I was wondering if an AC clamp meter could be used to approximate the current through a motor using this method:
Using a PWM-based motor controller, set the duty cycle to 50%. On a Jaguar, this should yield a square wave at about 30kHz.
Now, put the current clamp around ONE of the motor leads, and take your reading.
Will the device give an accurate measurement of AC current? Could this value easily be converted to DC current?
My biggest concern is that this signal will be well out of the operating range for the current-clamp device, since most are made for 50-60hz.
For those of you that are wondering “won’t the coils in the motor smooth out that AC to DC purely through inductance?”, I’ve looked at the waveform from a Victor to a fisher price motor. There’s definitely inductive ring, but the square wave is still very well defined.
I’d go with something that is designed for higher frequencies. I had some samples of http://www.amveco.com/pdf/AC1005.pdf a while back, they did ok for a similar frequency (totally different application though). Maybe find its bigger brother?
And remember, the victor’s chop rate is two orders of magnitude slower than the Jaguars. Check the waveforms.
PS: I’m no longer at TI - I’m at NI now . You may want to resubmit your query
Marshal,
The clamp type meters I have used in the past have a max frequency of about 400Hz. That is OK for the 150 Hz switching frequency of the Victor. The Jaguar is 15 kHz and although it may show some current on the meter, it will not be accurate and will show considerably less than actual. The type of current transformer Eric linked to also has a frequency response characteristic that will play into the readings. The sheet does not spec the response but I am guessing it is optimized for 60 Hz.
You are right, I linked the wrong part. That is the 50/60Hz I used on the AC mains. I’ll try to find the one I used for the 10kHz circuit later, but I’ll likely be without chiefdelphi access for a while…
What about using the power inputs to the victors/jaguars instead of the outputs? We have a DC clamp meter and use it on the inputs at least once a year with no apparent issues. In fact, two years ago it highlighted an alignment problem that was causing a lot of friction in one of our systems, and let us fix it before competition.
The input current still should reflect the output current switching waveforms. We used a current sense circuit many years ago called StangSense. The current waveform was then filtered and processed to provide a usable output. http://www.chiefdelphi.com/forums/archive/index.php/t-3302.html
We have not tried the circuitry on a Jaguar equipped robot. As there is considerable rise time issues at low throttle (small PWM pulses), an implementation would need some massaging to be accurate. It is my understanding that the current sense in the Jaguar is primarily looking at peaks not average currents. It is a simple resistor in one of the power supply leads. The current flow produces a voltage drop across the resistor which is then converted and used by the onboard processor.
Don’t forget about the capacitor. The input should be somewhat smoother than the output. Just based on the noise of the DC-DC power supplies on the PD board, I haven’t noticed a difference when the Jaguars are drawing a load compared to when they aren’t. I realize we’re looking at current and not voltage, but it seems the 5v regulators on the DSC and AB put more noise on the line than the Jaguars.
So the answer is no, a common AC clamp meter will not work. Thanks.
Now on the theory side, is 50 watts AC the same amount of power as 50 watts DC?
Does the shape of the waveform affect the accuracy of a reading? (Do most meters assume it’s a low-noise sine wave?)
That is a really cool project Al. I’m going to have to bring it to the Robettes attention… I’m already thinking of some situations where something similar could be very useful while prototyping/building.
50 watts of power is 50 watts of power in AC, DC, mechanical, etc. Though 10Amps @ 12VDC is NOT the same as 10Amps @ 12VAC. Assuming the AC is a sine wave, it contains about 71% of the power as the DC line. This comes from take the RMS of the line voltage (and/or current), this can also be done for the PWM square-wave pulses a Jag will output. Yay math!
The shape of the waveform should not affect the accuracy of a reading because the sensor does not care about the shape of the wave. What could be a problem is sampling at too low of a frequency. If the Jag outputs a 15KHz square wave and your current probe reads at 15Khz you will either see every reading at full voltage and current, or every reading at zero voltage and current. One can usually get away with sampling at 2X the signal frequency for sine-wave outputs, but a safe bet is 4x the signal frequency. That means you would be looking for a probe capable of 60+KHz to get good readings.
Don’t forget that the cRIO is designed to be a high-speed DAQ board, you could measure voltage drop over a known resistance. The old-school way to do this is to measure the differential voltage on either side of a precisely machine piece of electrical-grade copper. The NI modules are a little pricey, but you can get up to 1MHz sampling rates and they’d probably give a nice discount to FIRST teams.
Marshall,
The cap will take away some of the voltage fluctuations as the controller supplies current to the motor. The majority of the motor current comes from the battery. If you think in terms of impedance and where the current is flowing, it might help. The battery is a nearly infinite supply of current (compared to the cap in the controller) at a relatively low impedance, about .011 ohms. When you add in the controller, the cap now sits across the battery as a load. There is a small shot of current as the cap charges but then it simply idles at the current of the controller. When you turn on the motor, it is now in parallel with the cap. The cap will supply some current but the battery supplies the majority. However, there is the series resistance of the wiring. As this resistance increases (small wire, poor crimps, etc), some voltage drop will occur at the input to the controller. The cap is designed to supply current to the electronics to keep that voltage drop from affecting the regulators on board the controller.
DC and true RMS are the same. 12 volts AC RMS is 34 volts peak to peak or 17 volts peak. Please be sure to check your meters, most are calibrated to make an equivalent RMS reading on 60 Hz, sinusoidal voltages only. Change the frequency or the waveform and the meter will lie to you.
Al, you’re multiplying the AC RMS voltage by the square root of two to get the AC peak?
Or are you just dividing a memorized value by 10 (down from 120V AC)?
Marshal,
For sinusoidal waveforms, you multiply the RMS voltage by 1.414 to get peak and then the peak voltage times two to get peak to peak. The line voltage is 120 volts RMS at your house but the peak to peak voltage (what you will see on a scope) is 339 volts. The peak voltage is 169 volts and that is why a full wave rectifier in a standard switching power supply has about around 165 volts output. (line voltage peak less the voltage drop across the diodes, common mode filter, etc.) This is also the common primary power supply voltage in line operated TV sets. This is the reason line voltage operated devices have a voltage breakdown in the 500-1kV range. Typical house wire insulation has a 600 volt breakdown rating.
. You may want to resubmit your query 