I've some questions on LM1875 amps.
1>>>> http://sound.westhost.com/project72.htm , here the link to sound westhost project using the chip.
, niw see there has only two 1Amp fuse in both rails, now for my purpose, I'll use it as a 15 watt Centre channel Amp. Accoarding to the datasheet it'll gimme the 15/18W RMS at +/- 20V. So how much current will draw the circuit? 
can I use the same 1Amp fuse.
And can I use a 150VA trafo for that?
Again, I never made a PCB of the 1875 amp. I made it in a veroboard.
now I heard that there is a issue with the amp if some capacitor or resistor not placed near the IC. is ya true?
my veroboard layout given below.
now, go below the website http://sound.westhost.com/project72.htm, you'll see a power supply design. I wonder there is no any regulator, only two 10,ooo mfd caps, will it be sufficient for 1875?
1>>>> http://sound.westhost.com/project72.htm , here the link to sound westhost project using the chip.
, niw see there has only two 1Amp fuse in both rails, now for my purpose, I'll use it as a 15 watt Centre channel Amp. Accoarding to the datasheet it'll gimme the 15/18W RMS at +/- 20V. So how much current will draw the circuit? can I use the same 1Amp fuse. And can I use a 150VA trafo for that?
Again, I never made a PCB of the 1875 amp. I made it in a veroboard.
now I heard that there is a issue with the amp if some capacitor or resistor not placed near the IC. is ya true? my veroboard layout given below.An externally hosted image should be here but it was not working when we last tested it.
An externally hosted image should be here but it was not working when we last tested it.
now, go below the website http://sound.westhost.com/project72.htm, you'll see a power supply design. I wonder there is no any regulator, only two 10,ooo mfd caps, will it be sufficient for 1875?
100va (if ei core) is about 3 times larger than I've ever used for LM1875, so one would think its sufficient if it doesn't get warm.
They vary greatly in quality, so you never know if 100va is a peak rating or if it can do that all day long.
When they're really big, its popular to bolt them to the side of the case instead of the bottom of the case.
Looking at the board layout, one would suggest to relocate the NFB (gain loop) resistors to directly onto the chip pins (underneath the board).
It seems to be short a few capacitors. Recommended minimum power caps are on LM1875.pdf from national semiconductor's website. . . but, here's my guess: 220uF parallel with 0.1uF--one set each rail.
Check out the K50 kit at www.electronics123.com and click on the reference documents.
They vary greatly in quality, so you never know if 100va is a peak rating or if it can do that all day long.
When they're really big, its popular to bolt them to the side of the case instead of the bottom of the case.
Looking at the board layout, one would suggest to relocate the NFB (gain loop) resistors to directly onto the chip pins (underneath the board).
It seems to be short a few capacitors. Recommended minimum power caps are on LM1875.pdf from national semiconductor's website. . . but, here's my guess: 220uF parallel with 0.1uF--one set each rail.
Check out the K50 kit at www.electronics123.com and click on the reference documents.
No, don't do that. Mounting components off the PCB (except for the possible exception of bypass caps) is unnecessary, reduces reliability and looks bad. Unless daniel can provide you with a reason why you should do that, leave the resistor where it is.danielwritesbac said:
This obsession the "gainclone" crowd have with short feedback paths is not based in reality. If you do the sums, you will find that the effects of a little bit of PCB track on your feedback are well below the THD+N of the amp chip (and I mean *well* below). I suspect all this came from the original gaincard, where the NFB loop was made short because it was built extremely cheaply, without using good PCB design techniques.
Yes, you should definitely have two 0.1 uF ceramic capacitors (X7R dielectric type or worse; just probably not NPO or C0G type), one from each chipamp power pin to power ground.
They should be connected as close to the power pins as possible, since they are meant to divert the higher frequencies.
Putting them farther away from the pins increases the parasitic inductance and resistance between the capacitor and the power pin, which you don't want, since it will make them much less effective for shunting the high frequencies to ground, and will make the amplifier more likely to be able to oscillate at higher frequencies.
I would also want to solder a 0.1uF X7R ceramic cap directly between the two chipamp power pins, for basically the same reasons.
They should be connected as close to the power pins as possible, since they are meant to divert the higher frequencies.
Putting them farther away from the pins increases the parasitic inductance and resistance between the capacitor and the power pin, which you don't want, since it will make them much less effective for shunting the high frequencies to ground, and will make the amplifier more likely to be able to oscillate at higher frequencies.
I would also want to solder a 0.1uF X7R ceramic cap directly between the two chipamp power pins, for basically the same reasons.
As gootee says, these caps prevent the amplifier from oscillating. Basically, as the amp draws power from one of the rails, there is a small transient dip in the rail voltage due to the resistance and inductance of the tracks between the amp's pins an the power supply caps. These dips can cause the amplifier to turn into a large power oscillator - wrecking sound quality, heating up the amp and possibly damaging your speakers.Paswa said:
ok, i understood all the things, but can u tell me why the decoupling caps need to be placed near the ic?
I would recommend picking up a copy of The Art of Electronics by Horowitz and Hill. You can get one in India for fairly little money (Cambridge low price editions).
conclusion:
1> Put the decoupling caps near ic
2> Use a ceramic/film or any good decoupling capacitor
3> 220uF and 0.1uF caps, which are required to connect with the both rails, I placed them in the power supply board.
4> I'll use a 20V 100VA transformer to power up TDA2050 (25W) and LM1875(25W, but I'll use it as 15-18W). 100VA means 5Amp @ 20V!!!not it tooo much?
5> No power regulators necessary. only two 10,000uFs will do the job well.
6> As I'll use it as a center channel amp; I'll need a good 30 Watt Loudspeaker.
Thank You Alll
1> Put the decoupling caps near ic
2> Use a ceramic/film or any good decoupling capacitor
3> 220uF and 0.1uF caps, which are required to connect with the both rails, I placed them in the power supply board.
4> I'll use a 20V 100VA transformer to power up TDA2050 (25W) and LM1875(25W, but I'll use it as 15-18W). 100VA means 5Amp @ 20V!!!not it tooo much?
5> No power regulators necessary. only two 10,000uFs will do the job well.
6> As I'll use it as a center channel amp; I'll need a good 30 Watt Loudspeaker.
Thank You AlllThose caps need to be near the chip, on the same board. There are already 10mF caps on the PS board, which you can bypass (with 100nF films) and snubberize if you see fit. But seriously, you need to bypass the rails well on the same board as the chip (preferably within 1cm or so).Paswa said:
100VA @ 15-0-15Vac gives 3.3Aac on continuous duty.QUOTE] once you said
hey friend, how you counted it? since you know that I'm a newbie. Is it was done in that way -
+-15V trafo of 100VA means the amperage will be 100/15*2 = 3.3 A, after rectification and filtering caps the ampereage will be half. Is it the right way?
Make it Clear, how much current will be drawn by LM1875 at +-20V? Make sure I'm always saying about the split supply circuit.
According to the official datasheet's power supply vs output power graph at pg no. 3, at +-20Volt the IC will give 18W RMS power.
I dont know if this method is right or wrong -
The potential difference = +-20V = 40V
The RMS power = 18W
so, the current will be 40/18 A = 2.2A
so, I'll need a such trafo which will give me 2.5 Amp DC contonously.
According to my friend AndrewT's rule,
Trafo rating = 150VA, Voltage = +-20V
so, the current = 150/40 A = 3.75A
after rectification and filter the current will be = 3.75/2 = 1.875 A
SO 100VA TRAFO CANT BE USED TO DRIVE A LM1875
now if it's 200VA trafo @ +-20V, the current will be (after rectification and filter) 2.5A.
SO 200VA TRAFO CAN BE USED WITH LM1875
is it right.
Though once my friend andrew said,
I dont used any smile bcoz I'm too tooo toooo tooooo toooooo much confused
hey friend, how you counted it? since you know that I'm a newbie. Is it was done in that way -
+-15V trafo of 100VA means the amperage will be 100/15*2 = 3.3 A, after rectification and filtering caps the ampereage will be half. Is it the right way?
Make it Clear, how much current will be drawn by LM1875 at +-20V? Make sure I'm always saying about the split supply circuit.
According to the official datasheet's power supply vs output power graph at pg no. 3, at +-20Volt the IC will give 18W RMS power.
I dont know if this method is right or wrong -
The potential difference = +-20V = 40V
The RMS power = 18W
so, the current will be 40/18 A = 2.2A
so, I'll need a such trafo which will give me 2.5 Amp DC contonously.
According to my friend AndrewT's rule,
Trafo rating = 150VA, Voltage = +-20V
so, the current = 150/40 A = 3.75A
after rectification and filter the current will be = 3.75/2 = 1.875 A
SO 100VA TRAFO CANT BE USED TO DRIVE A LM1875
now if it's 200VA trafo @ +-20V, the current will be (after rectification and filter) 2.5A.
SO 200VA TRAFO CAN BE USED WITH LM1875
is it right.
Though once my friend andrew said,
I dont used any
smile bcoz I'm too tooo toooo tooooo toooooo much confused - Status
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