Help!
I built the amp on page 1 of
http://www.national.com/ds/LM/LM3886.pdf
using samples I ordered from National. I power it from a 42V transformer. The problem, though, is that no sound comes out of the speaker. I had to guess on a lot of the values, since National was not nice enough to give me the values. Has anybody built this same amp and know what the value of Cs, Rm, and what Ci is for, it says it is optional. Righit now, I am using a 1K for Rm and a 2uf for both Cs. I am using two 10uf caps parallel to each other for Cs, but since it didn't say the polarity for them, I guessed. What is the correct polarity? Also, I have a PCB that I made and checked to the schematic many times, so it shouldn't be an incorrect wiring. If it is needed, I am designing this for a simple, cheap, subwoofer for my computer. Can anybody help a beginner?
Thanks, Mike
LM3886 design
Re: LM3886 design
< HELP ! HELP !….me Rhonda.><p>Mike:
From the unitary components designators that you gave , sounds like U utilized the initial schematic provided instead of the meatier applications on down in the app note. Also if you used that xformer to get yourself split ~ +21-26 VDC and –21-26VDC supplies to feed the unit in a differential mode and U used that fig 1 schematic, check on the current thru that 1k res that U selected by inserting a milliameter in the line to see if you have, ideally, 1 ma passing thru, that, your mute circuitry IF not the unit is in the mute mode. In a single supply configuration, the mute circuitry grounds to common ground. Whereas on Figure 1 , the dual supply configuration, it has that mute switches bottom terminal going to the –Vcc supply as shown.in your Figure 1.Try both possibilities, according to which circuitry U are using…..mainly the inline current reading afortementioned should tell you that you are merely in mute mode which seems to be your no output difficulty.
1k for Rm figures about right for the V-2.8/18 they specified, if not trim with your milliameter being the reference to get that 1 ma.
The Cs capacitors, placed right AT the IC’s pins 1 and 4, with each at the proper polarity in respect to common ground provide bypass filtering of the + and - supplies at those points.
Ci is a hi-pass roll off cap in that feedback loop, since this is a bass woofer oriented application…I would use it.
That chip sounds like a lot in asmall footprint with protective features and impressive specs…and at the least, the price was right.<p>73's de Edd
[email protected] ............(Interstellar~~~~Warp~~~Speed)
[email protected]........(Firewalled-Spam*Cookies*Crumbs)<p>
From the unitary components designators that you gave , sounds like U utilized the initial schematic provided instead of the meatier applications on down in the app note. Also if you used that xformer to get yourself split ~ +21-26 VDC and –21-26VDC supplies to feed the unit in a differential mode and U used that fig 1 schematic, check on the current thru that 1k res that U selected by inserting a milliameter in the line to see if you have, ideally, 1 ma passing thru, that, your mute circuitry IF not the unit is in the mute mode. In a single supply configuration, the mute circuitry grounds to common ground. Whereas on Figure 1 , the dual supply configuration, it has that mute switches bottom terminal going to the –Vcc supply as shown.in your Figure 1.Try both possibilities, according to which circuitry U are using…..mainly the inline current reading afortementioned should tell you that you are merely in mute mode which seems to be your no output difficulty.
1k for Rm figures about right for the V-2.8/18 they specified, if not trim with your milliameter being the reference to get that 1 ma.
The Cs capacitors, placed right AT the IC’s pins 1 and 4, with each at the proper polarity in respect to common ground provide bypass filtering of the + and - supplies at those points.
Ci is a hi-pass roll off cap in that feedback loop, since this is a bass woofer oriented application…I would use it.
That chip sounds like a lot in asmall footprint with protective features and impressive specs…and at the least, the price was right.<p>73's de Edd
[email protected] ............(Interstellar~~~~Warp~~~Speed)
[email protected]........(Firewalled-Spam*Cookies*Crumbs)<p>
Re: LM3886 design
I don't know if I am using a split supply. Heres the schematic for my supply basicly<p>---)|(-----------|____|---------------+ to V+
)|( |AC + | +_
AC )|(DC 30V | |15,000uf - Out = 42VDC
)|( |AC__-_| | _
---)|(-----------| |-----------|--- to V-<p>There is a full connection to the cap on the top, I just couldn't draw it. SO are you saying that with my supply, I need to hook the switch to the ground instead of V-? Or do I need a bipolor schematic. I found one for this amp that uses 2 transformers.
The schematic for the amp and PS is at
http://home.pacbell.net/lengal/ip/lm3886.pdf
(its only 17K, so it takes no time to open.)
I noticed that it is the same as national's schematic on page 1 of the datasheet, but it gives the values of the components and adds a 250pf cap between the + and - inputs, and an inductor and 10ohm resistor at the end. Are those added parts needed? Thanks for your help.
)|( |AC + | +_
AC )|(DC 30V | |15,000uf - Out = 42VDC
)|( |AC__-_| | _
---)|(-----------| |-----------|--- to V-<p>There is a full connection to the cap on the top, I just couldn't draw it. SO are you saying that with my supply, I need to hook the switch to the ground instead of V-? Or do I need a bipolor schematic. I found one for this amp that uses 2 transformers.
The schematic for the amp and PS is at
http://home.pacbell.net/lengal/ip/lm3886.pdf
(its only 17K, so it takes no time to open.)
I noticed that it is the same as national's schematic on page 1 of the datasheet, but it gives the values of the components and adds a 250pf cap between the + and - inputs, and an inductor and 10ohm resistor at the end. Are those added parts needed? Thanks for your help.
Re: LM3886 design
Mike:
Well you found out just how unfriendly the posting of ASCII schematics on this site is if U try to use multiple spaces….I use periods. I couldn’t make out the schematic but since you just showed one electrolytic…you wouldn’t have a dual positive and negative supply with a common ground with that 43 v transformer and only one filter cap unless you had a centertap on its secondary and utilize 2 rectifier mediums and 2 filter caps.
Or if you were to use it to make 2 half wave rect circuits it would take a heap of filtering capacitance and the 45 V plus voltage level outputs from those neg and pos supplies would be crunching the max voltage spec for that chip of yours.
Just for schematic reference purposes see this site, for the xformer , diodes and filtering hookup, but just disregard the 3 term regs and anything on to the right of the schematic, as this only is a very low current power supply. But it is a dual supply for U to see it. http://www.uoguelph.ca/~antoon/circ/741-ps.html
Can U come up with( 2) 24V transformers ? Or does your 43V unit have a centertap ? for that is what you will need if you want to use the first circuit mentioned as well as the same circuit at the last URL U supplied .
Now if you want to use that T-former you have with a Full Wave Bridge for rectification and your 30k cap[ filtering. Go down to page 5 bottom figure 2 and duplicate that circuit as it is arranged to bias its input down to establish a common ground and operate from a single ended supply, so this is the logical route to go in order to make use of what you have on hand already.
Its 220 pf is for hi freq bypassing/bandpass…use it. Also the parasitic osc inductor and damping/isolation resistor on the output to the speaker…common practice on audio amp output ckts.<p>73's de Edd
[email protected] ............(Interstellar~~~~Warp~~~Speed)
[email protected]........(Firewalled-Spam*Cookies*Crumbs)
Well you found out just how unfriendly the posting of ASCII schematics on this site is if U try to use multiple spaces….I use periods. I couldn’t make out the schematic but since you just showed one electrolytic…you wouldn’t have a dual positive and negative supply with a common ground with that 43 v transformer and only one filter cap unless you had a centertap on its secondary and utilize 2 rectifier mediums and 2 filter caps.
Or if you were to use it to make 2 half wave rect circuits it would take a heap of filtering capacitance and the 45 V plus voltage level outputs from those neg and pos supplies would be crunching the max voltage spec for that chip of yours.
Just for schematic reference purposes see this site, for the xformer , diodes and filtering hookup, but just disregard the 3 term regs and anything on to the right of the schematic, as this only is a very low current power supply. But it is a dual supply for U to see it. http://www.uoguelph.ca/~antoon/circ/741-ps.html
Can U come up with( 2) 24V transformers ? Or does your 43V unit have a centertap ? for that is what you will need if you want to use the first circuit mentioned as well as the same circuit at the last URL U supplied .
Now if you want to use that T-former you have with a Full Wave Bridge for rectification and your 30k cap[ filtering. Go down to page 5 bottom figure 2 and duplicate that circuit as it is arranged to bias its input down to establish a common ground and operate from a single ended supply, so this is the logical route to go in order to make use of what you have on hand already.
Its 220 pf is for hi freq bypassing/bandpass…use it. Also the parasitic osc inductor and damping/isolation resistor on the output to the speaker…common practice on audio amp output ckts.<p>73's de Edd
[email protected] ............(Interstellar~~~~Warp~~~Speed)
[email protected]........(Firewalled-Spam*Cookies*Crumbs)
Re: LM3886 design
Sorry about the bad schematic. Basically, it is the one on my web site:
www.electronet.5u.com
Click on Schematics, Then go down to 'P' and then click on 'Power Supplies'.
Anyway, will not using the correct power supply fry the amp? also, will it not work at all, not ever badly? Also, does the center tap need to be hooked up anywhere? The Ground??? Thats where it connects in the URL i gave, but not in yours. Thanks again, Mike
www.electronet.5u.com
Click on Schematics, Then go down to 'P' and then click on 'Power Supplies'.
Anyway, will not using the correct power supply fry the amp? also, will it not work at all, not ever badly? Also, does the center tap need to be hooked up anywhere? The Ground??? Thats where it connects in the URL i gave, but not in yours. Thanks again, Mike
Re: LM3886 design
Mike:
The power supply you provided is a single polarity one, the one supplied for reference from Tony’s site is a dual polarity supply . It provides both a negative and a positive supply source, as referenced to their shared common ground. In this case it is the transformers center tap which is their common ground..Two series aiding wired separate transformers could have been utilized just as well if no transformer of the desired voltage was available.
Since you already have all of the power supply components required for a single polarity supply for the unit, you might as well use the App Notes page 5 fig 2 circuit, as it should fulfill your needs for the computer woofer amp circuit you needed. Move on up to the dual power supply circuitry utilization if you later need to reach on up to the max power capabilities of that IC power amp. The lower power should be a multi quantum leap up from your prior experience utilizing the small LM386.<p>73's de Edd
[email protected] ............(Interstellar~~~~Warp~~~Speed)
[email protected]........(Firewalled-Spam*Cookies*Crumbs)
The power supply you provided is a single polarity one, the one supplied for reference from Tony’s site is a dual polarity supply . It provides both a negative and a positive supply source, as referenced to their shared common ground. In this case it is the transformers center tap which is their common ground..Two series aiding wired separate transformers could have been utilized just as well if no transformer of the desired voltage was available.
Since you already have all of the power supply components required for a single polarity supply for the unit, you might as well use the App Notes page 5 fig 2 circuit, as it should fulfill your needs for the computer woofer amp circuit you needed. Move on up to the dual power supply circuitry utilization if you later need to reach on up to the max power capabilities of that IC power amp. The lower power should be a multi quantum leap up from your prior experience utilizing the small LM386.<p>73's de Edd
[email protected] ............(Interstellar~~~~Warp~~~Speed)
[email protected]........(Firewalled-Spam*Cookies*Crumbs)
Re: LM3886 design
I've got the parts to make a dual supply, so I'll make it, since I really want to squeeze out all of the power from the amp I can. Also, I have already etched a board and placed all of the parts on it, so I really want to stick with what I have now. Note that on my site, it shows a reg, but I didn't use one. I know that the center tap is the ground of the power supply circuit, but does it also need to be connected to the ground of the amp's circuitry? Also, did using the inproper supply fry anything? One more final question: I want to build an enclosure and make it look nice, so I would like to have an LED on the front light up when the amp is on, but the 20 or so V the transformer makes is too much to power a LED. is there any way to reduce the power w/o a reg, since the only one that I have is a +5 reg, which is too much. Also, can an LED be powered off of a dual power source? Or would I have to connect it right after the bridge, and before the caps on the supply? Thanks again, Mike
Re: LM3886 design
Mike:
Looks like you could go up to the full 68 Watts on that unit with a plus and minus ~42V supplies, that’s with their commonality at pin 7 of the IC and the pos supply to pin1/and/5 of the IC and the negative supply feeds to pin 4 of the IC . On the other end of the scale the minimum voltage requirement seems to be as low as plus and minus 12V supplies. As I understand you’re your original hook up with a single supply to 1/and/5 and 4 I suspect no damage to the IC due to no biasing turn on due to the absence of a grounding to the chip circuitry.
For your pilot lamp , easiest to use a line voltage rated unit across the same leads as the xformer primary, as will be energized by the power sw on turn on. Or alternatively, on the xformer secondary’s DC power source, just connect a lamp or LED with a proper value (as computed by Ohms law) of voltage dropping/current limiting resistor in series with it across your positive supply.
You mentioned regulation, and on this powerful and dissipative of a unit you will be using raw filtered B+.<p>73's de Edd
[email protected] ............(Interstellar~~~~Warp~~~Speed)
[email protected]........(Firewalled-Spam*Cookies*Crumbs)
Looks like you could go up to the full 68 Watts on that unit with a plus and minus ~42V supplies, that’s with their commonality at pin 7 of the IC and the pos supply to pin1/and/5 of the IC and the negative supply feeds to pin 4 of the IC . On the other end of the scale the minimum voltage requirement seems to be as low as plus and minus 12V supplies. As I understand you’re your original hook up with a single supply to 1/and/5 and 4 I suspect no damage to the IC due to no biasing turn on due to the absence of a grounding to the chip circuitry.
For your pilot lamp , easiest to use a line voltage rated unit across the same leads as the xformer primary, as will be energized by the power sw on turn on. Or alternatively, on the xformer secondary’s DC power source, just connect a lamp or LED with a proper value (as computed by Ohms law) of voltage dropping/current limiting resistor in series with it across your positive supply.
You mentioned regulation, and on this powerful and dissipative of a unit you will be using raw filtered B+.<p>73's de Edd
[email protected] ............(Interstellar~~~~Warp~~~Speed)
[email protected]........(Firewalled-Spam*Cookies*Crumbs)
Re: LM3886 design
Thank You so much for your help. It worked! It ends up, though, that 42V is how much the transformer puts out between the entire coil, not between one side and the center and the other side and the center. So, I really only get about 21V + and -, but 21V is barely meeting the power requirements for the 3886 (20V min). I found a transformer at Radioshack that is 25V CT. Does that mean 25V across the entire secondary, or between one side and the center and the other side and the center. Since 28V will put 60W into 4 ohm, and my speaker is 4ohm, a 25V transformer should work, since for some reason, the bridge rectifier I use boosts power a few volts. What I want to know is if giving more power (such as 35V), will provide more than 60W. I know that going from 28-35V boosts power in 8ohm, but what about 4 ohm. This amp is being used for a small sub, so it needs to be as powerful as possible. <p>Thanks, Mike
Re: LM3886 design
Mike:
That will be 24v across the total secondary winding or that amount halved in respect to the center tap and either end of the secondary…….so effectively you wouldn’t even be but a smidge better if you used two of those 24 V RS transformers.
The value of transformer you will need for max is not a commonly available unit. You might consider one leg of the supply being provided by a 12 and a 24VAC series aiding configuration to get you 36 vac for one pol of the dual supply and a duplicate pair for the other polarity of the PS. That should come closer to the max spec of the unit and leave a little hedge for protection. On your higher voltage value from the “magical bridge” it will pull down and dissapear when you load down the unit.
73's de Edd
[email protected] ............(Interstellar~~~~Warp~~~Speed)
[email protected]........(Firewalled-Spam*Cookies*Crumbs)
That will be 24v across the total secondary winding or that amount halved in respect to the center tap and either end of the secondary…….so effectively you wouldn’t even be but a smidge better if you used two of those 24 V RS transformers.
The value of transformer you will need for max is not a commonly available unit. You might consider one leg of the supply being provided by a 12 and a 24VAC series aiding configuration to get you 36 vac for one pol of the dual supply and a duplicate pair for the other polarity of the PS. That should come closer to the max spec of the unit and leave a little hedge for protection. On your higher voltage value from the “magical bridge” it will pull down and dissapear when you load down the unit.
73's de Edd
[email protected] ............(Interstellar~~~~Warp~~~Speed)
[email protected]........(Firewalled-Spam*Cookies*Crumbs)
Re: LM3886 design
First, you assume that the load is a 4 ohm resistance being driven with a continuous sine wave signal. These assumptions are probably false, but making them avoids permanent brain damage on the part of the designer.<p>A class B amplifier at maximum output is about 68 percent efficient at best, and probably a little worse. This means that the chip will have to dissipate at least half of the power that it is putting out in the form of heat. Efficiency is poorer at less than full output, but the heat dissipation is also less.<p>The temperature inside an enclosed piece of equipment can easily be in the 75 degree centigrade range on a warm day, depending on where you live. The junction temperature should be kept below 150 degrees centigrade (maximum rating) and possibly down around 125 degrees centigrade (more reliable). The LM3886 hase a junction to case thermal resistance of about 1 degree centigrade per watt, and the joint between the case and heat sink will be about .2 degrees per watt with thermal grease. An insulator may increase this to .7 to 1 degree per watt. The heat sink thermal resistance will probably be over 1 degree per watt. This depends on how much aluminum you want to use, but 2 degrees per watt may be reasonable. The temperature rise could easily be 4 degrees per watt when you add these all up, unless you use a large heat sink, or operate the heat sink at one of the supply potentials. If you assume a 75 degree temperature rise with a total thermal resistance of 4 degrees per watt, you can dissipate about 19 watts. This limits your full output to less than 38 watts, probably to about 30 watts when you consider that you can't get the output all the way to either supply voltage.<p>The LM3886 appears to limit the peak output current to 7 amps. There is no point to using supply voltages over 30 volts with a 4 ohm load, since the peak output voltage will be limited to 28 volts by the current limit.<p>Delivering 30 watts into a 4 ohm load requires an RMS voltage of about 11 volts. The peak voltage is higher by a factor of 1.414, or about 15.5 volts Add a couple of volts to allow for some drop in the output stage, and 18 volt power supplies are probably about right.<p>The RMS current for 30 watts in a 4 ohm load is about 2.7 amps. The peak current is about 3.9 amps, and the average current is about 2.5 amps. This is the current that the power supply should be designed for.<p>An 18 volt RMS winding will deliver about 25 volts peak. About 1 volt will be lost in the rectifier, and the line voltage might be a little low. It would appear that about 4 volts of ripple can be tolerated. In 8 milliseconds, 2.5 amps should discharge the filter capacitors about 4 volts. This means that the filter capacitors should be about 5000 microfarads. A value of 4700 microfarads should be fairly easy to get.<p>The current rating of the transformer winding should be higher than 2.5 amps, since the current through the transformer secondary flows in high, short pulses whose average value is 2.5 amps. A high value of filter capacitance makes this worse. On the other hand, the current only flows in half the secondary on any given cycle. Overall, a 36 volt center tapped winding rated at 3 amps or more should be adequate.<p>You can design for higher power if you can guarentee that the unit will never be operated with full output sine waves, even during testing. Music tends toward higher peak powers and lower average power than a sine wave. How far this process can be carried depends on the conscience of the designer.
Re: LM3886 design
Stephen: That heat sink stuff got me lost. I'm just a beginner, and don't know degrees per watt yet. All I know is that I have the largest heatsink I had sitting around on it, and the chip is never hot. The heatsink is, but the 3886 is cool to the touch, so I would assume my heatsink is doing the job. my transformer says 28.6V CT on the secondary, but in reality, it produces around 30, which my 'magic rectifier' brings up to 41V. When using the split supply, it puts out + 21VDC and -21VDC, while the transformer produces about 15V. The transformer is a 2amp model, so is that not enough? I know that my rectifier can only handle 4amp at 100V. The following is an image of the heatsink and two of the entire amp, if anybody is interested: [img]file://f:\dcim\100_pana\p1000149.jpg[/img]
[img]file://F:\DCIM\100_PANA\P1000148.jpg[/img]
[img]file://F:\DCIM\100_PANA\p1000147.jpg[/img]<p>Thanks, Mike
[img]file://F:\DCIM\100_PANA\P1000148.jpg[/img]
[img]file://F:\DCIM\100_PANA\p1000147.jpg[/img]<p>Thanks, Mike
Re: LM3886 design
Sorry, i typed the wrong location for my images. The correct pictures are:
Re: LM3886 design
Stupid free web hosting!! 
They dont let me show images on other sites from their site. Well, if you want to see them, go to:<p>www.electronet.5u.com/p1000147.jpg (entire amp)
www.electronet.5u.com/p1000148.jpg (entire amp)
www.electronet.5u.com/p1000149.jpg (heatsink)<p>Sorry, Mike
They dont let me show images on other sites from their site. Well, if you want to see them, go to:<p>www.electronet.5u.com/p1000147.jpg (entire amp)
www.electronet.5u.com/p1000148.jpg (entire amp)
www.electronet.5u.com/p1000149.jpg (heatsink)<p>Sorry, Mike
Re: LM3886 design
Heat flows between points that are different in temperature. The path between two points can be described as having either a thermal resistance or a thermal conductance. The thermal resistance (degrees centigrade per watt) is the difference in temperature between the two points divided by the amount of heat (watts) flowing between them. Thermal conductance (watts per degree centigrade) is the inverse of this. It is the amount of heat (in watts) which will flow between two points with a 1 degree temperature difference. Thermal resistance is the easiest to use, since successive thermal paths add directly.<p>In an integrated circuit or transistor, heat is generated in a thin layer on the top surface of the chip or die. For normal and reliable operation, the temperature in this small area must be kept below a maximum of 150 degrees centigrade for integrated circuits or possibly 200 degrees centigrade for a power transistor. When the device is operating, the temperature of this area rises until the flow of heat away from this area is equal to the amount of heat generated in this area.<p>This heat must flow from the junction to the case of the device. At this point it may be radiated, or a heat sink may be attached, which carries the heat to a heat sink. The heat sink, in turn radiates the heat to the surrounding environment. The highest temperature point will be the junction and the lowest temperature will be the surrounding environment. The intermediate points will be at intermediate temperatures depending on the thermal resistances between them.<p>The greek letter theta is sometimes used as an abbreviation for thermal resistance on data sheets. In the case of the LM3886, the text indicates that the thermal resistance from the junction to the case is about 1 degree centigrade per watt. If the LM3886 is dissipating 30 watts, this indicates that the junction temperature will be 30 degrees higher than the case temperature. Notice that the case temperature is the temperature of the metal tab (usually measured as close to the plastic package as possible), not the temperature of the plastic surface of the package.<p>Each joint in the thermal path also has a thermal resistance. A joint between two flat surfaces using thermal grease may be about .2 degrees per watt. Most insulators, with the exception of beryllium oxide (toxic) or diamond (expensive) are very poor thermal conductors. A thin kapton or mica sheet may have a thermal resistance of .5 degree per watt or more, depending on its area and thickness.<p>The heat sink heats up until the heat transferred to the environment is equal to the heat received from the semiconductor device. This can also be described as a thermal resistance. Large aluminum extrusions may have thermal resistances close to a degree per watt, most heatsinks are a somewhat higher.<p>These thermal resistances are in series, and may be added directly together. If we assume that the heat sink is 2 degrees per watt, the total thermal resistance will be: 1 degree/watt (junction to case) plus .2 degree/watt (case to insulator joint) plus .5 degree/watt (through insulator) plus .2 degree per watt (insulator to heat sink joint) plus 2 degrees/watt (heat sink to ambient) for a total of 3.9 degrees per watt. This means that a power dissipation of 30 watts will raise the junction temperature 117 degrees above the ambient temperature. The case temperature (2.9 degrees/watt) will be 87 degrees above the ambient. The heat sink (2 degrees/watt) will be 60 degrees above the ambient temperature. Your actual numbers will vary, but I hope that this shows the method.<p>Data sheets like to talk about a 25 degree centigrade ambient temperature. This is unrealistic, since it corresponds to 77 degrees farenheit. A warm summer day in most of the United States far exceeds this. Many semiconductor devices are rated based on a 25 degree centigrade case temperature, which is probably achieved by bolting the package directly to a water cooled block of metal cooled below 25 degrees. Using methods like this can give you ratings of 4 watts for some TO-39 packaged transistors. The free air rating for these parts is usually .6 watt, and they are hot to the touch under these conditions.<p>On a warm day, inside a piece of operating equipment, it is reasonable to expect ambient temperatures in the 65 to 70 degree centigrade range. Adding the 117 degree temperature rise previously estimated gives a junction temperature of over 180 degrees, which is excessive for most devices.<p>Using the total thermal resistance allows estimating the maximum power which can be dissipated. The total temperature change would be 80 degrees (150 degree junction minus 70 degree ambient). Dividing this by a thermal resistance of 3.9 degrees/watt gives a maximum power dissipation of 20.5 watts.<p>The thermal resistance from the junction to the plastic case of the LM3886 is so high that an insignificant amount of heat flows along that path. The thermal resistance of most plastics is probably several hundred times that of copper or aluminum, so most of the heat flows out the metal tab. The significant point is that your heat sink is hot. The metal tab is the "case" that the data sheets refer to, and not the physical surface of the plastic case.<p>With regard to your "magic" rectifier. There are three ways of specifying the voltage of a sine wave: root-mean-square (RMS), peak, and average. The RMS voltage is the unvarying (DC) voltage which would transfer the same power to a resistive load. It is .707 times the peak voltage of the sine waveform. The peak voltage is the maximum voltage reached by the waveform at any time. It is 1.414 times the RMS voltage. The average voltage is the average value of the absolute value of the sine wave. It is .637 times the peak voltage, or about 90 percent of the RMS voltage.<p>The AC power line, and the voltage on your transformer secondary are specified as RMS voltages. The 28.6 volt secondary on your transformer will produce a peak voltage of 40.4 volts with its specified load current. A lightly loaded transfer will produce a higher peak voltage, possibly by as much as 10 or 20 percent higher. The peak voltage would then be about 44.5 volts. The voltage across half the secondary would be 22.25 volts. The voltage drop in the rectifier will be about 1 volt or less, which would give a 21 volt peak output. A rectifier charges the filter capacitor to a voltage near the peak value, and if the capacitor is large, this voltage remains throughout the full cycle of the input voltage. These voltages may be slightly higher if the line voltage is above its nominal value of 117 or 120 volts. This is often the case. So your measured voltage is about what would be expected.<p>I hope this makes a little more sense. Estimating temperature rise can get very complicated when the thermal path is two or three dimensional.<p>[ June 28, 2003: Message edited by: stephen ]</p>
Who is online
Users browsing this forum: No registered users and 59 guests