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Constant, High-Current Source By Ross Tester Whether it’s for charging batteries or in more esoteric applications like stepper motors, a source of reasonably high level constant current is a handy little device to have around. This one’s simple, cheap and about ten minute’s work with a soldering iron! L ast month, you will recall we presented a mini stepper motor driver. (Incidentally, our apologies for the gremlin which got into the system and caused most of the earth symbols and one resistor to disappear. No, we don’t know why either!) That stepper motor driver operates from about 8-35V DC but as we pointed out, a stepper motor really likes to have a constant current source so that the motor current (and therefore power/torque) remains constant throughout the stepper’s speed range. This, then, can be regarded as a companion to the Stepper Motor Controller. It is capable of delivering more than 10A with suitable heatsinking – and we cover that shortly. However, there are a lot of other applications for a constant current source. Nicad battery charging is one which immediately springs to mind. Anything where the constant colour temperature of a globe is important (such as phototographics) is another. And in electronics, there are countless occasions where constant current circuits are used. 72  Silicon Chip So while we’re presenting this specifically for the Stepper Motor Controller, it could be used in a raft of projects and circuits. How it works Let’s get REG1, a 7812, out of the way first of all since it has nothing to do with the constant current source. Its is obviously a constant voltage source and its sole task is to supply 12V DC to the heatsink fan. OK, back to the main circuit. It’s actually two circuits in one – the first is based on the LM317 adjustable regulator. As you can see, the “ADJ”, or adjustment, terminal is connected to the output via a resistor. The voltage between the adjustment terminal and the output terminal is www.siliconchip.com.au always 1.25V, so a constant current of 1.25/R is established. The LM317 is rated at a maximum output current of 1.5A, so in theory this resistor could be as low as about 0.83Ω (1.25/.83 = 1.5). But that’s sailing pretty close to the wind, despite the LM317’s ability to shut down if it gets too hot under the collar. A much better approach is to add a current ‘amplifier’ to increase the output. That’s the purpose of Q1 & Q2 which are ‘slaved’ to the LM317 so that it does not have to work so hard. The circuit works in the following way: say the LM317 was carrying 500mA as its share of the load current. Ignoring the base current of Q1 & Q2 for the moment, that would mean we have 500mA passing through the 3.9Ω resistor at its input. With that current, we must have 1.95V (3.9 x 0.5) across the resistor and it is this voltage which controls Q1 and Q2 which effectively work as emitter followers, applying 1.3V across their 0.22Ω resistors. This forces each transistor to carry 5.9A, giving a total of 12.3A for the circuit. In practice, we have to allow for the base currents drawn by the two transistors but the result will still be a total current of around 12A when REG1 is carrying 0.5A. Whether the circuit can actually supply 10A will depend on the overall dissipation and this is the product of the difference between the input and output voltages and the desired current. This shot shows the two halves of the project opened out – the Pentium II heatsink with its integral fan on the left and the controller itself on the right. The lower two resistors are chosen according to the output current. For example, if you have an input voltage of 25V and you are using the circuit to supply 10A to a load which requires 12V (eg, two 12V 50W hal- Fig.1: this circuit is ideal for stepper motors but could also be used in a variety of other applications. ogen lamps), the difference voltage across the circuit would be 25-12 = 13V and therefore the total dissipation would be 13 x 10 = 130W. Would Q1 MJE2955 7812 LM317 0.22 5W + IN 0.22 5W REG1 7812 15 - 35VDC INPUT 3.9 1W + OUT IN COM REG2 LM317 IN FAN MOTOR 100F 16VW – 100F 35VW ADJ 2002 CONSTANT CURRENT SOURCE www.siliconchip.com.au E C OUT + R1* R1a* OUTPUT TO LOAD 100F 35VW – SC  B IN OUT ADJ COM 100F 35VW OUT C OUT COM Q2 MJE2955 MJE2955 – * R1 & R1a ARE 5W RATED & CONNECTED IN EITHER SERIES OR PARALLEL. THEIR VALUES ARE CHOSEN TO SET CURRENT LEVEL: R1 (TOTAL) = 1.25/CURRENT IN AMPS — SEE TEXT June 2002  73 Parts List – Constant Current Source Capacitors 4 100µF 35VW electrolytics Resistors 2 0.22Ω, 5W 1 3.9Ω, 1W 2 5W resistors to suit output current – see text & tables the circuit be able to cope with this, even with the fan-cooled heatsink? Highly unlikely, so you see that if we want 10A, we need to reduce the input voltage (or increase the output voltage) to get the overall power dissipation down. However, the beauty of this circuit is that it can’t overheat because the LM317 is on the same heatsink as the two transistors, so if they start to get really hot, so does the LM317 and it then shuts down before damage can occur. So there it is. A handy constant current circuit but you have to make decisions about input voltage, output voltage and current to get the best out of it. Q2 MJE2955 100F 100F REG2 LM317 Fig.2: assembly should take no more than about 10 minutes if you follow this component overlay. + FAN – REG1 7812 + OUT – 100F + MLG R1 SEE TEXT R1a SEE TEXT 0.22 5W IN + 0.22 5W 3.9 1W + + 100F – Semiconductors 2 MJE2955 PNP power transistors (Q1, Q2) 1 7812 12V positive voltage regulator (REG1) 1 LM317 adjustable voltage regulator (REG2) Q1 MJE2955 + 1 PC board, 75 x 30mm, coded K-142c 1 Pentium II-type heatsink and 12V fan assembly 2 2-way PC-mount terminal blocks 4 M3 10mm screws & nuts 4 sets TO-220 insulating washers & bushes FLAT SIDE FLAT SIDE HEATSINK & FAN ASSEMBLY (MOUNTS OVER INVERTED TO-220 DEVICES) Next, solder on the four electrolytic capacitors and the two PC-mounting terminal blocks. The two 5W resistors at the other end must be chosen for the output current required. As shown in the tables, they can be series or parallel connected. If you are going to parallel them, great – that’s the way the board has been set up. Simply choose the two Fig.3: connecting R1 and R1a in series is a bit more tricky . . . SOLDER SOLDER SOLDER Construction Everything – the components and fan-cooled heatsink – are mounted on a PC board measuring 125 x 40mm and coded K-142c. In fact, the heatsink is not actually connected to the PC board – it is screwed to the two power transistors and two regulators which of course are soldered to the board. Start by checking the board for any defects (rare these days, but possible) and solder on the 3.9Ω 1W resistor and the two 0.22Ω 5W resistors at one end. 74  Silicon Chip The upside-down view of the completed assembly. The heatsink is held onto the PC board by the four screws and nuts through the transistors and regulators. www.siliconchip.com.au resistor values you want and solder them in. If you have to series them, you’ll need to be a bit cleverer! Only one (opposite) end of each resistor is soldered to the PC board; the other ends must connect together across the top of the board. Either way is fine but the parallel arrangement is just a bit neater. The downside of parallel resistors is that when they are unequal, they have different power dissipations. Ideally, they should be fairly close values. Now we come to the tricky bit – soldering in the two transistors and two regulators. First of all, note carefully their positions on the PC board. The second thing to note is that they are actually soldered in “upside down” compared to normal. If you lay the devices flat on their backs, all legs have to be bent up 90° to go through the PC board. The exact position of the bend depends on where the holes are in the heatsink – you have to be pretty accurate to get them all to line up. See Fig.4 for more details on the way the heatsink and transistors mount together. And make sure you get the right one in the right place. They’re all TO220 packages so it’s easy to get them mixed up! Ideally, all should be fitted with insulating washers – the tabs should not be connected together. Well, to be truthful, the tabs of the LM317 (OUT) and the MJE2955 (C) are all connected together anyway (via their pins) so they can all be shorted together via the heatsink without any particular concerns. But the tab of the 7812 must be insulated from the other three devices. (Note that the 7812 in the Oatley kit has an isolated tab so no washers are required on any of the devices and none are supplied in the kit.) You’ll find it easier to fit the heatsink before you fit the fan – there’s not much room between the fan and fins to fit a screwdriver. The fan screws to the heatsink with four long self-tappers. It matters little which way up it goes – one way sucks air through the heatsink, the other pushes air through the heatsink. However most fans are polarised – you must get the red wire on the +12V pin and the black on the –ve pin. And, apart from mounting the assembly in a suitable case, that completes the construction side. www.siliconchip.com.au Fig.4: this sectional diagram shows how to mount the PC board to the heatsink/fan assembly. Take special care with the bends on the regulators and transistors. FAN HEATSINK INSULATING SLEEVE INSULATING WASHER M3 x 10mm SCREWS FLAT WASHER REGULATOR & TRANSISTOR LEADS BENT UP AT 90° (AWAY FROM TABS) PC BOARD (COPPER SIDE DOWN) M3 NUT In use Wheredyageddit? We’re not even going to attempt to go there: if you are building a constant-current supply, you know what you are going to do with it and how to connect it! Just bear in mind the limits we placed on the output current. Of course, if you wanted industrial-strength muscle, there would be nothing to stop you adding some more MJE2955s in parallel (with their load-sharing resistors) mounted on an even bigger heatsink (also fan assisted). But you’re very quickly going to reach the point where the tracks on the PC board won’t handle the current without significant thickening. (You could solder wire over the tracks). The design and PC board pattern is copyright © Oatley Electronics. A complete kit of parts including PC board, components and the Pentium II fan/heatsink assembly is available from Oatley Electronics for $29.00. This includes the two 0.22Ω 5W resistors and 1Ω and 0.47Ω 5W resistors for R1 & R1a, selected to give an output current of about 3.8A with the resistors in parallel (0.32Ω). Oatley Electronics are at PO Box 89, Oatley NSW 2223, phone (02) 9584 3563, fax (02) 9584 3561, email sales<at> oatleyelectronics.com; or they can be contacted via their website: www. oatleyelectronics.com SC TABLE 1: Values for SERIES combinations of resistors R1 & R1a R1 R1a 0.1 0.1 0.22 0.47 1.0 1.2 1.5 2.2 3.3 4.7 5.6 0.2 0.32 0.57 1.1 1.3 1.6 2.3 3.4 4.8 5.7 0.44 0.69 1.22 1.42 1.72 2.42 3.52 4.92 5.82 0.94 1.47 1.67 1.97 2.67 3.77 5.17 6.07 2.0 2.2 2.5 3.2 4.3 5.7 6.6 2.4 2.7 3.4 4.5 5.9 6.8 3.0 3.7 4.8 6.2 7.1 4.4 5.5 6.9 7.8 6.6  8.9 9.4 10.3 0.22 0.47 1.0 1.2 1.5 2.2 3.3 4.7 5.6 11.2 How easy is this: these tables give various likely combinations of R1 and R1a in series and parallel – simply divide the figure in black into 1.25 to get the output current you want! TABLE 2: Values for PARALLEL combinations of resistors R1 & R1a R1 R1a 0.1 0.22 0.47 1.0 1.2 1.5 2.2 3.3 4.7 5.6 0.1 0.22 0.47 1.0 1.2 1.5 2.2 3.3 4.7 5.6 0.05 0.069 0.082 0.09 0.092 0.094 0.096 0.097 0.098 0.098 0.11 0.15 0.18 0.186 0.192 0.20 0.206 0.21 0.212 0.235 0.320 0.338 0.358 0.387 0.411 0.427 0.434 0.50 0.545 0.60 0.688 0.767 0.825 0.848 0.60 0.666 0.776 0.88 0.956 0.988 0.75 0.892 1.03 1.137 1.183 1.1 1.32 1.499 1.579 1.65 1.94 2.076 2.35 2.555 2.80 June 2002  75