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Old 12th September 2012, 02:28 AM   #21
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Quote:
We can calculate the loss. Firstly, at audio frequencies C1 is effectively a short circuit (look up the formula for capacitive reactance, and calculate it for this cap at 20Hz and 20kHz), so the bottom side of this "L pad" is actually R2 in parallel with R3. Since R3 is so much lower than R2 (1/10th its value), the effective value is only slightly lower than the lower value, or the value of R3. With R1 this forms a voltage divider of R3/(R1+R3), or 22/23, or just slightly less than 1. Using the voltage-ratio-to-dB formula (I'm too lazy to go through it, but it's not too complicated) it would probably come out in the ballpark of 0.1dB, or (for the purpose of figuring out the gain of this circuit) practically no loss.
Now , that's exacly what i call as the clear, perfect and upto the point explanation of the problem! Actually, this was the toughest of all the questions for me and i was not expecting a truly satisfying answer to this question. However,it came out to be totally different and opposite of my expectations.

Well done and well explained Mr.benb! Thanks a lot! You cleared lots of other questions tinkering in my mind, regarding this input circuit.

Capacitive reactance (Xc) for C1(1uf) is as follows:

Xc = 1/(2*pie*f*C) , where,

Xc = Capacitive Reactance in ohms.
2*pie = 6.284.
f = frequency in Hertz,
C = capacitor value in Farads.

So, at audio frequency, of 20 Hz(Maximum Xc), Xc will be:
Xc = 1/(6.284*20*1*(10^-6)) = 7957 ohms or 7.9K(also neglegible?).

Similarly at audio frequency, of 20Khz(Minimum Xc), Xc will be:
Xc = 1/(6.284*20000*1*(10^-6)) = 7.957 ohms(neglegible)

Capacitive reactance(Xc) is higher at lower audio frequencies and lower at higher audio frequencies. So the maximum capacitive reactance is at lower frequencies.

Similarly, for C1 as 4.7uf (if we use) Maximum Capacitive reactance(Xc) at 20Hz will be = 1693ohms ~ 1.7K(neglegible).

And again, for C1 as 10uf (if we use), Max. Capacitive reactance(Xc) at 20Hz will be = 796ohms ~ 0.8K(neglegible).

And that's what really accounts (decrement of the capacitive reactance, as we increase the capacitor value for lower frequencies) for more Bass effect in the sound as we increase the value of C1 capacitor.

So, as according to the explanation given by you, we can neglect all of these capacitive reactance(Xc) values for C1, in particular.

Formula, for voltage ratio to db is:

av = 20 log(base10) Av, Hence,

av = (20 log 22/23)
av = (20 log .95652173913043478260869565217391)
av = (20 * -0.019305155195386642903838246283672)
av = (-0.38610310390773285807676492567345) db.

which is again neglegible value. Negative sign shows here the phase inversion, if i am not wrong.

Now,

1. Where would i place the 10k or 22k(value not more than this) volume taper/linear pod So, that the other parameters of this input circuit won't get disturbed at all? I will manage the source impedance accordingly, or will use the voltage follower(unity gain buffer), don't worry about that.

2. Can i replace R3(22k) with a volume control? Becoz, i really wanna put volume control here right at the input of the amplifier.Would'nt it disturb the DC bias/ground path for the chipamp's input or DC output offset? And, how will i manage the input impedance if i replace R3 with a volume control(becoz R3 is a fixed value and the pot is variable)?

Many-Many thanks for suggesting me textbooks on DC circuits and AC circuits analysis. I think this is the domain where most of my problems lie. Actually, i repeat, i am not an electronic guy.

Thanks.

Last edited by noddy55; 12th September 2012 at 02:46 AM.
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Old 12th September 2012, 02:35 AM   #22
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Mr. Loudthud, Thanks for the reply,

Quote:
Otherwise the gain and GBW product will determine the HF rolloff
That's exactly i wanted to hear about it. Thanks! you clear a doub't into a positive confirmation.

Quote:
R4/C3 sets the low frequency rolloff and reduces DC offset at the output because it reduces the gain to 1 (from pin 1 of the IC) at DC. C1 prevents DC on the input from getting to the to the speaker.
Can we determine this DC gain with respect to the R4/C3 values? I mean to say do we ever know what the DC gain is with a particular combination of values(R4/C3) at a particular moment?or, is that the dc gain is still unity or not even after changing R4/C3 values? If i am not wrong, then measuring the DC offset at the output of an Op Amp is the only possible answer to this question?

And, How does the different values of C3 affects the DC offset? If i change it from 22uf/63v to 100uf/63v. What will happen to the DC offset? Will it still be fine? Offcourse, -3db cut of point will change, no doubt about it.

Quote:
R3 and R5 set the closed loop gain by the formula Av=1+(R3/R5).
Accordingly, Av = 1+(22K/180K)

Av = (1/1) + (22000/180000)
Av = (180000+22000)/180000
Av = 202000/180000
Av = 1.1222222222222222222222222222222(near unity gain)

You mean to say that this circuit has a closed loop unity gain?

What for then Av=1+(R5/R4) is used? This will give a gain(v/V) of 19.
Av = 1+(180K/10K)
Av = 1+18
Av = 19.

Quote:
These should be chosen so as the capacitance at pin 2 of the chip and any parasitic capacitance of the PCB do not adversely affect the frequency response.
Please, elaborate this point(if you can), How we choose(or calculate) the values and how the frequency reponse(u mean bandwidth here?) is adversely affected with different values? This is what i exactly want to know about.

I repeate the question for you, where should i keep the 10K or 22K volume taper/linear pod in the circuit? Just need your advice and suggestion.

Thanks.

Last edited by noddy55; 12th September 2012 at 03:04 AM.
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Old 12th September 2012, 02:42 AM   #23
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Thanks! Mr.Abraxalito,

I got your point. Input high pass filter is more important and prescious then the Negative Feedback(NFB) high pass filter. And, the latter has its limitations after which it drop its weapons for a white flag and surrender to the former.

Thanks.

Last edited by noddy55; 12th September 2012 at 02:53 AM.
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Old 12th September 2012, 08:29 AM   #24
AndrewT is offline AndrewT  Scotland
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Quote:
Originally Posted by noddy55 View Post
Formula, for voltage ratio to db is:

av = 20 log(base10) Av, Hence,

av = (20 log 22/23)
av = (20 log .95652173913043478260869565217391)
av = (20 * -0.019305155195386642903838246283672)
av = (-0.38610310390773285807676492567345) db................... Negative sign shows here the phase inversion,...............
-xdB ratios indicate a gain of <1
+ydB ratios indicate a gain of >1

Remember than dB ratios are "ratios of two values". If the two values are equal then the ratio is 1 and the corresponding dB value is 0dB

Your result of -0.4dB indicates a ratio of slightly less than 1, yes 22/23rds is slightly less than 1 (23/23rds).

Quote:
Originally Posted by noddy55 View Post
.........................Accordingly, Av = 1+(22K/180K)

Av = (1/1) + (22000/180000)
Av = (180000+22000)/180000
Av = 202000/180000
Av = 1.1222222222222222222222222222222(near unity gain) {gain = +1.0016dB}

You mean to say that this circuit has a closed loop unity gain?

What for then Av=1+(R5/R4) is used? This will give a gain(v/V) of 19.
Av = 1+(180K/10K)
Av = 1+18
Av = 19. {gain = +25.6dB} .....................
Have you got the first ratio 22k/180k upside down? Or did you mean 22k/180k?

Moderators:
I am reporting my own post for clarification on adding my edit shown in bold to the Members post extract quoted here.
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regards Andrew T.

Last edited by AndrewT; 12th September 2012 at 08:38 AM.
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Old 12th September 2012, 12:41 PM   #25
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Quote:
Negative sign shows here the phase inversion, if i am not wrong.

-xdB ratios indicate a gain of <1
+ydB ratios indicate a gain of >1
I am really sorry about that. Actually, the inverting configuration of Op-Amp was ringing bells at the time when i was writing this post, so i got mix with and confused with the negative sign of output voltage gain of Op-Amp in an inverting configuration where it really means the phase inversion.

And, yes here in my calculations above, negative sign means an atteunation of signals rather than the increment or we can say that the gain is slightly less then one.

Thanks, Mr.Andrew for correcting me.


Quote:
gain = +1.0016dB
gain = +25.6dB
Yes, these values are perfectly correct. We need to find the log of V/V gain ratio and multiply it then with 20, as according to the formula 20 log V/V. Just need to remember here that the log value should be taken as base 10 instead of base2 or base e.


Quote:
Have you got the first ratio 22k/180k upside down? Or did you mean 22k/180k?
No, I got the first ratio as according to the formula (Av=1+(R3/R5) given by Mr.Loudthud in point 6 above and i am really confused about it too. I seek Mr.Loudthud for this clarification about Av=1+(R3/R5) instead of Av=1+(R5/R4).

Av=1+(R5/R4) is the correct formula, what we really use to find the gain.

Thanks.

Last edited by noddy55; 12th September 2012 at 12:46 PM.
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Old 16th September 2012, 02:04 AM   #26
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Quote:
"Negative sign shows here the phase inversion, if i am not wrong."
Please, consider this line in post 21 as:

Quote:
"Negative sign shows here the atteunation of signals, if i am not wrong."
Thanks.
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Old 16th September 2012, 02:09 AM   #27
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Mr. Loudthud, where are you?

Can you please, help me clear the confusion created above in your post? Specially, about the formula given by you as: Av=1+(R3/R5).

Thanks.
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Old 17th September 2012, 05:23 PM   #28
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Sorry, the formula should be Av=1+(R5/R4), roughly 25.6dB.
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Old 18th September 2012, 09:57 PM   #29
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Thanks to all of you in this thread,

You answered almost all of my questions. And, i am 100% satisfied. You guys are really genius in your work, knowledge and experience.

Let me see what i can do about placing the volume at the right place, myself.

I will surely come up for more questions later. Be happy until.

Thanks a lot to all of you!
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