Is it sorcery to make an ammeter with zero resistance? Not really. This kind of design has been used for ammeters in corrosion science where any voltage drop would influence the experiment.
Most commonly, current is measured by passing it across a shunt resistor, then measuring the voltage drop. You can even get specialized amplifiers for that purpose (INA210-series, INA199, ZXCT1110). But such a technique creates the necessary voltage drop. What if you want not to drop any voltage and yet measure the current? There are Hall-effect probes which measure the magnetic field created by flowing current. They are good especially for larger currents. For everything else, there's Mastercard Zero Resistance Ammeter.
The ZRA is an active device; it needs a current source equal to the current being measured. For low currents, only a single op-amp is needed for maintaining equal voltages on the two input terminals. For larger currents you can boost its capabilities by a single transistor. A second op-amp (or current shunt monitor) is needed to scale the voltage drop for external measuring device or ADC.
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| Complete schematic of the zero resistance ammeter |
My design uses a single 1.2V NiMH battery as the power source. Because the op-amp needs at least 2.2V, I put a small DC switching boost regulator to increase the voltage. A switched-capacitor type would work likewise.
The positive terminal is connected to ground. Current is sinked and for the negative terminal it is sourced from the battery. The current passes through a NPN transistor, a Schottky diode (to block reverse polarity) and a shunt resistor to the negative terminal. I used a dual op-amp, OPA2376, which has a very small input offset voltage. It would work well with a cheaper type though. One op-amp senses the voltage at the negative terminal and adjusts the current to the base of the NPN in such a way that the voltage at the negative terminal is kept zero. The NPN works in linear mode, lowering the voltage from the battery. The second op-amp only amplifies the voltage at the shunt resistor by a factor of 100 to have 1V = 1A. Capacitors C1 and C2 are needed for stability.
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| Assembled on breadboard |
In the picture above, the shunt resistor is actually 0R02 and the gain was thus only 50.
From left: Trimmer for gain, OPA2376, shunt resistor, Schottky diode SS14 (bottom), FZT853 NPN, TLV61220 boost regulator with capacitors and inductor soldered directly to adaptor board, Sanyo Eneloop 1.2V NiMH battery.
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| 1.003 V (= 1.003 A) and true zero voltage drop |
The maximum current is mostly affected by heat sinking of the NPN. With a FZT853 in a rather small SOT-223, it is good for 1A. With beefier transistor, it would go somewhere around 4A, where both the current to base and the current from the battery will limit further expansion. On breadboard, keep it under 1A. I learned the hard way that at 2A, the board sometimes melts :-)
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ReplyDeleteHello , could you tell me where arranged the schematic of tlv61220 , or it was you who did it? It would be possible to send me the schematic or tell me where you downloaded ? Thank you, greetings .
ReplyDeleteOK, so this is what I did: I took the Eagle drawing from Farnell for TPS61070DDCR and just changed the device name. Both devices have the same footprint and are pin compatible. I hope this helps.
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ReplyDeleteNice arrangement. Still the connections of the JP1 is not much clear. Can you describe how to connect a current source to the circuit?
ReplyDeleteThe connection is in the "value" of JP1: 1 = IN- (negative), 2 = IN+ (positive), 3 = measured current (use multimeter or ADC to read it).
DeleteIf you have a circuit where you want to measure current, just split the current path and insert IN- and IN+ into it. The conventional flow of current should be from IN+ to IN-. It looks counter-intuitive to ground the positive (IN+) terminal, but that's how it works. Is this explanation sufficient?
Nicely explained indeed. . :) . But, what I do not understand is that, once I split the current path and connected it to the IN+ (GND), the path is directly grounded. Right? So, how come the circuit can measure a grounded current? then the IN- becomes useless. I actually tried your circuit. But, it did not give the expected results. I wonder what went wrong.
ReplyDeleteAlso, can we use this arrangement for smaller currents by increasing the gain? Such as micro and milli ranges?
Well, important note: The two circuits (measured one and measuring one) have to be galvanically isolated. That is why I use a battery as a clean and galvanically isolated power source.
DeleteAlso, depending on your choice of components, the op-amp may need a different compensation. But that is only in case it oscillates.
You can use even easier arrangement for lower currents. The transistor can be dropped and the circuit can work using only the op-amp. Of course, just as you said, then you increase the gain by adjusting the resistor value to match the voltage range of the op-amp.
I am not so good on the concept, can you explain why would a voltage drop influence the measurement of an electrochemical cell? Also, how is this circuit better than a regular ammeter in terms of accuracy.
ReplyDeleteWell, if you have voltage drop in cables, then you don't know the exact voltage of the cell. But maybe the biggest advantage is that you can drive the voltage down to zero, which is nice to have in fuel cell research. A potentiostat can do that too.
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