LM 1875 questions

[gootee]

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.

[cabbagerat]

Originally posted by Paswa


ok, i understood all the things, but can u tell me why the decoupling caps need to be placed near the ic?

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.

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).

[Paswa]

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

[cabbagerat]

Originally posted by Paswa
conclusion:

1> Put the decoupling caps near ic

3> 220uF and 0.1uF caps, which are required to connect with the both rails, I placed them in the power supply board.

Those 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).

[AndrewT]

Originally posted by Paswa
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?

is this a 20+20Vac or a 20Vac transformer?

[Paswa]

it is a transformer that will gimme +-20 V after rectification.

Oh, I find the 1875 on veroboard is filled with too much error. so I decided to make ya on a PCB. There is many PCB's, but I'll make one my own. I will post it here next. hope you'll see

[AndrewT]

A 20Vac transformer will not give +-20Vdc after the rectifier.
What have you got/planned?

[Paswa]

Vdc = Vac*1.41
or, Vac = Vdc/1.41
or, Vac = 20/1.41
or, Vac = 14.18 (app.)

ok, so I'll use a 15-0-15 trafo which will give me approx. 20Vdc after full wave rectification

[AndrewT]

Hi,
100VA @ 15-0-15Vac gives 3.3Aac on continuous duty.
But, when rectified and fed into a capacitor input filter (the smoothing caps) the maximum continuous DC output current is de-rated to about half the AC current. i.e. 100VA gives ~1.7Adc
The peak currents come from the smoothing caps.

[Paswa]

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,

150VA can supply sufficient for between 75W and 150W of output power.

This could power two or even three 1875s if you wish.
You do not need a regulator.
.

I dont used any smile bcoz I'm too tooo toooo tooooo toooooo much confused