(I killed it twice actually. The first time I hooked it up with the polarity reversed in the input. That blew the LT1170. Later, after replacing the LT1170, I shorted the output and burned up some of the other components. A 3 amp fuse on the input would be a good idea.)
This circuit is very similar to the one provided on the data sheet for its main component, the LT1170. The component values are as calculated by Linear Tech's SwitcherCAD program, whose cyptic output Ikufumi provided. I believe therefore that it is in the public domain.
|L1||inductor||100 µH||Hurricane HL-AK210Q
|must handle at least 3 amps|
|Nichicon x1 UPL1E330MAH||25V ESR=.800 Irms=.155A|
|Nichicon x2 UPL1V391MPH||35V ESR=.028 Irms=2.440A|
|LT||IC||switching regulator||Linear Technologies LT1170CT||data sheet|
|Any high speed diode that can handle the power would do. Schottky type is best but not absolutely required.|
|R1||resistor||16.7k; 1%||R1 and R2 set the output voltage; The ratio of R1 to R2 should be 13.5 to 1.|
The supply works by shorting the inductor across the 12 volts through the regulator. This stores energy in the inductor. When the internal switch in the LT1170 goes off, the inductor is placed in series with the 12 volts, adding to it. This voltage pulse is stored in the output capacitor and smooths the output. The diode is used to keep the output capacitor from discharging during switching.Richard Johnson says:
It contains a 100 khz current based oscilator whose output is controled by feedback provided by R1 and R2. These make a voltage divider such that at the wanted output voltage there is 1.24V at the junction of these two resistors. They carry only 1 ma of current and are there to provide a reference voltage to an op amp in the IC. As you draw more current from the output the total voltage would drop as would the 1.24v reference signal. The op amp feeds this to the oscillator telling it to provide more current to fill the demand. This brings the voltage up to 1.24v reference or 18 output. Reduce the load and the opposite happens.The guts of this circuit is (obviously) the LT1170; you can see it's data sheet at Linear Technology's Web site; the PDF version of the sheet has much more information. Here is a block diagram of the LT1170; the pin configuration is at right.
Now you are wondering where the transformer is. Well this works on a bit different principle. At 60Hz you need a transformer and a pulsing current to step up voltage. At the much higher frequency this works at L1 and D1 do the same thing as well as rectify the current. Remember instead of the transformer windings ratio controlling voltage it is the feedback of your two (fried) resistors that establish the voltage.
Ric Ecker has a similar circuit on his WWW site.
Brian Bond has a version of this site with a little extra info about the inductor.
Roger Doering says:
I'm an electrical engineer. After spending a few hours looking over the stuff on your page and the leinear data sheet, I can make a couple of important additions.
1. The criptic output from the linear program claims that both the LT1170 chip and the Schottky diode REQUIRE a heat sink! Basically a large chunk of metal with fins that the part is screwed/clamped to that will carry away the heat produced by the parts. The IC heat sink is specified at less than 8 degrees C/Watt. Since the IC package has its own thermal resistance which is only 2 DegC/W in the LT1170CT, the TO-220 package, if you get good coupling from the package to the heat sink (not real likely unless you use heat coupling grease) you'll end up with about 10 DegC/W and since the IC will be dissipating about 3 Watts that will raise the temperature inside by 30 degrees over ambient. -The good news is that ambient is usually not all that high while doing field work with a telescope- even 104 F is only 40 degrees C which should keep the temperature low enough! Note that with no heat sink, the thermal rating on the IC is 75 DegC/W which will raise the temp by 225 degrees, easily exceeding the 100 degree C maximum rating on the part. Be aware that if you enclose the heat sinks inside a box, the ambient temperature inside the box will rise. A metal project box, or one that has at least a metal lid might provide enough cooling, if the parts are well connected to the metal, but I don't have the numbers handy. Be aware that the heat tab on the IC is connected to its ground pin, and that the diode therminals must not be accidentally connected to the IC through the heat-sink.
2. The data sheet specifically states that the IC cannot provide protection from short-circuits on the output because current can pass from the battery straight through the inductor and the diode to the output. The implication here is that a Fuse would a real good idea to prevent a total melt down. Most batteries are capable of Huge currents, and with nothing to limit them, you're likely to melt the insulation right off the wires. The output of the converter is a better place to put a fuse in this step-up type of converter, although either place is better than none. I would recommend a 3 amp fuse. A fuse on the input should be rated for higher current, say 5 amps. Any fuse will cause a slight decrease in overall efficiency.
3. One user commented that they burned out their unit by connecting the input voltage backwards. That will certainly burn out the IC and pop C1, the input capacitor. This could be prevented by adding another Schottky diode to the +12V input which would prevent any current from flowing if the inputs were reversed accidentally. This diode would also require heat sinking and would lower the overall efficiency by a couple of percent- but would prevent an accident with battery clips from destroying the converter. A fuse on the input side would also prevent the cables from melting, but would not save the IC from certain death.
John Weber says:
Over the last few weeks I have done considerable research with regard to the 1812 converter.
All the components are available from Digi-key, they have a $25 minimum order policy - however, if your order total is less than $25 they will only charge a $5.00 handling fee. All the parts they carry are equivalent to the values listed in your components table, except for the following: L1 is rated at 2.8 amps, capacitors C1 and C2 are only available in 56 uf 35v and 390 uf 35v respectively, R1 is available in 16.5k or 16.9k - not 16.7k. As for my particular problem with the 1812; as I mentioned before my converter output was 18 volts without a load, but apparently below that under load, because the motors ran considerably slower than normal. The only thing that I remember happening that could have caused this problem is that, one night, I left my scope running and went to bed. The next day, I found it with the optical tube pointing almost straight down and my battery totally dead. Anyway, I replaced the inductor with one from Digi-key (part number DN 4512 - ND), only because white smoke residue on the inside of the case was most concentrated around this component, and my converter is operating just fine so far! I also unscrewed the LT1170CT from the circuit board, stood it up, and and attached a larger heat sink to it (it had been hot enough to burn the circuit board). Now I had to place the board in a deeper project box, which I got from Radio Shack.
As a result of this project I have collected much more additional information, some general, some specific in nature. If you receive other inquiries about the 1812 please feel free to pass along my e-mail address, as I may be of some help to others.
I have tried to be as accurate as possible, but I'm not electronics expert. I haven't even actually built this thing yet. If you fry something because of a stupid mistake I made, it's not my fault :-)