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Look after your Lithiums! By Nicholas Vinen 2x 12V Battery Balancer Two 12V batteries are often significantly cheaper than one equivalent 24V battery but you need to be careful connecting batteries in series as their voltages and state-of-charge may not be identical. The difference in voltage can increase over time, leading to battery damage from overcharging and/or under-charging. This compact, low-cost device keeps them balanced so that they last a long time. O cause they tolerate overcharging much less than a similar n page 28 of this issue, we describe a high-perforlead-acid battery would. mance Uninterruptible Power Supply (UPS) you This design incorporates a low-voltage cut-out which can build yourself, that uses two 12V LiFePO4 batprevents the batteries being discharged too far if it is unteries wired in series to form a 24V battery. able to keep them balanced and its very low quiescent This was a much cheaper solution than buying a 24V batcurrent of under 0.02mA means it will have virtually no tery with equivalent performance, even taking into account effect on battery life. the $100 or so we paid for a commercial battery balancer. It also incorporates a LED to show when it is monitoring You can build this balancer for a lot less than that and the battery voltages and two more LEDs to show when one it will do a similar job. or the other is being discharged or shunted. Our version can’t handle quite as much By default, the low-voltage cut-out is set up so that the current, because it lacks the large batteries are only balanced when they are being charged, heatsink. however, there are definitely situations where you might But you can want the batteries to be balanced during discharge, too. easily parallel In that case, you just need to several of our change a resistor or two in balancers if you order to adjust the cut-out need a higher curthreshold so it is near the rent capacity and minimum battery voltthe cost would still age. In this case, the cutbe quite reasonable. out will still act to proIt can be used with prettect the batteries but will ty much any battery chemallow balancing during istry, as long as the battery charging and discharge, voltages will stay within the right down to that lower range of 5-16V. threshold. Balancing is most critical Shown rather significantly oversize for clarity (the PCB It’s a compact unit at with lithium-based rechargemeasures only 31.5 x 34.5mm) – see the $2 coin for reference – just 31.5 x 34.5 x 13mm, able batteries, though, beall components mount on this single board. 70 Silicon Chip Celebrating 30 Years siliconchip.com.au Fig.1: the circuit for the Battery Balancer, shows the balancing section at top and low-voltage cut-out at bottom, based around dual micropower op amps IC1 and IC2 respectively. IC1 drives dual Mosfets Q1 & Q2 to perform balancing while necessary; IC2 drives the indicator LEDs and disables IC1 using Mosfet Q3 when the battery voltage is low. so you can tuck it away inside just about any device. And if the 300mA balancing current is not sufficient for your purposes, all you need to do is wire two or more units in parallel and they will operate in concert to keep the batteries balanced. Balancing operation There are two sections to the circuit; the balancer and the low-battery cut-out. The entire circuit is shown in Fig.1, with the balancing circuitry in the top half and the lowvoltage cut-out below. Starting with the balancing section, schottky diodes D1 and D2 are connected in series with the two batteries so that no damage should occur if they are wired up incorrectly. These diodes are then connected to Mosfets Q1a and Q2b at the right-hand side of the circuit diagram, via a pair of 27Ω 3W resistors. These Mosfets are normally switched off and no current can flow through them. If the voltage across one battery rises by more than 100mV above the other, the Mosfet across that battery is switched on. siliconchip.com.au If the battery is being charged, this has the effect of shunting some of the charge current around that battery so that it receives a lower charging current than the other, decreasing the voltage differential over time, as the battery with the lower voltage is then receiving more charging current. If the unit is operating while the battery is not being charged, the effect is to slightly discharge the battery with the higher voltage until they are closer in voltage. It’s a linear circuit so the shunt current is proportional to the difference in voltage. As the imbalance rises, so does the shunt current until the limit of around 300mA is reached. This is to prevent the Mosfet and resistor from overheating. Detecting a voltage difference A resistive divider comprising two 10MΩ resistors and 200kΩ trimpot VR1 is connected across the battery, before diodes D1 and D2 so that their forward voltage does not affect the calculation of the difference in voltages. VR1 is adjusted so that the voltage at its wiper is exactly half that of the total battery. This half-battery voltage is buffered by voltage follower op amp IC1a. Celebrating 30 Years May 2018  71 ing current, all the dissipation would be in this resistor and none in the Mosfet, meaning the maximum current • Minimum battery voltage: 5V would be 200mA [14V ÷ 68Ω]. • Nominal battery voltage: 12-13V We realised we could increase this • Maximum battery voltage (fully charged): 16V by 50% by splitting the dissipation be• Battery voltage difference for balancing to start: approximately 100mV tween the Mosfet and its series resis• Battery voltage difference for maximum balancing current: approximately 130mV tor. The resistor has a 3W rating while • Maximum balancing current: approximately 300mA (multiple units can be paralleled) the Mosfet has a 2W rating, giving the • Maximum balancing power: approximately 4.5W (multiple units can be paralleled) possibility of a total of just under 5W. • Maximum recommended charging current: 10A per unit With a battery voltage of 29V and a • Quiescent current: < 20A balancing current of 300mA, dissipa• Low-voltage cut-out threshold: 27V (can be changed) tion is around 2.7W in the resistor and • Low-voltage cut-out hysteresis: 0.25V 1.7W in the Mosfet. We achieve this dissipation sharing This op amp has a very high input resistance of around by preventing the Mosfet from turning on fully and using 40GΩ, resulting in a low input bias current of approxi- a lower value limiting resistor. This is the purpose of Q1b mately 250pA, so the high values of these resistors (cho- and the three resistors between TP2 and TP3. These resistors bias the gate of Q1b at a voltage that’s sen to minimise the quiescent current) will not result in a initially about halfway between the negative and positive large error voltage. The other half of the dual op amp, IC1b, compares the terminals of the upper battery (ie, at a voltage between that voltage at the junction of the two batteries (from pin 2 of of pins 1 & 2 of CON1). However, as the balancing current CON1) to the output voltage from IC1a. If the upper bat- for the upper battery increases, the voltage at the junction tery has a higher voltage than the lower battery then the of the 27Ω resistor and Q1a drops and therefore so does half-battery voltage will be higher than the voltage at pin 2 Q1b’s gate voltage. Q1b is a P-channel Mosfet and so it switches on when of CON1. That means that the voltage at non-inverting input pin 5 will be higher than at the inverting input, pin 6. its gate is a few volts below its source terminal. The source As a result, IC1b’s output will swing positive. The ra- terminal is connected to the gate of Q1a, which is about 2V tio of the 390kΩ feedback resistor to the 10kΩ resistor that above pin 2 of CON1 when Q1a is in conduction. So as the current through Q1a builds and Q1b’s gate voltgoes to the battery junction (ie, 39:1) means that the output will increase by 40mV for each 1mV difference in bat- age drops, eventually Q1b begins to conduct, pulling the gate of Q1a negative and cutting it off. This forms a negatery voltages. Once the voltage at output pin 7 has risen by a couple tive feedback path and due to the gate capacitances, the of volts, N-channel Mosfet Q1a will switch on as its gate circuit stabilises at a particular current level. With 300mA through the 27Ω resistor, the voltage across is being driven above its source, which connects to pin 2 it will be 8.1V [0.3A x 27Ω] and this translates to a gateof CON1 via a low-value shunt resistor (47mΩ). So current will flow from the positive terminal of the up- source voltage for Q1b of around -2V, ie, just enough for it per battery, through diode D1, the 27Ω 3W resistor, Mosfet to conduct current. The 4.7kΩ resistor between output pin Q1a and then the 47mΩ resistor to the negative terminal 7 of IC1b and the gate of Q1a prevents Q1b from “fighting” the output of the op amp too much. of the upper battery. Note that 8.1V is slightly more than half the typical voltOnce this current starts to flow, it will also develop a voltage across the 47mΩ resistor which will increase the age of one 12V battery and this is why the resistor dissipates voltage at pin 6 of IC1b, providing negative feedback. This slightly more than the Mosfet, in line with their ratings. feedback is around 1mV/20mA, due to the shunt value. This prevents Q1a from switching fully on. Rather, its Balancing the other battery The other half of the balancing is a mirror-image; for balgate voltage will increase until the current through the 47mΩ resistor cancels out the difference in the two voltages. ancing the lower battery, Mosfet Q2b is a P-channel type and Hence, the maximum shunt current of 300mA will thus switches on when its gate is driven below its source. be achieved with an imbalance around 130mV (100mV + As with Q1a, its source is connected to the junction of the two batteries via the 47mΩ resistor. 300mA x 0.047Ω ÷ 2). When the lower battery voltage is higher than the upThe 10MΩ resistor between pin 3 of IC1a and pin 2 of CON1 serves mainly to prevent the balancer from operating per battery, output pin 7 of IC1b goes negative, switching should the junction of the batteries become disconnected Q2b on. And the same current-limiting circuitry is present but from CON1. It also makes setting the unit up and adjusting VR1 easier. It has a negligible effect on the voltage at this time, Q2a is an N-channel Mosfet, so that as current pin 3 since there’s normally such a small voltage across it. builds through the lower 27Ω resistor and the voltage at the junction of it and Q2b rises, Q2a switches on and limCurrent limiting its the current to a similar 300mA value, with roughly the Had we specified 68Ω resistors in series with Q1a and same dissipation split between the two components. A 10nF capacitor across IC1b’s 390kΩ feedback resistor Q2b (rather than 27Ω), there would be no need for additional current limiting circuitry since the resistors would naturally slows down its action so that it doesn’t react to any noise limit the balancing current within their dissipation ratings. or EMI which may be present at the battery terminals (eg, However, this would mean that at the maximum balanc- due to a switchmode load). Features & specifications 72 Silicon Chip Celebrating 30 Years siliconchip.com.au Fig.2: use the PCB overlay diagram at left and matching photo at right as a guide to assembling the PCB. Only one SMD component (a 10MΩ resistor)is soldered to the bottom, the rest go on the top as shown. The main aspects to pay attention to during constructon are that the semiconductors are correctly orientated and that you fit the resistors and capacitors in the correct locations. It also prevents the circuit from oscillating due to the negative feedback and the action of the current limiters. Under-voltage cut-out Commercial battery balancers tend to only operate when the battery voltage is near maximum, as this is when they are being charged. That avoids the possibility of the balancer discharging the batteries when they are under load. However, we’re not convinced this is a good idea. It’s possible to have a sufficient initial imbalance that one battery could be over-charged before the balancer even activates. And full-time balancing also has the advantage that it can start re-balancing the cells as soon as an imbalance occurs, which also avoids over-discharge and gives it more time for re-balancing. There is one other advantage to having a higher undervoltage lockout threshold and that is that it will prevent the balancer being triggered due to differing internal resistance of the batteries when under heavy load. This could create a voltage difference between the batteries even when they are at an equal state of charge. If you want the balancer to be active even when the bat- teries are not being charged, you still need the under-voltage lockout circuitry to prevent the balancer from over-discharging either battery. But in that case, you would change its threshold to be close to the fully-discharged voltage of your combined battery. For a pair of lithium-based 12V rechargeable batteries, this would normally be around 20V total. That’s to protect against the case where one battery has a failure (eg, shorted cell) which causes its voltage to drop dramatically. The under-voltage detection circuitry will then prevent the balancer from over-discharging the other battery in response, and potentially destroying it. See the section below on how to change the cut-out threshold if you want to take this approach. The increased battery drain of the low-voltage cut-out section is only about 10µA. As a bonus, it drives the three LEDs to indicate when the balancer is operating and which battery is being shunted. This is implemented using IC2a, another LT1495 op amp. Its positive supply is the same as for IC1a but its negative supply is connected directly to the negative terminal of the bottom battery, allowing it to sense the total battery voltage Many years ago, long before the days of smartphones and computers, even before the days of television, it was considered a “right of passage” for dads to sit down with the sons (or daughters) and help them as they built their own radio receiver. FM? Not on your life no such thing! DAB+? Hadn’t been invented yet! No, it was all good, old reliable AM Radio. And they could listen to stations hundreds, perhaps thousands of miles away! The beauty of it all was that they were building something that actually worked, something they’d be proud to show their friends, to their school teachers, to their grandparents! Enjoy those days once again as they build the SILICON CHIP Super-7 AM Radio See the articles in November & December 2017 SILICON CHIP (www.siliconchip.com.au /series/321) SUPERB SCHOOL PROJEC T! • • • • • • • Covers the entire AM radio broadcast band. Has on-board speaker ... or use with headphones. SAFE! –power from on-board battery or mains plug-pack Everything is built on a single, glossy black PCB. All components readily available from normal parts suppliers Full instructions in the articles including alignment. See-through case available to really finish it off! IT LOOKS SO GOOD THEIR FRIENDS WON’T BELIEVE THEY BUILT IT! siliconchip.com.au Celebrating 30 Years May 2018  73 (ie, between pins 1 and 3 of CON1) more easily. This is done using a string of three resistors (390kΩ, 6.8MΩ and 1MΩ) connected across the batteries. These form a divider with a ratio of 8.19 [(390kΩ + 6.8MΩ) ÷ 1MΩ + 1]. The divided voltage from the battery is applied to inverting input pin 2 of IC2a. A 3.3V reference voltage is applied to the non-inverting input at pin 3. This is provided by micropower shunt reference REF1, which is supplied with around 2A via a 10MΩ resistor. The voltage at pin 2 of IC2a is therefore above the voltage at pin 3 when the battery voltage is above 27V [3.3V x 8.19]. When this is the case, output pin 1 of IC2a is driven low, pulling the gate of N-channel Mosfet Q3 to the same voltage as the negative terminal of the bottom battery. As the source of Q3 is connected to the junction of the two batteries, Q3 is off and so does not interfere with the operation of the balancer. However, should the total battery voltage drop below 27V, the output of IC2a goes high, switching on Q3 and effectively shorting input pin 3 of IC1a to the junction of the two batteries. This means that the voltages at pins 5 and 6 of IC1b will be equal (with no current flow through the 47mΩ resistor, as will quickly be the case), therefore preventing any balancing from occurring. When the output of IC2a goes high, this also causes a slight increase in the voltage at its pin 3 input, due to the 10MΩ feedback resistor. This provides around 1% or 250mV hysteresis, preventing the unit from toggling on and off rapidly. In other words, the battery voltage must increase to 27.25V to switch the balancer back on. When the output of IC2a is low and the balancer is active, IC2a also sinks around 0.25mA through LED1 and its 100kΩ series resistor, lighting it up and indicating the balancer is operating. And when one or the other battery is being shunted, IC2b amplifies the voltage across the 47mΩ shunt by a factor of 2200 times. So if there is at least 20mA being shunted, that results in around 1mV across the 47mΩ resistor which translates to 2.2V at output pin 7 of IC2b, enough to light up either LED2 or LED3. LED2 is lit if it’s the upper battery being shunted and LED3 if it’s the lower battery. dissipate up to around 4.5W. If you’re using a 3A charger, that means it can handle a ~10% imbalance in charge between batteries (which would be unusually high). However, with a 10A charger, it will only handle a ~3% imbalance, with a 20A charger ~1.5% etc. A greater imbalance could potentially lead to over-charging as the balancer can’t “keep up”. So if your charger can deliver more than 5A, you may want to consider paralleling multiple balancers and we would strongly recommend it for a charger capable of 10A or more. When properly adjusted, the balancers will share the load. Realistically, one of them will start balancing first but if it’s unable to keep the imbalance voltage low, the others will quickly kick in and shunt additional current. Since the only external connections are via 3-way pin header CON1, you could simply stack the boards by running thick (1mm) tinned copper wire through these pads and soldering them to each board in turn. You can then solder the battery wires to these wires. Changing the cut-out voltage Construction To change the cut-out voltage, simply change the values of the 6.8MΩ and 390kΩ resistors using the following procedure. First, take the desired cut-out voltage and divide by 3.3V. Say you want to make it 24V. 24V ÷ 3.3V = 7.27. Then subtract one. This is the desired total value, in megohms. So in this case, 6.27MΩ. This can be approximated a number of ways using standard values. For example, 3.3MΩ + 3.0MΩ = 6.3MΩ which is very close. So use these values in place of the 6.8MΩ and 390kΩ resistors. Keep in mind there will still be around 1% hysteresis, so the switch-on voltage will be about 24.24V. Two more examples would be a 22V cut-out, which would require 5.67MΩ total; you could use 5.6MΩ + 68kΩ. Or for a 20V cut-out, you would need 5.06MΩ which could be formed using 4.7MΩ + 360kΩ. The 12V Battery Balancer is built on a small doublesided PCB measuring 31.5 x 34.5mm and uses mostly surface-mounted parts. These are all relatively large and easy to solder. Refer to the overlay diagram, Fig.2, to see where each component goes on the board. Some of them (the ICs, Mosfets, diodes and trimpot) are polarised so be sure to fit them with the orientation shown. There are two small SOT-23 package devices, Mosfet Q3 and voltage reference REF1. Fit these first. They look almost identical so don’t get them mixed up; only the tiny coded markings on the top of each set them apart. Tack solder the central pin to the pad in each case then check that the other two pins are centred on their pads and that all pins are in contact with the PCB surface. If not, reheat the initial solder joint and nudge the part into place. Then solder the two remaining pins and add a little extra solder to the first pin (or a bit of flux paste and heat it) to ensure the fillet is good. Next, solder IC1, IC2, Q1 and Q2. They are all in eight- Paralleling multiple boards As stated, one board can handle around 300mA and will 74 Silicon Chip Sourcing the parts The PCB is available from the SILICON CHIP Online Shop – simply search for the board code 14106181. All the other parts are available from Digi-Key. While they are based overseas (in the USA), you can pay using Australian dollars and they offer free courier delivery for orders of $60 or more. You can find the semiconductors on their website by searching for their part number and then narrowing down the list (eg, ignoring listings which are out of stock or only sold in large quantities). For the other, more generic parts like SMD resistors, you can find them by searching for (for example) “SMD resistor 1206 4.7k 1%” and then sorting the result by price. The cheapest part which matches the specifications should do the job just fine. But be careful because sometimes the search results include parts with different properties than you are expecting. You will need to skip over those. Mouser, another large electronics retailer based in North America, will almost certainly have all the required parts too. And if you don’t want to order from overseas, chances are that you can get most of them from element14 (formerly Farnell; http://au.element14.com). Celebrating 30 Years siliconchip.com.au pin packages and must be orientated correctly. Identifying pin 1 can be a bit tricky. For IC1 and IC2, you have to find the chamfered edge which is quite subtle. Pin 1 is on that side. Q1 and Q2 have pin 1 marked by a much more clear divot in the corner of the package. But you can also orientate IC1 and IC2 by matching the position of the markings up to our photo. In each case, make sure the device is positioned correctly and tack solder one pin, then as before, check the locations of the other pads are correct and solder them before refreshing the first joint. If you accidentally bridge two pins with solder, use a little flux paste and some solder wick to clean it up. The only remaining SMD parts which are polarised are diodes D1 and D2. Fit these now, ensuring the striped end goes towards the top edge of the PCB, as shown in Fig.2 and marked with “K” on the PCB. Then solder the two 3W resistors in place. Follow with the remaining SMD ceramic capacitors and chip resistors as shown in the overlay diagram. For the two-pin devices, make sure that you apply the soldering iron long enough so that the solder adheres to the PCB and the component. Adding a little flux paste to the PCB pads before positioning the part will make this easier. There is a single component on the underside of the board, a 10MΩ resistor positioned between CON1 and VR1. Solder it in place but use a minimal amount of solder, so that you don’t plug the through-holes underneath. You can add more solder later after CON1 and VR1 are in place. All that’s left then is to solder trimpot VR1 with the adjustment screw orientated as shown, and a pin header for CON1. We used a normal pin header but a polarised header would be a good idea if you’re going to use a plug to make connection to the batteries so that it can’t be accidentally reversed. If it is reversed, D1 & D2 should prevent damage but the balancer won’t work! Or you can solder the battery wires directly to these three pads. They only need to be rated to handle 300mA per board; medium duty hookup wire should be more than sufficient, even if paralleling multiple boards. Testing & set-up Connect your batteries in series, then connect the negative-most terminal directly to the negative terminal on CON1. Do not connect the junction of the two batteries to the Balancer just yet. Ensure that the total battery voltage is well above the threshold and that they are reasonably close to being balanced. You can ensure they are balanced by charging both independently and then connecting them in parallel via a low-value, high-power resistor (eg, 1Ω 5W) and leaving them for a few hours. The voltage across the resistor should drop to a very low level once their voltages equalise. Now connect the most positive terminal to the positive pin of CON1 via a 1kΩ resistor and check that LED1 lights up. LEDs 2 & 3 should remain off. Measure the voltage across the 1kΩ resistor. It should be under 20mV. If it’s under 5mV or over 20mV, disconnect the battery and check for errors in your PCB assembly or battery wiring. Assuming the voltage is within the specified range, remove or short out the 1kΩ test resistor and then connect the junction of the two batteries to pin 2 of CON1. LED2 and LED3 may light up. If so, rotate the adjustment screw siliconchip.com.au Parts list – 2 x 12V Battery Balancer 1 double-sided PCB, coded 14106181, 31.5 x 34.5mm 3 3-way right-angle or vertical pin header (CON1) Semiconductors 2 LT1495CS8 dual micropower op amps, SOIC-8 (IC1,IC2) 1 ZXRE330ASA-7 micropower 3.3V reference, SOT-23 (REF1) 2 DMC3021LSDQ dual N-channel/P-channel power Mosfets, SOIC-8 (Q1,Q2) 1 2N7000 N-channel signal Mosfet, SOT-23 (Q3) 1 green LED, SMD 3216/1206 (LED1) 1 red LED, SMD 3216/1206 (LED2) 1 blue LED, SMD 3216/1206 (LED3) 2 S1G 1A schottky diodes or similar, DO-214AC (D1,D2) Capacitors (all SMD 3216/1206 X7R ceramic) 2 100nF 50V (measure value before installing!) 1 10nF 50V (measure value before installing!) Resistors (all SMD 3216/1206 1%) For tips and tricks 6 10MΩ (Code 1005) on soldering SMD 1 6.8MΩ (Code 6804) components, refer to the 2 5.6MΩ (Code 5604) SILICON CHIP articles “How to Solder 1 2.2MΩ (Code 2204) Surface Mount Devices” 1 1MΩ (Code 1004) in March 2008 2 390kΩ (Code 3903) www.siliconchip.com.au/ 2 100kΩ (Code 1003) Article/1767 2 10kΩ (Code 1002) and 2 4.7kΩ (Code 4701) October 2009 1 1kΩ (Code 1001) www.siliconchip.com.au/ 2 27Ω 3W (SMD 6331/2512) Article/1590 [eg, TE Connectivity 352227RFT] 1 47mΩ [eg, Panasonic ERJ-L08KF47MV] 1 200kΩ 25-turn vertical trimpot (VR1) in VR1 until they are both off. Now check that there is no balance current flowing by measuring the voltage between TP1 and TP2, and between TP3 and TP4. In each case, the reading should be zero. If you get a non-zero reading between TP1 and TP2, current is flowing through Q1a. And if there’s a voltage between TP3 and TP4, current is flowing through Q2b. Since you started out with balanced voltages, this should not be the case, so adjust VR1 further until you get a zero reading across both pairs of test points. Ideally, VR1 should be adjusted to halfway between the point where the voltage starts to rise between one pair of test points, and the point at which the voltage rises across the other pair of test points. This ensures the balancing will be, for lack of a better word, balanced! The maximum reading you should get between one pair of test points should be 8.8V. Any more than that and you risk the resistor dissipation rating being exceeded. In this case, disconnect the batteries and change the 10MΩ resistor right next to VR1 on the top side of the board with a slightly lower value (eg, 9.1MΩ or 8.2MΩ) to reduce the current limit. If that doesn’t fix it then it’s likely that the current limiting circuitry is not working so you should check for soldering problems or faulty components. SC Celebrating 30 Years May 2018  75