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#1 |
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diyAudio Member
Join Date: Nov 2010
Location: A Coruna, Galicia
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Hello to all of an apprentice of the profession.
Show up with my first post, with a circuit I'm trying to accomplish. This is a 2x22W stereo amplifier using a TDA1554 and tone control circuit. The truth is that I have my doubts whether I'm riding well, while still "studying" the situation, I leave the circuit to see what you think of it. Will it work? I'm working on .. Saludoss.
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#2 |
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diyAudio Member
Join Date: Oct 2010
Location: UK
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Welcome aboard!
The biggest problem I see straight away is that your tone control stage has a loss of around 20dB when the controls are set centrally. This means that you're going to need roughly ten times as much signal to give you the same output signal level as if they weren't there. I haven't simulated the circuit to check anything out, but I have a feeling that the volume pot may be placing a very large load on the tone controls and may cause them to behave quite non-linearly in respect to slider position. By the way, don't forget that you need a logarithmic pot for the volume control to give a perceived linear increase in sound level as you turn up the volume control. It would make more sense to build a more conventional tone control circuit where the tone control network is in the feedback loop of an opamp. This makes the performance of the tone controls independent of load and would give you a gain of 1 with the tone control pots set centrally and the volume pot at maximum. Ideally, the input of the tone control stage also needs to be driven by a buffer to provide a very low drive impedance in order to maintain linear performance in relation to pot position and tone control centre frequencies that don't change with drive impedance. Take a look at this PDF to see how this is all normally put into practice.
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#3 |
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diyAudio Member
Join Date: Nov 2010
Location: A Coruna, Galicia
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Thank you very much for your reply! Now I am waiting to try it. Meanwhile, following your advice, I'm looking at other options like making control tones with operational amplifiers or with a single integrated "TA7630P". Later, I'll comment my news about it.
Regardss! |
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#4 |
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diyAudio Member
Join Date: Nov 2010
Location: A Coruna, Galicia
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I was testing the following circuit:
![]() Uploaded with ImageShack.us Now it has more volume than before, but it has noise too, a noise shouldn't be listen. I will continue testing to see if I can find the error. Meanwhile, I'm thinking of doing otherwise .. How about a LM1036 + TDA1554? Any thoughts? |
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#5 |
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diyAudio Member
Join Date: Apr 2007
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How do you feed the non-inv pin of the preamp IC ?Is it derived from + and - with two (30-50 K ohm ) resistors and a 100 uF cap ? Is the voltage of the PSU clean ?
Also...maybe , a GND reference resistor before and after the input cap would help . |
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#6 |
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diyAudio Member
Join Date: Nov 2010
Location: A Coruna, Galicia
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Feed of the non-inv of the preamp is by voltage divider from 12Vdc and two 100k resistors, this is 6Vdc (more or less). Where should put a 100uF cap?
Thank you for your reply! Regars! |
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#7 |
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diyAudio Member
Join Date: Oct 2010
Location: UK
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Please try to understand that your proposed circuit is never going to work for the reasons I outlined in my first reply. Just placing an opamp before it doesn't fix any of the problems apart from the loss of gain.
I respectfully refer you back to the datasheet linked to at the bottom of my first post for a fully working implementation. |
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#8 |
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diyAudio Member
Join Date: Nov 2010
Location: A Coruna, Galicia
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#9 |
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diyAudio Member
Join Date: Nov 2010
Location: A Coruna, Galicia
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Anyway, I'm looking at a heatsink for the TDA1554 (2x22W stereo amplifier), I don't understand good the datasheet about part thermal.
According to the Datasheet I have this data: - Total power dissipation 60W (I will use about 10W?) - Junction temperature Tj = 150 º C (For security, I will use 100ºC) - Power Derating (0.67 ° C / W?) I want to use the formula: Rda = ((Tj - Ta) / IDPs) - Rjc - Rcd I initially had thought of: Rda = ((100 - 50) / 7.5) - (0.67 º C / W) - (0.12 º C / W) = 5.88 º C / W My questions to you experts: Am I wrong? What am I doing wrong? Thanks!! Regards!! |
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