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LEECH666

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  1. No, this is the most recent parts list / circuit diagram: http://www.electronics-lab.com/forum/index.php?topic=19066.msg1010703#msg1010703 I really think someone should update the frontpage to the most recent version. That way everyone would find the most recent version of the circuit and it would also stop the people from asking the same question over and over again.
  2. The latest circuit diagram can be found here: http://www.electronics-lab.com/forum/index.php?topic=19066.msg1002666#msg1002666 According to this application note from Fischer Elekronik (heat sink manufacturerer) http://www.fischerelektronik.de/pim/upload/fischerData/datasheet/base/technischeerlaeuterungen_d.pdf (sorry it's in German, there used to be an english version of this but it's gone) you calculate it like this. Florian
  3. More pictures. Not quite sure which variant I want to use. 2N3055 on a L-profle mounted on the heat sinks or TIP3055 mounted flat on the heat sinks.
  4. I've got four TIP3055 (TO-218 / SOT 93) here, and two 2N3055 (TO-3). I've decided to use silent CPU coolers. I am uncertain if a single BD139 will be enough to drive four TIP3055s. I am only aiming for 30V / 3A output, so the additional TIP3055s would just be there to compensate for the lower power rating and to dissipate the heat more easily. I have two options now ... (see attached pics):
  5. Very nice. Is this the latest REV2 version of the PSU or the original version? The shematic seems to be different, and you're only using one output power transistor.
  6. In any case a smaller Kelvin per Watt value is always safer. A heat sink with 5.24 K/W is most likely more expensive than one with 11.9 K/W. Florian
  7. Nothing wrong really. Audioguru either made a small typo and actually wanted to type (130°C - 76.4°C)/4.5W = 11.91K/W or he substracted another 30°C of safety margin from the maximum junction temperature value of 130°C. (130°C - 30°C - 76.4°C)/4.5W = 5.24K/W Florian
  8. I noticed that only the original Philips datasheet of the BD139 specifies a value for fT = 190 Mhz. So far I haven't found a transistor in the TO-220 package that's that fast. I think the fastest one I've found so far is the 2SD1762 @90Mhz or so, but the dc current gain doesn't seem to be as high. MJF44H11 seems to have a similar minimal dc current gain and fT=50Mhz. *sigh* Not as easy as I thought. ;) Guess I'll just stick to the BD139 for now and see how that transtor works out on a heat sink. Thanks again for your help audioguru. Florian
  9. Ah thanks. I'll have to adjust my calculations then. I've got a 160 VA transformer with 1x 230V primary and 2x 15V (5,33A) secondary windings. Im going to put the secondary windings in series so the resulting RMS voltage should be 30V. My rectifier bridge drops 1.1V for each diode inside of it. So I guess my total unregulated supply voltage should be: Us = 30V * sqrt(2) - 2 * 1.1V = 42.43V - 2.2V = 40.23V If the TO-126 package of the BD139 is so bad couldn't I just use some other general purpose NPN transistor in a TO-220 package? Or maybe drive some Darlingtons directly from the opamp instead of the 2N3055? Florian
  10. Hello once again. I still haven't managed to build this supply even though I designed my own PCB, which probably is still crap. Who cares. The reason I post here today is because I finally want to finish the PSU and I am not sure what type of heat sink I should pick for Q2 (BD139). It only says "on a pretty big heat sink" in the parts list, which is a kind of vague assertion. Some toughts on this topic (not necessarily correct or complete): Q2 is used to drive the two power transistors Q4 and Q5. At the nominal 3A output current (IOUT=3A) the current gets shared between the two transistors and each transistor has to handle 1.5A of current. I took a look at the OnSemi datasheet for the 2N3055 and the minimum DC current gain (B) seems to be 20 (worst case). IB = IC / B IB = 1.5A / 20 = 75 mA Since we're driving two 2N3055s we have to supply two times the base current that I just calucated. IBx = 2 x IB = 2 x 75mA = 150 mA Since I sort of suck at electronics I have no idea how much voltage the transistor Q2 will see (worst case). However I need the voltage to calculate the total power dissipated by Q2, which then would allow me to calculate an appropriate heat sink. Looking into the output transitor circuit more closely I think it kinda works like this: Supply Voltage: US = 30V Q2 Emitter Resitor Voltage: UR16 = UR24 + UBE_Q5 = (IOUT/2 * R24) + UBE_Q5 = (1.5A * 0.33R) + 0.7V = 0.495V + 0.7V = 1.195V Q4 and Q5 Collector Emitter Voltage: UCE_Q4 = UCE_Q5 = US - UR24 = 30V - 0.495V = 29.505V Q2 Collector Emitter Voltage: UCE_Q2 = US - UR16 = 30V - 1.195V = 28.805V Power dissipated in Q2: P_Q2 = UCE_Q2 * IBx = 28.805V * 150mA = 4.32W Thermal resistance of (heat) sink to ambient: RthSA = to be calculated Thermal resistance of case to (heat) sink (thermal grease layer): RthCS = 0.3K/W (don't know where this number comes from, http://sound.westhost.com/heatsinks.htm#s7 suggest 0.25 for beryllium oxide which is kind of grease like (?)) Thermal resistance of transistor junction to case: RthJC = 10K/W (from ST datasheet) Maximum ambient temperature: Ta = 55°C (selected, ambient temp in summer in north germany hardly ever gets over 35°C, 20°C safety margin included) Maximum operating junction temperature of transistor: Tj = 150°C (from ST Datasheet) Tj = P_Q2 (RthJC + RthCS + RthSA) + Ta (equation from http://irf.custhelp.com/app/answers/detail/a_id/91/~/heat-sink-selection-and-thermal-calculation.) Solve for RthSA: (Tj-Ta/P_Q2) - (RthJC + RthCS) = RthSA RthSA = ([150°C - 55°C] / 4.32W) - (10K/W + 0.3K/W) = 11.69K/W RthHS = RthSA * 0.80 = 11.69K/W * 0.80 = 9K/W (another 20% safety margin) So the heat sink should have a thermal resistance of around 9K/W or lower. Hope my calculations are ok, not entirely sure. I probably forgot to take the rectification factor into account and US is probably higher than 30V. What type of heat sink did you guys use in your build of this PSU? Florian //Edit: Few typos corrected.
  11. Thanks for the hint regarding Q3. That sure saved me a lot of trouble. I revised my schematic so that the ref names of the components in the schematic are matching the original schematic. Also it should be a little more easy to read it now as I used bigger font sized in most places. I also worked hard and redesigned the whole PCB (not fully done yet) but here is my progress (see attached files). The Transistor Q3 is still connected the other way around in the PCB, in case you wonder. Gonna fix that tomorrow. Also might put a bit more thought into the track widths. The thick one is 5mm and the traces branching from the thick one are either 2 mm or 1 mm. Anyway it's pretty late and i need to get up in 4 hours. Good night, Florian //EDIT Updated the schematic for even better readability. WIP_PCB_LP0002.pdf SCM_LP0002_2010_12_07.pdf
  12. Sorry about the popups. I usually use some ad blocking meassures (like Adblock+ or Proxomitron) to get rid of ads. So I didn't even notice the pop up infestation of that page. I will re-upload the files once i get my main PC back to work (currently using my gaming PC). Um ... I recon that not all tracks need to be 3-5 mm in width. I think the tracks connected to the collectors / emitters of the 2N 3055 transistors for sure and and pretty much ever track connected to the output connector and also every track regarding the connector through which I want to connect the big filter caps (12.000 LP0002_SCHEMATIC.pdf
  13. Note to self: Values of RV3 and RV1 are switched in the shematic. RV1 needs to be the 5K poti and RV3 needs to be the 100K poti.
  14. Oh yeah you're right, I actually leared that stuff at technician school, but I am a little rusted. I think there is also a spreadsheat by Fischer Elektronik that explains the process. I'm going to look it up. Found it: >> PDF <<
  15. Hello again, I got a question regarding the heatsink of the BD 139 transistor. I think "a pretty big heatsink" is a bit vague/relative. Could someone translate "pretty big heatsink" into a
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