walid Posted January 12, 2006 Report Share Posted January 12, 2006 I understand the 555 timer IC when used in its astable mode as an oscillator, but the one shown below i need some help in:(1) how this oscillator work?(2) the function of the diode?(3) how to choose the components values to achieve some freq.?(4) What the formula used to calculate the freq.?(5) Can I replace Q1 with a NPN one?Note that bat = 6 v and IC = LM393Thanks to everybody Quote Link to comment Share on other sites More sharing options...
audioguru Posted January 12, 2006 Report Share Posted January 12, 2006 Hi Walid,An LM393 isn't an opamp, it is a dual comparator. A comparator oscillates at a high frequency if it has negative feedback.Its pin 4 needs to be grounded.It is a Schmitt-trigger oscillator. Its pin 5 is pulled up to about +2.7V and down to about -2.7V when the output toggles. The voltage at pin 2 is delayed by C2.You calculate its frequency by figuring how long it takes for R4 and R5 to charge and discharge C2 to the +2.7V and -2.7V thresholds, for each half-cycle, then add the half-cycles.The diode makes C2's negative charging time short, because then the diode is forward biased and R5 is put into the circuit.An NPN transistor can be used if the comparator's output has a pull-up resistor for its base bias current. The comparator doesn't have a push-pull output like an opamp, just the collector of an NPN transistor that can turn-off the base of the external NPN transistor. Quote Link to comment Share on other sites More sharing options...
walid Posted January 12, 2006 Author Report Share Posted January 12, 2006 Hi audioguru, thank u very much for your kind answer.Its pin 5 is pulled up to about +2.7V and down to about -2.7V when the output toggles. (1) pin 5 always connected to the voltage divider formed by R1 and R2 and so it is always at 3v potential, correct me.(2) How u reach these numbers, +-2.7v(3) What the values of the o/p at pin 7?The diode makes C2's negative charging time shortIt is the first time for me to hear about negative charging!An NPN transistor can be used if the comparator's output has a pull-up resistor for its base bias current. What is a pull-up resistor and what it do?Than u very much Quote Link to comment Share on other sites More sharing options...
audioguru Posted January 13, 2006 Report Share Posted January 13, 2006 (1) pin 5 always connected to the voltage divider formed by R1 and R2 and so it is always at 3v potential, correct me.No. When the comparator's output is at +9V, R3 pulls pin 3 to a positive voltage. When the output of the comparator is at -9V, R3 pulls pin 3 to a negative voltage.(2) How u reach these numbers, +-2.7vActually, it is +3.0V and -3.0V because the output transistor of the comparator doesn't have an emitter-follower loss like opamps. R2 goes to -9V. Both R1 and R3 are at +9V. The 18V is divided by 3 so pin 3 is 6V less than the positive supply. The opposite occurs when the comparator's output is at -9V.(3) What the values of the o/p at pin 7?Pin 7 swings almost to +9V and almost to -9V.It is the first time for me to hear about negative charging!You have a negative supply. I didn't notice that C2 connects to the negative supply instead of to 0V which is more common, then C2 is charged positively slowly by R4 and is discharged quickly by R5 in parallel with R4. An NPN transistor can be used if the comparator's output has a pull-up resistor for its base bias current. What is a pull-up resistor and what it do?An NPN transistor with its emitter connected to -9V needs a positive current feed to its base for base current. The output of the comparator doesn't have anything to provide a positive current, so a resistor is needed from the base to the positive supply. It is called a "pull-up" resistor. Quote Link to comment Share on other sites More sharing options...
walid Posted January 14, 2006 Author Report Share Posted January 14, 2006 Hi guru, thank u very muchYou have a negative supply. I didn't notice no it is 6 volt (4 x 1.5V batt), the +ve terminal is 6V and the -ve terminal is 0 V.Pin 7 swings almost to +9V and almost to -9V.From where u get the + and - 9 voltAn NPN transistor with its emitter connected to -9V needs a positive current feed to its base for base current. The output of the comparator doesn't have anything to provide a positive current, so a resistor is needed from the base to the positive supply. It is called a "pull-up" resistor. u want to say that a "pull-up" resistor connect the point to +batt and the a pull-down resistor connect a point to ground Quote Link to comment Share on other sites More sharing options...
audioguru Posted January 14, 2006 Report Share Posted January 14, 2006 Sorry, Walid, I didn't notice your 6V battery. Here is my reply corrected for a 6V battery. The negative of the battery is the circuit's "ground".Quote from: walid on January 12, 2006, 12:01:46 PM(1) pin 5 always connected to the voltage divider formed by R1 and R2 and so it is always at 3v potential, correct me.No. When the comparator's output is at +6V, R3 pulls pin 3 to a more positive voltage than 3V. When the output of the comparator is at ground, R3 pulls pin 3 to a more negative voltage than 3V.Quote(2) How u reach these numbersActually, it is +4.0V and +2.0V because the output transistor of the comparator doesn't have an emitter-follower loss like opamps. R2 goes to ground. Both R1 and R3 are at +6V. The 6V is divided by 3 so pin 3 is 2V less than the positive supply. The opposite occurs when the comparator's output is at ground, pin 3 becomes 1/3 of the supply above ground, which is 2.0V.Quote(3) What the values of the o/p at pin 7?Pin 7 swings almost to +6V and almost to ground. Quote Link to comment Share on other sites More sharing options...
walid Posted January 15, 2006 Author Report Share Posted January 15, 2006 hi audioguruR3 pulls pin 3 to a more positive voltage than 3VI think that u mean pin 5 not pin 3because the output transistor of the comparator doesn't have an emitter-follower loss like opamps. You may mean the internal (inside the IC) transistor.When the comparator's output is at +6V, R3 pulls pin 3 to a more positive voltage than 3V. When the output of the comparator is at ground, R3 pulls pin 3 to a more negative voltage than 3V.This is OK, i calculate and understand it but i have question:now let the o/p at pin 7 is 6volt then the voltage at pin 5 = 4v, if pin 6 is at ground, the o/p at pin 7 will remain at 6v. to change this o/p, pin 5 voltage must be less than pin 6 (5 more -Ve or 6 more +ve).can u, please, tell me more about this mechanesim of oscillation.thank you. Quote Link to comment Share on other sites More sharing options...
audioguru Posted January 15, 2006 Report Share Posted January 15, 2006 I think that u mean pin 5 not pin 3Oops, yes.You may mean the internal (inside the IC) transistor.Yes. The output transistor inside the comparator is an NPN transistor with its emitter at the negative supply and its collector at the output pin.now let the o/p at pin 7 is 6volt then the voltage at pin 5 = 4v, if pin 6 is at ground, the o/p at pin 7 will remain at 6v. to change this o/p, pin 5 voltage must be less than pin 6 (5 more -Ve or 6 more +ve).can u, please, tell me more about this mechanesim of oscillation.Pin 6 never gets to ground. C2 charges and discharges between 2V and 4V.R3 causes positive feedback for a snap-action of the comparator with hysterisis.With pin 7 at 6V then pin 5= 4V. Pin 6 is somewhere between 2V and 4V and C2 is charged by R4 until its voltage is 4V. When pin 6 tries to go more positive than 4V, the high gain of the comparator causes its output at pin 7 to drop. With pin 7 dropping, R3 also causes the voltage at pin 5 to drop, which speeds-up pin 7 dropping until pin 7 is at ground, pin 5 is 2V and R5 in parallel with R4 begin quickly discharging C2. When pin 6 tries to go lower than 2V then the high gain of the comparator causes pin 7 to rise. With pin 7 rising, R3 also causes the voltage at pin 5 to rise, which speeds-up pin 7 rising until pin 7 is at 6V, pin 5 is 4V and R4 slowly begins charging C2 again. ;D Quote Link to comment Share on other sites More sharing options...
walid Posted January 17, 2006 Author Report Share Posted January 17, 2006 Hi audioguruFirst thank you very much for your time with us Pin 6 never gets to groundI disagree, long time ago we have taught that caps are approx. short circuit during the charging or discharging process, so it connects pin 6 to ground... R3 causes positive feedback for a snap-action of the comparator with hysterisis.What u mean by snap action and hystreisis, i know that snap is a very quick and high voltage pulse.=====================================Now I have more questions like:(1) Why the designer did not use another diode in series with R4 to make the discharge process only through R5, since R4//R5 approx = R5(2) From your point of view, what the modifications that must be done to the circuit to work more efficient with less components?thank you Quote Link to comment Share on other sites More sharing options...
indulis Posted January 17, 2006 Report Share Posted January 17, 2006 Without going into the differential equations, the solution and general form for charging is:Vcap=Vapplied*(1-e^-t/RC)and for discharge is:V=Vcap*(e^-t/RC)on the discharge, don't forget to include the forward voltage of the diode. Quote Link to comment Share on other sites More sharing options...
audioguru Posted January 17, 2006 Report Share Posted January 17, 2006 I disagree, long time ago we have taught that caps are approx. short circuit during the charging or discharging process, so it connects pin 6 to ground...You were taught wrong. Even the main filter capacitor following the rectifier in a power supply connects to ground and is continuously charging and discharging. It doesn't short to ground.A capacitor has a low reactance at high frequencies, not a low resistance.What u mean by snap action and hystreisis, i know that snap is a very quick and high voltage pulse.Snap action is quick (without a high voltage). Look-up hysteresis in a dictionary or on Google.=====================================(1) Why the designer did not use another diode in series with R4 to make the discharge process only through R5, since R4//R5 approx = R5Another diode won't make any difference and isn't required.(2) From your point of view, what the modifications that must be done to the circuit to work more efficient with less components?I might use a Cmos 555 timer IC to replace the comparator but it costs more. Then R1, R2, R3 and R6 aren't required. A Cmos Schmitt-trigger inverter would also make a simple and cheap pulse generator. Quote Link to comment Share on other sites More sharing options...
indulis Posted January 17, 2006 Report Share Posted January 17, 2006 He wasn't "exactly" taught wrong, you just have to remember the conditions under which it is true!! Take a simple series RC circuit (no initial conditions i.e. 0 volts on the cap)with a DC source and a switch. If you close the switch at time "zero" the capacitor at that instant indeed looks like and behaves like a short!!!!!!! If you have a charged capacitor and you apply a transient, you would model it as a discharged capacitor in series with a voltage source that is equal in value to the capacitor voltage prior to the transient!!! In this case such a lage value for R4 it doesn't matter much, but a diode in series would block R4 from being in the discharge equation. If R4 was 100K, IT WOULD MAKE A BIG DIFFERENCE!!! Quote Link to comment Share on other sites More sharing options...
audioguru Posted January 17, 2006 Report Share Posted January 17, 2006 Hi Indulis,Of course the capacitor's voltage is at ground the moment power is applied to the circuit, then the comparator waits an extra amount of time for its voltage to reach the Schmitt-trigger threshold voltage. It is like the famous 555 type of 1st half-cycle time extension.Speaking of the 555, I forgot to mention that it wouldn't need a diode for it to have a long capacitor charge time and a short discharge time. Quote Link to comment Share on other sites More sharing options...
indulis Posted January 18, 2006 Report Share Posted January 18, 2006 "Of course the capacitor's voltage is at ground the moment power is applied to the circuit...."That's not the point.... you said, "You were taught wrong." He wasn't, and I just gave a simple example of how a capacitor can "appear" to look like a "SHORT" (short being the key word) to ground. While applying principals of transient circuit analyisis, capacitors become "SHORTS" with a series voltage source and inductors become "OPENS" with a parallel current source. Walid's application of those principals just wasn't 100% correct. Quote Link to comment Share on other sites More sharing options...
audioguru Posted January 18, 2006 Report Share Posted January 18, 2006 Whatever teaching-about-capacitors method Walid's teacher used that made him think that capacitors are a DC short is wrong.Maybe the teacher was completely mixed-up and was really talking about inductors.Maybe Walid is completely mixed-up about capacitors and inductors. ??? Quote Link to comment Share on other sites More sharing options...
indulis Posted January 18, 2006 Report Share Posted January 18, 2006 Mr KlampferExplain to me what you don't understand about the following statement?"While applying principals of transient circuit analyisis, capacitors become "SHORTS" with a series voltage source and inductors become "OPENS" with a parallel current source. " Quote Link to comment Share on other sites More sharing options...
audioguru Posted January 18, 2006 Report Share Posted January 18, 2006 Indulis,I understand that a transient contains high frequencies and that the reactance of a capacitor is low (like an AC short) a high frequencies. It isn't a DC short.I don't understand how an inductor can be "open" when fed from a DC current source.An inductor resists the flow of high frequency current due to its reactance rising at high frequencies. At DC an inductor is a short. Quote Link to comment Share on other sites More sharing options...
walid Posted January 18, 2006 Author Report Share Posted January 18, 2006 Hi audioguruYou were taught wrong. yes this is true, i hope that you always teach me.I forgot to mention that it wouldn't need a diode for it to have a long capacitor charge time and a short discharge time. Quote Link to comment Share on other sites More sharing options...
indulis Posted January 18, 2006 Report Share Posted January 18, 2006 First, let me reiterate that the concept I am referring to, is applicable to transient circuit analysis (not steady-state), and by definition only holds true for the instant in time that you are interested. Again, take a simple series RC circuit with a source and switch. To make it simple, let Quote Link to comment Share on other sites More sharing options...
audioguru Posted January 18, 2006 Report Share Posted January 18, 2006 I forgot to mention that it wouldn't need a diode for it to have a long capacitor charge time and a short discharge time.Please explain this more.Here's a sketch: Quote Link to comment Share on other sites More sharing options...
audioguru Posted January 18, 2006 Report Share Posted January 18, 2006 Remember also, that AC can be considered infinitely small DC steps!!!I agree with your explanation and your math.Why does this site mess-up certain English symbols? I see a lot of open squares in your text, and in the text of my projects.My extremely low distortion sine-wave generator uses a CD4018 Johnson Counter as a DAC to make a stepped sine-wave with very low 2nd to 8th harmonics. You could say it is an over-sampled sine-wave.It is filtered by an 8th-order switched-capacitor Butterworth lowpass filter IC that also has a stepped output but at a very high frequency. It is followed by a 2nd-order (or 3rd?) analog Butterworth lowpass filter circuit. It has steps all over the place but its output is perfect! ;D Quote Link to comment Share on other sites More sharing options...
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