audioguru Posted September 26, 2004 Report Posted September 26, 2004 Hi Guys,In another post we have discussed why this part of a project doesn't work as a low-pass filter that is described in the project's text, and produces severe distortion and noise.The original part of the schematic is on the left side, and I re-drew it on the right side to make it easier to see.This audio amplifier's input is near line level and is DC-coupled from the previous preamp opamp. It doesn't have any feedback, so its voltage gain of about 200,000 will amplify its own and the previous opamp's offset voltage of up to 6mV, so that its output will be resting near a supply voltage, instead of its output being biased near half-supply voltage. Therefore when an input signal is applied, its output will half-wave rectify it, causing severe distortion.The high gain will also amplify the high internal noise level of the 741 and the noise from the previous stage, so that its output will be full of noise.The low-pass single-stage action of R5 plus R6 and C4 is cancelled by the high-pass single-stage action of R5 and C3, so that the output frequency response will be wideband. Quote
audioguru Posted September 26, 2004 Author Report Posted September 26, 2004 A simple error was made on the original schematic since the junction of C3 and R7 should be connected to pin 6 of the opamp. Then it will be a Sallen and Key Equal Component Butterworth Low-pass Filter.Do you think that C3 should have twice the value of C4? Only if the opamp has a gain of 1.The feedback from R7 and R8 create a voltage gain of 1.6, so the RC networks can have equal values. Quote
MP Posted September 26, 2004 Report Posted September 26, 2004 The theory section is the perfect place for you to post this, since:a. You have not attempted to build this projectb. You do not know the intentions of the original author for the separate stagesc. The circuit works fine in several different simulator programs (with simple schematic correction)d. anything that you cannot figure out must be redesigned with theoryIt should also be noted that:a. Others have told me they have built this circuit (with simple schematic correction)b. You have some kind of issues with Aaron Cake and everything that he posts to the web. Interesting since you are both from Canada.No need to reply audioguru. I only posted this for those who might be new and are not familiar with your past on this forum.This reminds me of when you attacked the author of the temp controlled fan until I asked him to join the forum and confront you. Then you backed down and quit telling everyone the project would not work. Must the author of every project come to the forum and answer to you? I think not. As I recall, you were beginning to redesign that one as well.MP Quote
audioguru Posted September 27, 2004 Author Report Posted September 27, 2004 Sure MP,a) My heart is fine and I don't need this Quote
MP Posted September 27, 2004 Report Posted September 27, 2004 Yeah, if you can't get it to work, much easier to design another one from scratch than find the fix.Hostile attacks on you? Whatever ::)Sounds like a little paranoia there.MP Quote
audioguru Posted September 27, 2004 Author Report Posted September 27, 2004 Which project did I redesign from scratch?Just simple little fixes, that's all.The power supply project has only 1 additional output transistor added, plus 2 emitter resistors to balance them. Same circuit, but with proper parts selection. No redesign.What about the 500W inverter project, is that a complete redesign?Same oscillator, but with its short-circuit removed and a proper choice of its RC network.Same driver IC, but a suitable dual instead of the original quad.Twice as many output transistors and their short-circuit removed plus balancing emitter resistors.Protection diodes also added.Now there isn't a melt-down anymore.My Plants Watering Watcher circuits are blinking brightly, so I better water the plants soon. They work a lot better than the original project, are much brighter and their smaller batteries last much longer. Same circuit, but with a modern (only about 18 years old) more suitable version of the same IC, plus a few more parts to make it better.The simplest fix that I made was to the "Transistor Tester" that had the clock input pin of its flip-flop grounded by mistake. Nobody else noticed that error. Quote
jam Posted September 28, 2004 Report Posted September 28, 2004 i wnt to know about that inverter of yours?! so how is that? can you give me a design of that thing?i have this thng in mind coz i wanna make it. here's my specs in mind: 12vdc - 230vac, 500w inverter. pure sinewave output. okie... btw, im mark from phil.you? is the parts of your design available here in phil? Quote
audioguru Posted September 28, 2004 Author Report Posted September 28, 2004 Hi Mark,Welcome to our forum.Are you from St. Phillips Island, The Phillipines or Philadelphia?I'm up here in Canada, no snow here, yet.The 500W inverter that is mentioned above is a fix to a defective one in our Projects section. A member has reported that the fixed version works very well and gives 720W output. It has a square-wave output. Do you want a link to its schematic?I am designing a sine-wave inverter that will start as 300W but can go higher if you add more Mosfets. It is not finished yet. Quote
jam Posted September 29, 2004 Report Posted September 29, 2004 im from phillipinesand i wanna know the link of that thingpls help me sois the parts of that inverter available here in phillipines?what is the link to its diagram?is your sinewave inverter done? Quote
audioguru Posted September 29, 2004 Author Report Posted September 29, 2004 Hi Mark,It is good that you are from The Phillipines because Ronnie (called Rhonn, and his handle here is Kasamiko) is also there. He helped me fix this 500W inverter and he has many inverter circuits. His e-mail address is on this schematic of the fixed inverter that he drew. Quote
Kevin Weddle Posted November 27, 2004 Report Posted November 27, 2004 What a good example of a how a transformer works. I normally write the voltage like this :10 -10 10 0 and this is equal to this 0 0 10The second example shows the voltages as they come off the circuit posted. You can read this using ground for reference. We assume ground to be a 0 volts constant but it does not have to be. If we look at the second example, we can set the lower part as our ground reference. Then we say we have 10 with respect to 0. Keeping in mind that we can simulate the effect both ways, we can say that it is 0 and -10. We could also say that because there is no reference coming from the transformer output that we have 20 and 30. It is your option as to how you want to draw it out. Sometimes it makes it easier to see. Quote
audioguru Posted November 27, 2004 Author Report Posted November 27, 2004 Hi Kevin,A 20VAC center-tapped transformer secondary winding is usually written like this: 10-0-10. The "0" is usually the ground, and the "10" at each end of the winding are of opposite phase.You can make a full-wave rectifier with a center-tapped winding by adding only two diodes, in series with each end with the same polarity and the free end of each diode are joined together for the unfiltered DC output.In the inverter project the center-tap of the transformer is connected to +12V. When an output transistor drives one end of the winding to ground, transformer action drives the other end of the winding to 24V. Therefore the entire winding has a square wave of 24V across it. Quote
Heey Posted April 16, 2005 Report Posted April 16, 2005 Hi,Can I use any other transistor than C1062?I don't find it in any of the shops here in norway, so......Thanks! Quote
audioguru Posted April 16, 2005 Author Report Posted April 16, 2005 Hi Heey,Welcome to our forum. ;DThere is nothing special about the 2SC1061 transistor except it is available in the orient. Any 40V, 3A to 6A medium power transistor will replace it. I would use a TIP31 but maybe you could find a BDxxx equivalent. Quote
Kevin Weddle Posted April 17, 2005 Report Posted April 17, 2005 Something worth noting is that there is a difference between low pass filtering and active filtering. I am not saying that they are wrong, only loose in how they describe the circuit. Whenever you use active filters, you are using the reactive component as the gain element. So your voltage is really going to take off and you will realize a db gain that more than a regular filter which is 20 db per decade.And incidentally I see the reactive components are in the positve feedback portion. This will directly affect the value of input voltage which is not gain and therefore is in fact just a low pass filter in conjunction with an opamp. Notice how they will use capacitors just about anywhere. You can only speculate what the db attenuation is. Probably somewhere between 20 and 40. It reminds me of the phase shift oscillator where you have a series of capacitors that combine to give you 180 degrees. Amazing. It is for this reason you can have a 20 to 40 db attenutation from just a simple RC circuit. Quote
audioguru Posted April 17, 2005 Author Report Posted April 17, 2005 Kevin,A Butterworth active filter is the same as discrete RC filters except is much better.Two cascaded RC filters affect each other because the 2nd one loads-down the 1st one. If you buffer the input of the 2nd one so that it doesn't cause loading, the combination of the two filters cause the response at their calculated cutoff frequency to be down 6dB and the response to be very droopy. Long before their cutoff frequency their response will begin dropping. At a frequency that is far from their cutoff frequency the response will eventually drop at exactly 40dB per decade.A Butterworth two-pole active filter uses a form of positive feedback to hold the output up just before the cutoff frequency. This results in flat response up to near the cutoff frequency, exactly 3dB down at the cutoff frequency, a very sharp response corner at the cutoff frequency and the response dropping off immediately at exactly 40dB per decade.They don't use capacitors "just about anywhere". Mr. Butterworth, Mr. Sallen and Mr. Key planned where the capacitors go long ago. You don't have to speculate what the attenuation will be, it has been documented.A phase-shift oscillator doesn't use a series of capacitors. It uses cascaded RC stages that each provide some of the phase shift. Three stages at 60 degrees each equals a total of 180 degrees.I don't know of any circuit that gives more than 20dB per decade with only a single RC stage. Quote
Kevin Weddle Posted April 20, 2005 Report Posted April 20, 2005 Take a low impedance RC circuit, tag on a high impedance RC circuit, and there you have a 40db circuit. I believe they use this concept in the phase shift oscillator. Quote
audioguru Posted April 20, 2005 Author Report Posted April 20, 2005 Kevin,You can't use only two RC stages in a phase-shift oscillator, it needs at least three stages.Since the output load and the input impedance of the input to the inverting amplifier are low impedance, a buffer is needed and you might as well use RC stages having the same value. Quote
joseph@glolab Posted November 29, 2005 Report Posted November 29, 2005 Hi! im joseph I just want to know what IC is used in the 500W low cost power inverter Quote
audioguru Posted November 30, 2005 Author Report Posted November 30, 2005 There are two ICs used in the fixed and improved circuit.Go to www.google.com or www.datasheetarchive.com and look at their datasheets. Quote
arteka Posted December 6, 2005 Report Posted December 6, 2005 can someone post a corect scheme of this inverter with sinusoidal output plz? Quote
audioguru Posted December 7, 2005 Author Report Posted December 7, 2005 This is a simple square-wave inverter. Its output transistors switch fully-on then fully-off to reduce their heating. A sine-wave inverter is completely different and much more complicated. If you use a linear amplifier to drive the transformer then the transistors would heat nearly as much as the load. We talked about sine-wave inverters on another post. Quote
indulis Posted December 7, 2005 Report Posted December 7, 2005 Show me the math.................A Butterworth filter, irregardless of order, only has poles in it's transfer function, so could you explain...."A Butterworth two-pole active filter uses a form of positive feedback to hold the output up just before the cutoff frequency." Quote
audioguru Posted December 7, 2005 Author Report Posted December 7, 2005 A Butterworth filter has a Q of 0.7 to provide its flat response, then its sharp drop at cutoff. If you connect two RC networks in series (buffered, so they don't interact), then they are a Bessel filter, the Q is only 0.5 and the response is sloppy and droopy. ;D Quote
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