MP Posted March 26, 2004 Report Share Posted March 26, 2004 Just a quick look at the circuit above...looks like you could use 2N3055 in place of the MosFets if you use a resistor to ground as discussed earlier. I think you could also get around the SG3525 with a 556 timer IC. One would have to compare the data sheets to get the correct pin out and frequency for the clocking.This is just a quick look at the schematic and I have not looked at any data sheets on this. Any comments welcome.MP Quote Link to comment Share on other sites More sharing options...
audioguru Posted May 2, 2004 Report Share Posted May 2, 2004 Hotwaterwizard,Your inverter design is coming along, but I have some suggestions:1) D3 and D4 don't appear to do anything. I suggest replacing them with 100 ohm, 1/4W resistors so that the bases of the output transistors have a pull-down instead of a float-down.2) Add diodes in series with the multivibrator transistors' bases with the cathode towards the base, and add 1K ohm resistors from the bases to ground. Most silicon transistors have a maximum reverse-bias voltage of only about 7V for the base-emitter junction due to avalanching. But the avalanching causes damage to the transistor's hFE. In your circuit, the bases are driven to almost -12V.3) Add 2A diodes across the transformer windings with the cathodes to the switched +12V center-tap and the anode to the output transistors' collector. These diodes protect the output transistors from a flyback voltage spike that would occur if the load was removed with the circuit operating.4) Add 8 ohm, 10W resistors in series between the emitter of the driver transistors and the bases of the ouput transistors. These resistors will reduce the voltage across the driver transistors when conducting, and reduce their unlimited current (less E times less I = less heat). Quote Link to comment Share on other sites More sharing options...
Guest Kasamiko Posted May 2, 2004 Report Share Posted May 2, 2004 Just a quick look at the circuit above...looks like you could use 2N3055 in place of the MosFets if you use a resistor to ground as discussed earlier. I think you could also get around the SG3525 with a 556 timer IC. One would have to compare the data sheets to get the correct pin out and frequency for the clocking.This is just a quick look at the schematic and I have not looked at any data sheets on this. Any comments welcome.MPHave you revised the circuit yet?? A 556 looks interesting.. Quote Link to comment Share on other sites More sharing options...
hotwaterwizard Posted May 2, 2004 Report Share Posted May 2, 2004 Zapco Used no MOSFET in their old Power supplies. Here is their Diagram. The problem with this one is it uses PNP and I already have about 150 Transistors 2N3055. The transformer is special made and hard to find. Quote Link to comment Share on other sites More sharing options...
hotwaterwizard Posted May 2, 2004 Report Share Posted May 2, 2004 And here is a Schematic from a Pyramid PB300 Amplifier.PB300G_Schematic.pdf Quote Link to comment Share on other sites More sharing options...
MP Posted May 2, 2004 Report Share Posted May 2, 2004 audioguru, in your comments to hotwaterwizard you mentioned that he should change the 2 diodes in the first transistor stage to resistors in his design. These transistors are connected as a darlinton pair. Adding such a resistor would disrupt the whole purpose of using a darlington pair. It sounds as if you are changing the design to an audio power amplifier. MP Quote Link to comment Share on other sites More sharing options...
ante Posted May 2, 2004 Report Share Posted May 2, 2004 Hi MP, hotwaterwizard,Just a comment on the Zapco diagram; there are no emitter resistors here!Ante ::) Quote Link to comment Share on other sites More sharing options...
audioguru Posted May 3, 2004 Report Share Posted May 3, 2004 Hotwaterwizard,I am sorry, but I made errors when I suggested modifications to your project:1) DO NOT add diodes across the transformer's primary.2) DO NOT add 8 ohm resistors in series with the bases of the output transistors.My reasons are explained in the other inverter post here:http://www.electronics-lab.com/forum/index.php?board=2;action=display;threadid=659;start=15 Quote Link to comment Share on other sites More sharing options...
audioguru Posted May 3, 2004 Report Share Posted May 3, 2004 MP,I did not see any function for diodes D3 and D4 since they are always reversed-biased. Pull-down resistors having only about 20mA were suggested to replace them for reasons as follows:1) To turn-off the output transistors, fast. Without pull-down resistors, the bases of the output transistors just float down.2) To bypass the leakage current of the output transistors and ensure that they actually turn off. The leakage current increases with temperature (thermal runaway). 3) To discharge the capacitance (the miller effect multiplies it) at the bases of the output transistors quickly. Without the pull-down resistors then the output transistors become integraters with their own substantial capacitance.Darlingtons are emitter-followers with their collectors tied together. The miniscule 20mA of current into the pull-down resistors won't affect the amps of current that are fed to the bases of the output transistors.Darlingtons are not used for high-speed-switching nor high quality audio because they are too darn slow. Quote Link to comment Share on other sites More sharing options...
hotwaterwizard Posted May 3, 2004 Report Share Posted May 3, 2004 The diodes are for spike protection to prevent the transistors from burning when the field colapses in the transformer windings. If you look at a good circuit involving a relay you will see a protection diode across the coil for this reason. Quote Link to comment Share on other sites More sharing options...
MP Posted May 3, 2004 Report Share Posted May 3, 2004 looks like the collectors are tied together to me. What schematic are you looking at? Perhaps I am on the wrong page.MP Quote Link to comment Share on other sites More sharing options...
MP Posted May 3, 2004 Report Share Posted May 3, 2004 BTW, I think you are misinformed about darlington pairs. I have attached a pdf document with information regarding them. Darlington pairs are in nearly every IC on the market.MPDarlingtonCircuit.pdf Quote Link to comment Share on other sites More sharing options...
audioguru Posted May 3, 2004 Report Share Posted May 3, 2004 Hotwaterwizard,Good point about spike protection!I was trying to figure out how a positive spike at the collectors of the output transistors can get through their turned-off collector-base junction, and how a reversed-biased diode can arrest this spike. Now I realise that the positive spike produced at the collectors of the output transistors that are turning-off, causes a negative spike at the collectors of the output transistors that have not yet turned-on, due to center-tapped transformer action. Since this negative spike forward-biases the latter's collector-base junction, then it also forward-biases the protection diode, which arrests the spike.Those diodes are good protection.Another concern: Perhaps you should change your multivibrator to the 4047 type that is used in the other post:1) The 4047 has a divide-by-two to ensure that its outputs have exactly a 50-50 mark-space ratio. Yours depends on resistor/capacitor values, parts tolerance and lack of drifting.2) I have seen it happen: If your values are the same then both transistors may "latch" during power-on, when both transistors turn-on and stay-on. Then they won't oscillate.3) The 4047 needs only 1 resistor and 1 capacitor, compared to all the parts for your multivibrator, and costs much less.But if you use a 4047, you will need to add a pre-driver transistor, like the other post, to boost its output current.These two projects are getting to be more and more identical! Quote Link to comment Share on other sites More sharing options...
audioguru Posted May 3, 2004 Report Share Posted May 3, 2004 MP,Thanks for posting the article about darlingtons.There it is, in the schematic on the 1st page, a darlington showing pull-down resistors, and exactly the same description for their function that I gave.Sorry to continue this argument, but ONSemi (Motorola), in their MC34071 opamp, use pull-down current sources on the "darlington" (their wording) input transistors. See the chip's schematic. Most if not all opamps that use darlingtons have this arrangement. The MC34071 data sheet is here:http://www.onsemi.com/pub/Collateral/MC34071-D.PDF Quote Link to comment Share on other sites More sharing options...
MP Posted May 3, 2004 Report Share Posted May 3, 2004 The darlington article would seem this way if you only looked at the schematic and read the first few lines. If you read the article close enough you will learn that these resistors are not necessary and that they can be added to speed up the transition. Not to keep the supply from floating. This is the reason ON Semi uses this method. Speed of transition. I gave some thought to posting a schematic with both versions with and without the resistors, but elected to post the complete article and let everyone read instead.MP Quote Link to comment Share on other sites More sharing options...
audioguru Posted May 4, 2004 Report Share Posted May 4, 2004 MP,Don't you agree that those cheap resistors have nothing to lose and everything to gain? Quote Link to comment Share on other sites More sharing options...
MP Posted May 4, 2004 Report Share Posted May 4, 2004 No. I see no gain here and a question as to whether the circuit will function as originally intended with the voltage divider network that it will create. This is not the same as the resistors in the darlington configuration from the article. Note how you will be adding voltage dividers in combination with the other resistors already in the circuit.MP Quote Link to comment Share on other sites More sharing options...
ante Posted May 4, 2004 Report Share Posted May 4, 2004 MP,What is protecting the transistors in the Zapco circuit from destruction if the current starts to flow unequally through the emitters ?Ante ::) Quote Link to comment Share on other sites More sharing options...
MP Posted May 4, 2004 Report Share Posted May 4, 2004 ante, I am not sure what you refer to the zapco circuit?If you refer to the darlington pairs in the inverter project, the emitters are not directly connected to ground on the first transistor.This is only a problem when you have parallel transistors sharing a common ground on the emitter.MP Quote Link to comment Share on other sites More sharing options...
audioguru Posted May 5, 2004 Report Share Posted May 5, 2004 MP,With the 8 ohm resistors gone, there is no voltage divider. The pull-down resistors turn-off the output transistors and discharge capacitances, when the driver transistors turn-off. Without pull-down resistors, the bases of the output transistors gradually (as their capacitances slowly discharge though their leakage) "float" down to a voltage level that is determined by their substantial leakage current (which increases with temperature) and so the output transistors don't completely turn-off.The Zapco PSU schematic is in Reply #28, and shows 4 paralleled transistors on each side, without emitter resistors.Ante,Zapco either matched the paralleled transistors, or had good luck.Think of emitter resistors as a form of negative feedback. When a high-gain transistor attempts to conduct a large collector current, then that current creates a voltage-drop across the emitter resistor, which reduces the base-emitter voltage and therefore reduces base-drive.A transistor with less gain will attempt to conduct less current and therefore will have less base-drive reduction. So the gains of the transistors are equalized.Without emitter resistors, when a high-gain transistor is paralleled with a low-gain one, then the high-gain transistor will conduct more current than the low-gain one, which results in unbalanced current sharing. Without balanced sharing, the high-gain transistor may exceed its maximum current and/or thermal rating and blow-up. Quote Link to comment Share on other sites More sharing options...
audioguru Posted May 5, 2004 Report Share Posted May 5, 2004 For all,In another web-site's projects forum, there was a request for an inverter to provide all the power for an electricity-heated home.Numbers such as 12,000 Amps at 12 Volts (144,000 Watts) were discussed! Enough batteries to fill a large garage!But they didn't think of charging them by dragging them along the ground behind a motorcycle. Quote Link to comment Share on other sites More sharing options...
MP Posted May 5, 2004 Report Share Posted May 5, 2004 I went back through the posts and found the zapco post. There are quite a few schematic posts in this thread. You are trying to redesign an amp that is already on the market? Build a new amp? What is the purpose of this subject?This thread does not seem to have any direction.MP Quote Link to comment Share on other sites More sharing options...
ante Posted May 8, 2004 Report Share Posted May 8, 2004 MP,Now that you found the Zapco circuit, this is only one of many in fact most of the circuits I have seen and repaired have no emitter resistors. Question is how do they get away with it? All this equipment is out there working day after day and they are all designed wrongly?? Ante ::) Quote Link to comment Share on other sites More sharing options...
MP Posted May 8, 2004 Report Share Posted May 8, 2004 ante, I have been researching this question since you and I had several discussions on the matter some time ago. You will have to have "some" resistance. You cannot have a direct short in an AC or DC circuit without ending up with a problem. However, what I have found is that these circuits are working on a different principle.1. The quick switching action is not considered a direct short such as any other alternating current that crosses the zero path over and over again. Please note that some digital circuits on the market work entirely on the principal that switching capacitors which are connected to ground on and off with different duty cycles makes a variable resistor which is digitally controlled.2. The circuit is providing resistance in other ways. There is also resistance in the transformer as well as the capacitive/reactive resistances in the circuit.With this in mind, one does not need added resistance if the circuit is designed correctly. Therefore, I will have to say that both methods can be correct. You will notice that this is a different view than I had when we discussed this before. As I am researching this more, I am leaning more toward your view on this, since I can see where the resistance comes from.Also, I have had to brush up on what I know in this area of electronics as I am now working more with solar design than I have been before.MP Quote Link to comment Share on other sites More sharing options...
ante Posted May 8, 2004 Report Share Posted May 8, 2004 MP,With all respect; in the inverter a short never occur because there is a transformer winding in series, which is the resistance. The winding is an inductor so it has a time constant and it opposes to a change in current, this prevents a short. This phenomena is what we use in a converter. I am very happy that our opinions are getting closer! ;DAnte ::) Quote Link to comment Share on other sites More sharing options...
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