Subsequent poking around the the internet, looking at tons of schematics, I seem to be on the right track when it comes to connecting REF LO to IN LOW and will try that out once I get back to my lab. I am hopeful enough that I can continue with other modules. I was hoping to make it to A1 Parts since I am conveniently in Etobicoke today, but I feel like I’d rather not deal with public transit more than I have to in favour of a nice relaxing afternoon with my lovely lady. I’d probably be more likely to go if I drove. I intended to pick up a number of things including a 24V transformer (see last post on the control voltage board), an automatic wire stripper, and a panel cutout nibbler tool. Then I got to thinking about current shunts.
Of course, the voltmeter will need it’s current complement – the ammeter. I’ve been having a hell of a time not only finding good parts for one locally, but even figuring out the best components to get and the best configuration for it.
I’ve been following Dave Jone’s great videos on the EEVBlog YouTube channel, and on many posts he suggests using a parallel array of 10 1% resistors instead of a honking high-wattage single resistor for greater accuracy. While shopping for other parts last week I was disappointed that the values I needed were out of stock in my usual haunts (10 x 10Ω 1% resistors, which yield 1Ω in parallel). Though I could get the parts easily through DigiKey and the like, it had me take a step back and consider my other options. A search on DigiKey revealed lots of current shunt options including fancy chassis mount 4-wire temperature corrected ones. Sounds interesting.
The issue is that I want to place as little resistance and capacitance on the output as possible. Dave’s designs include a 1Ω shunt which makes the math easy but would dissipate 9W of power under maximum load, which is a bit high. For safety and to reduce the temperature induced error, my 10Ω resistors would need to be at least 2W (1W would give me only 10% headroom and would heat up like hell). The easy solution of course is to reduce the resistance to 0.1Ω or even 0.01Ω. Though this is easily possible, it would involve a complete recalibration and I would lose out on my 1V = 1A constant current control. For this reason, I think I will leave the 1Ω (ESR) resistor array there on the high side for the constant current sense and order acceptable parts from DigiKey. For this I have the option of getting my 10x 10Ω 1% 2W resistors, or a single 1Ω 1% resistor in a TO-220 package which has the advantage of me being able to heatsink it, reducing the temperature drift. A bit of a toss up. With the 10 resistor array, I have an opportunity to achieve a higher overall accuracy since the ESR would essentially be the average of the 10 values. The single one I can heatsink and will involve soldering two terminals instead of 20. Decisions, decisions.
The other issue comes in as to where I place my two shunts or even if there should be two shunts. Dave’s design in his power supply video (check his channel, most informative and helpful there), he places his current sense on the high side upstream from the voltage regulator and seems quite keen to avoid any sort of series resistance on the output side (no doubt for a good reason). I will follow him here as I barely have an idea of what I’m doing and prevents a lot of nasty problems I’d rather not deal with. That’s easy.
The only stick in the mud with that is that it’s position would also sense any power consumption of the voltage regulator and pass transistor as well as any losses in the circuitry so my 1V Iset will equal 1A … but not directly on the output. I can trim the control voltage to cancel out this modest difference and get it close to being what happens (effectively) on the output, but I am doubtful that it will track linearly with voltage and current changes. I highly suspect this is something I will just have to live with as it prevents a lot of other nastiness I most definitely don’t want to deal with. So it goes. I also kind of wanted to have a display for my Iset rather than some crusty, inaccurate panel markings for the knobs. The same issue applies – I would have to compensate for the current drawn by the regulators even if they are small.
Using CircuitLab, I simulated this up and it indeed indicates that I would have to set Iset at least 10mV higher to achieve the correct cutoff to the load, not even counting the limiting curve of the darlington transistor as I have it set up. So, if I metered the Iset voltage, I would have to trim the input to the meter as well to compensate. Though I just had the bright idea of building in a dummy load to use with Iset, which would actually be useful for calibration and other uses, that’s just too complicated a solution. I want to be able to set (and to see my setting) whether or not the load is connected to the output. It seems I will have to settle for trimming Iset a bit higher, and the input to the meter lower, to compensate. This way, I could use a 4T rotary switch to select the voltmeter display input for V+/0, V-/0, V+/V- and Iset adjusted to cancel the power consumption of the regulator etc. This is good so I can see Iset and Iout at the same time. Some kind of panel indicator for indicating this reading is in A instead of V would probably help too, if I am feeling that anal about it.
So that’s pretty clear. Though not ideal, it will have to do for my first power supply. I have to be mindful that I can’t go too crazy with it or I will never finish the project.
The next issue, however, is monitoring of the actual output current. This display of course must only display the current delivered (or sinked from) the load. This means putting the sense resistor (shunt) on or very near the output. Now, this is where I really do not want a significant resistance. Even 1Ω means not only 9W of power dissipated, but a 24Ω load becomes a 25Ω load as far as the constant current sense sees. This would necessitate me keeping this series resistance as low as possible, at least by 1 order of magnitude. This will also mean I have to make sure the input configuration on the meter reads at the correct scale as well. This seems a solvable problem and just a matter of simple arithmetic. How well and how accurate the display will be reading a lower input range needs some investigation (especially before I go and buy expensive shunts).
the ICL7107 seems to directly take a range of 200mV in 3.5 digits or 199.9 giving us a resolution of 100µV. Using a 100mΩ shunt resistor would mean that 100µV = 1mA which is expected. Leave the decimal point disconnected and the display will read directly in mA and go up to 1999mA. However, this is really pushing it when it comes to the ICL7107. I can’t find anywhere in the datasheet where it specifies it’s accuracy, but one can assume from the scale that it would be ±0.5mA (±50µV) under ideal conditions.
Applying the same simple calculations, if I change the shunt to 1Ω give me 1mV = 1mA a nice round number. Now the display will read up to 199.9mA. If I go the other way and use a 10mΩ shunt, this would mean 10µV = 1mA knocking the mA off the scale and giving me a max of 19.99A.
In going through the calculations I realize that I will probably be all right with the output resistance of this shunt using the three above. For higher currents, there will be lower resistance. For the maximum output possible on the lowest setting, which is 200mA (199.9mA actually), I get P=I2*R= 40mW which is peanuts. At full output and highest range, 3A and 20A (19.99A) respectively, the power dissipated would be 90mW. In the middle range, 1A load will cause shunt to dissipate 100mW. Far more acceptable that my back of the envelope (and as it turns out, unrealistic) calculation of 9W!
This did bring up an interesting issue here. I wrote in previous posts that the voltmeter schematic had a flaw on it that was causing non-linear readings. As it turns out, I got the Ammeter schematic from the same place and I would have to perform the same fix on the ammeter as the circuit beyond the input is identical. What the calculations above did show me was that the stated resolutions for each of the three ranges seems to be off by a factor of ten. The article attached to this schematic states that using the 1Ω shunt give you a display resolution of 1mA. This would mean that the IC would “see” 100µV/mA. My calculations show that it actually sees 1000µV/mA (1mV/mA) and since the 7107 reads directly in 100s of µV, the display should read “001.0″ for 1mA, not “0001″. Unless I’ve missed something or messed up. If you are reading this and can fault my math above – let me know.
Obviously, I will have to build and test it to see if my assertions hold water. Like anyone, my calculations could be wrong and his right. The sucky thing is, if I am right then I will have to find a way of switching between three shunts – something that can take a max of 3A of DC current running through it. If he’s right, I just use a DPDT switch no worries. If I am right, I will need a 3T (ON-ON-ON) switch which will not only be harder to find, but more expensive I’m sure. One thing I’m absolutely sure of is not wanting to add yet more relays to the system or have cascading DT switch which is just silly.
Regardless, instead of following the schematic and using 5W resistors, considering my maximum possible power dissipation of any of my shunts could be 9W if I have the wrong range selected, I’d much rather spring for 20W shunts (either single or parallel array) just to make sure that if I have the wrong scale setting on the Ammeter, I don’t melt anything. Makes sense as that sounds like exactly the sort of mistake I would make.
Here’s fun, as it turns out – we are both right! I had a hunch and checked the datasheet again to be sure the 7107 does indeed read 200.0mV full scale directly. The answer is it does, provided the voltage reference is set to 100mV. In his application (and most likely mine) the voltage reference is set to 1V which explains the order of magnitude difference in our calculations for the scale factor. With that in mind, I think I will opt for having two shunts, switchable by a SPDT switch which means less parts and less cost (hurrah!). Actually, this would have been my final choice in either case since I do not need a display of less than 1mA or greater than 3A so the only practical ranges I need are the 2A (1.999A) and the 20A (19.99A). If I set Vref to 1V then it follows I will need a 1Ω shunt for the 2A scale, and 0.1Ω shunt for the 20A scale.
But wait! I can do one better. Actually, it answers the initial question and problem I started with before digressing. If I do it my way, and set the reference voltage to 100mV then I can use a 0.1Ω shunt for 1.999A and 0.01 shunt for 19.99A. This accomplishes my original goal of minimizing the output resistance and power dissipation. I can even revise my power calculations, under maximum load of 3A, the 0.1Ω and 0.01Ω shunts will dissipate 900mW and 90mW respectively. Much better than 9W and this allows me to opt for 2W resistors instead of 20W ones. Not only is this a far cheaper solution, but there are more parts available in that range. This might also have applications for the scale dividers for the voltmeter, as mentioned in previous posts.