ATX PSU Modified Into a Car Battery Charger-Part 3
For those readers of my previous article, about that modification shown in two parts here below,
especially for those who asked me specific information in order for them to build their own charger, making the same modification, I have prepared: a) a basic drawing showing the components’ values I used and b) some essential instructions for their comfort.
You can see the basic drawing below:
First of all, when you start with the procedure, make sure that the PSU you have available is in good working condition. Otherwise you must first repair it and after the required verification that it is all right you can move on with the modification. This is obvious…
In the above schematic, most of the component values around the IC are typical values. Perhaps there will be some variations with the one you have available but this doesn’t matter. One way or another you don’t touch these components. All you have to do at this stage is to find the location of the resistors forming the resistive divider for the feedback voltage and then replace these resistors with the resistive “trio” consisting of R1, R2 and R3. Normally the divider network should be nearest to the output.
In Parts 1 and 2 I have explained that I used carbon film resistors instead of metal film ones. Add to this that just because my stock consists of the ordinary E12 series resistors, in order to have that R4, of 5KΩ in value, I connected two pieces of 10KΩ in parallel. The theoretical value of the resistor is 4,986KΩ exactly. This affects the maximum current set point causing a variance, but this variance as you already saw is not that significant for this application.
This resistive “trio” (R1, R2 and R3) will be your new voltage sensing and feedback circuit for the output voltage stabilization. The switch shorting the middle 1K2 resistor switches the output voltage from “maintenance” level (13,2V) to “boost charge” level (14,7V). I strongly suggest you to test the function of the voltage selection right after installing the resistive “trio”, before continuing further.
To those readers who are interested in having very good compensation of the output voltage against load variations I suggest them to install and connect this resistive “trio” exactly on the output terminals of the PSU, perhaps using a tiny perforated PCB to hold the resistors tight in place. This will also make things much easier for the modification procedure.
As regards the current sensing and limiting circuitry, things are a little bit difficult but nevertheless quite manageable. Please keep in mind that just because the life expectancy of the PSU will depend absolutely upon that circuit for the rest of its operational life, you must first study the above drawing carefully and work accordingly. Furthermore allow me to give you the basic information about it in order for you to know exactly what you should do explaining why you should do so.
The current sensing shunt resistor is that 0,1Ω/10W one shown in the schematic above, at the system’s output return path. It transforms the current flowing through it into a voltage drop across its terminals. This works in turn as a mirror, reflecting the current’s image as a voltage now for the IC to “see” it. So, as long as current flows at the output, a voltage develops at the terminals of this shunt resistor, the magnitude of which, of course, depends directly upon the magnitude of the flowing output current.
Given that the shunt is located at the current return path, this voltage (measured against the ground) has negative polarity on the node of the shunt resistor and R5. On the other hand, R4 and R5 form a resistive divider which predefines the maximum output current. This is the divider you have to install therein because it does not exist in your PSU.
Let’s take a closer look now on how this works. It is quite interesting. As long as there is no current flow in the output, the down side resistor of this divider (R5, in reference to ground) is at ground potential (through the shunt resistor). At the node of R4, R5 there is always a voltage of 600mV, because R4 is permanently supplied with the 5V reference from pin 14 of the IC. As the current flowing at the output is getting higher and higher depending on load increase, the voltage developed on the shunt resistor at some point (at 6A approximately as calculated) outweighs the 600mV stable reference with an equal voltage of reverse polarity and when it just exceeds it, the voltage at pin 15 of the IC becomes more negative than the null voltage of pin 16, against which this comparison takes place all the time. This reverses the status of the second (current sensing) op amp, the output of which jumps now from its “Low” to the “High” level. It’s the moment where the current op amp takes control of the PWM reducing the width of the output pulses down to a minimum in order to keep the predetermined current value stable. The PWM enters thus the current mode of operation. Voltage cannot be any longer stabilized to the predetermined level, until the output current drops below the limiting set point. When this happens, the voltage op amp takes control of the PWM, stabilizing the voltage. The current op amp is in relaxing status now.
Now there are some prerequisites for this circuit to function properly. First of all, you have to break the ground return path in order to install the shunt resistor. Therefore you must use a sharp knife or cutter and after finding the place where you intend to install your shunt cut the wide foil trace at a point a little bit back from the point where all those black cables were coming out. Your shunt resistor must bridge this broken path when installed. Quite understandable, isn’t it? We cut the foil in order to break the circuit and install a bridging resistor in order to measure the return current through it. That’s all. This is your second step of the modification.
Next to it, the installation of the current sensing resistive divider comes into play. Perhaps you will be forced to remove some components in order to obtain useful space for your new ones. If you do that, before accidentally removing anything vital for the operation of the IC, keep in mind that the above drawing contains all the necessary parts for the IC to operate properly. First do your verification, secure all those components which under any circumstances should not be removed from their place and only then continue removing the rest of them which you don’t need. Not before, because you will be lost in there. But even if this happens, don’t give up. Relax, prepare yourself a cup of coffee, follow the drawing and identify the existence of the vital components once again. Anyway it would be wise to check the function of the PSU after every intermediate step you follow.
Some more general info:
Pins 1 and 2 of the IC are connected to the first op amp within this PWM IC. Voltage control is achieved through them. Perhaps in the PSU you have in hand pins 15 and 16 are used for this function (that is, the second op amp). You cannot know that in advance, but it’s easy to identify it. The key to do this is to measure the voltages on the terminals 1, 2, 15 and 16 of the IC. If you find the 5V reference at pin 2 and the feedback 5V at pin 1, the connections will be as shown in the drawing above and you will know that op amp 1 controls the voltage. Some designs use the half of the reference voltage here, that is, 2,5V.
If you have a power supply available which uses the KA 7500 of Fairchild, be informed that this IC is pin to pin compatible with the TL 494. So you can use these instructions for that PSU as well. Nothing of this description changes for it.
The second op amp is (usually) not used. You will verify this using the same method again. Measure the voltages on 15 and 16. If you measure the 5V reference on pin 15, while pin 16 is grounded, then you will know that the second op amp is not used (in this way it is forced to a neutralized, “Low” output state). For using this op amp, you must first of all free its pin 15, in order to connect there your current feedback according to the drawing. During the installation of the current sensing resistive divider remember to install also that 10nF capacitor at the negative feedback connection of the op amp, exactly like that one shown connected between pins 3 (comparator output) and 15 (inverting input of the 2nd op amp). If you forget to install this cap, when current limit occurs during ordinary operation you will get an immediate reminder whistling noise…
Keep also in mind that the common output of both op amps appears at the comparator output (pin 3). Their individual outputs are connected together, each with an isolating diode at its output, with that pair of diodes forming a logic OR function. The voltage on pin 3 is therefore very useful information of what is happening within the controller circuitry, in connection to what we see at its final outputs.
As regards the cooling fan, it is also wise to change its supply connections in order to secure its longevity. Disconnect it from the 12V main output and connect it to the auxiliary output of the standby transformer. If this voltage is much higher than 12V, say 22V, use a 7812 three pin regulator in order to get a stabilized 12V voltage for the fan. Do not change this voltage, because it feeds the PWM IC as well (pin 12) and you will probably cause unpredictable behavior of the controller.
Moreover: pin 13 is the output control pin. When tied to Vref, as in our case, the IC works in push-pull mode dividing the clock frequency by two, for feeding each driver transistor with the relevant pulses. Pin 14 is the 5V reference out. Pin 5 is the CT (timing cap) terminal and works with RT, pin 6 (which is the timing resistor terminal). Pin 7 is the IC’s ground and pins 9 and 10 are the respective emitters of the two internal bipolar driver transistors. CT is the only polypropylene cap. All the rest (except the DT cap) are ceramic disk caps.
The tricky pin is No 4, the DT (dead time control pin). Most designers use this pin in order to neutralize the function of the IC against abnormal conditions. According to the data sheet, their choices in order to achieve that neutralization are three, by imposing the 5Vref signal a) on pin 3 (comp), b) on pin 4 (DT) or c) on the inputs of the op amps, depending on each case respectively.
In any case keep in mind that the 10μF cap (the only electrolytic one of the basic circuit) and the 10K resistor connected there, form a “time constant” circuit. During start up, the cap behaves as a short circuit until it is charged up. This happens within 100ms, which is the time constant of this circuit. This time equals a delay of 5 cycles of 50Hz mains voltage before normal operation of the IC. During this time the 5Vref is applied to pin 4, inhibiting the output of the PWM. As the cap charges, the voltage at pin 4 drops almost to ground level, the IC starts to operate smoothly and keeps on running normally thanks to this soft start R-C circuit attached to the DT terminal.
This means that in case of occurrence of strange problems of the type “it’s not starting although power is there”, first check the dead time control pin voltage. It should be near to ground potential. Its working range is 0V to 3,3V. Above that level there are no pulses at the output…
When everything related to the modification is finished the time for testing has come. As you intend to examine the overall performance of the unit, don’t be tempted to short its output terminals in order to see the effect of the current limit protection.
First load the equipment increasing gradually the load while you watch the current feedback voltage to the controller op amp. It should increase accordingly. Confirm as well that your reference (600mV) is in place and it gets less and less as the load increases.
If both voltages are present, increase the load above the 6A limit. (You can do this easily by paralleling car lamps. For example, for these 6A you need 72W for a nominal 12V operating voltage. This means a connection in parallel of a high beam filament of a 60W car head light lamp with a brake indicator lamp rated 21W. This will do the job). The output voltage should drop when you exceed this 6A set point.
If you see this effect, you are done. You can do whatever you like with your charger except one thing only. To (accidentally) connect the output terminals of the PSU in reverse polarity to that of the poles of the battery to be charged. In such a case you will be forced to perform serious, extended level repair in order to bring the damaged charger in working condition again.
Finally if you want to include a reverse polarity protection in it, you need to buy a relay having a 12V coil, most preferably one of those for car use, including a single pole single throw switch capable to withstand the rated current. For additional safety alone, chose a withstanding intensity of 20A or more for its contacts.
Now connect the positive output cable of the PSU to the C (common) terminal of the relay and its NO (normally open) contact to the output terminal of the PSU (connection in series, using the normally open contact of the relay’s internal switch).
The non grounded terminal of the coil should be supplied with power via a small, say 1A, ordinary rectifier diode. The anode of this diode takes power from the plus terminal of the output positive connector of the PSU, being connected there. The cathode of it feeds the relay coil. This means that if no battery is connected to the PSU, the relay will not be energized and there will be no output voltage. If the battery to be charged is connected with the right polarity, the relay gets energized taking power from the battery and its previously open contact closes now and supplies the battery with charging current. On the contrary, if the battery is connected in reverse, the relay will not be activated (due to the reversely biased diode feeding its coil) and it will stay inactive. No voltage output at the terminals of the PSU.
The drawbacks of using this protection scheme are two. A) There is no output if no battery is connected to it. This restricts the use of this PSU as an ordinary, general purpose power supply, limiting it to a battery charger exclusively. B) There is always a possibility that the battery to be charged can be totally flat. In such a case it will not have enough power to energize the relay and therefore it will not be charged…
Perhaps (for the very demanding readers) you can also use a bypassing switch to override the protection…which makes things rather complicated. However you can implement whatever you like in order to cover your own specific requirements and needs. Do as you wish anyway! It’s your decision exclusively!
Hopefully these guidelines will work in the best way for you. If you try it, I wish you success in your modification effort and further enjoyment of your bulletproof charger after that, whenever you use it!
This article was prepared for you by Paris Azis from Athens-Greece. He is 59 years old and has more than 30 years’ experience in electronics repairs, both in consumer and industrial electronics. He started as a hobbyist at the age of 12 years and ended his professional carrier as a senior electronics technician. He has been a specialist in the entire range of consumer electronics repairs (: valve radio and BW TV receivers, transistorized color CRT TV, audio amps, reel and cassette tape recorders, telephone answering and telefax devices, electric irons, MW cooking devices e.t.c) working in his early stages at the official service departments of National-Panasonic first and JVC afterwards, at their premises in Athens.
Then he joined the telecoms industry, working for 20 years as field supporting technician in the sector of DMRs (: Digital Microwave Radio transmission stations), ending his carrier with this subject. Now he is a hobbyist again!
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