V and the MCU will see 4.97V, easily within spec.You can do calculations on how much burden voltage you're able to tolerate, and do experiments to see how noisy your oscilloscope is. This way, you can figure out if a simple resistor is a suitable solution, or if you need the uCurrent.

author: Brumby
author_rank: Unknown
timestamp: January 12, 2016, 04:35:44 am
content: Quote from: rs20 on January 12, 2016, 03:20:30 amI repeat this statement, because it is key to proper understanding of current measurements.

author: Hobby73
author_rank: Regular Contributor
timestamp: April 02, 2016, 06:43:32 am
content: Great thread!To phrase my question, let's assume I'm planning to purchase uCurrent to measure low current in low power DC circuits. And I also need to purchase a DMM.If I want to get precise uA and mA current readings, would the best match of the DMM with uCurrent be a DMM unit which has high accuracy and precision for lowvoltageranges since uCurrent connects to the voltage input of the DMM? In this case, the DMM specs for DCcurrentare unimportant to me because I wouldn't be using the DMM to measure current directly. So I don't even have to be concerned about the current measuring accuracy or voltage burden of the DMM since I'm relying on uCurrent. Do I understand this correctly?

author: Dave
author_rank: Super Contributor
timestamp: April 02, 2016, 09:43:10 am
content: Yes, that's correct. You're only interested in the accuracy of the voltage range that covers +/- 1V.

author: paulgdonnelly
author_rank: Newbie
timestamp: June 08, 2016, 10:54:13 am
content: uCurrent Gold sounds like just what I've been looking for when measuring current consumption of BLE 4 devices running of CR2032 cells. However, I'm having trouble tracking down a supplier. Anyone still shipping them?RegardsPaul

author: alin_im
author_rank: Regular Contributor
timestamp: June 08, 2016, 11:01:19 am
content: Quote from: paulgdonnelly on June 08, 2016, 10:54:13 amAn UK supplier is this one:http://www.sparkylabs.co.uk/shop/product_info.php?cPath=30&products_id=146

author: vidavidorra
author_rank: Newbie
timestamp: July 21, 2016, 08:59:22 pm
content: The UK is out of stock. Is there any news whether there are new uCurrent being made (anytime soon) and if they're reselled in the UK?

author: 3db
author_rank: Frequent Contributor
timestamp: July 21, 2016, 09:17:19 pm
content: There's seems to be a long lead time for the uCurrent@bitshift I think the shunt resistor and using the scopes math function was a good solution for a "young player"