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chore: update README contents to match actual project implementation
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Signed-off-by: Nenad Radulovic <nenad.b.radulovic@gmail.com>
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nradulovic committed Aug 31, 2021
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Expand Up @@ -18,8 +18,7 @@ The amplifier architecture consists of the following sections:
* Power amplifier
* Power supply

**Input circuit** and **Power amplifier** sections are located on single PCB
while the **Power supply** is located on separate PCB board.
All sections are located on separate PCB boards.

.. image:: images/pcb_from_top.png

Expand All @@ -28,14 +27,12 @@ while the **Power supply** is located on separate PCB board.
.. image:: images/pcb_render_from_top.png


Input circuit
============
Input circuit
==================================

The input circuit contains:

* Input EMI suppression
* Input low pass filter
* The ground loop breaker resistor

Input EMI suppression
---------------------
Expand Down Expand Up @@ -69,6 +66,13 @@ Also, a 100n X7R capacitor shall be placed between SGND and chassis right at the
input connector. This capacitor will shunt radio and other interfirence signal
into the Chassis Ground potential.

Power amplifier
===============

* Input low pass filter
* The ground loop breaker resistor
* Output EMI suppression

Input low pass filter
---------------------

Expand Down Expand Up @@ -107,9 +111,6 @@ The ground loop breaker resistor
A ground loop breaker resistor is located between SGND and GNDPWR grounds. The
value of this resistor should be around 10 ohms.

Power amplifier
===============

Output EMI suppression
----------------------

Expand Down Expand Up @@ -151,36 +152,6 @@ Output shunt resistor should be between 2 Ohm and 5 Ohms. See
*Douglas Self - Audio Power Amplifier Design Handbook, 3rd Ed., Output networks,
chapter 7* for effect on power amplifier transfer function.

Paralleling multiple modules
----------------------------

Ballast resistor
````````````````

Each amplifier will connect to output bus via ballast resistor. The ballast
resistor is made of three 1 Ohm resistors wired in parallel, which gives
``Rb=0.33 Ohm``.Maximum output current of the power amplifier is:

.. math::
Io(max)=Uo(max)/Zload(min)
With Uo(max) approx 30V and Zload(min) equal to 2 Ohms we get:

.. math::
Io(max)=15A
This current is divided by the number of modules in the amplifier, given by the
variable ``N=3``. Maximum power dissipation in ballast resistor is therefore:

.. math::
Pbdiss(max)=((Io(max)/N)**2*Rb)/3=2.75W
Resistors with power dissipation of 3 Watts is a good and very conservative
choice.

Power dissipation
`````````````````

Expand All @@ -196,101 +167,69 @@ is approximately 50W we get that the transformer supports crest factor of 4
(see:
*https://www.neurochrome.com/taming-the-lm3886-chip-amplifier/power-supply-design*).

This means that effective output power is around ``50W/4 = 12.5W``.

Maximum voltages at:

* Instantenious dissipation for LM1875 is ``Pdiss=30W``.
* Load phase is ``LoadPHI=60degrees``.
* Including quiescent current dissipation.
* Case temperature is 60C degrees.
* Taking into account OPS SOA.

+-------------+-------------+-----------+--------------+
| Zload [ohm] | Vsupply [V] | Vdrop [V] | Pdiss [W] |
+-------------+-------------+-----------+--------------+
| 16 | 26 | 2.2 | 19.1 |
+-------------+-------------+-----------+--------------+
| 12 | 24 | 2.6 | 21.1 |
+-------------+-------------+-----------+--------------+
| 8 | 23 | 4.4 | 26.8 |
+-------------+-------------+-----------+--------------+
| 6 | 21 | 4.6 | 28.9 |
+-------------+-------------+-----------+--------------+
| 4 | 16 | 5.2 | 22.6 |
+-------------+-------------+-----------+--------------+

This table tells us that if we want to drive 4ohm load at 26V we need 4 pieces
of LM1875 in parallel. This is quite a number of ICs, but fortunately, the
table presumes that the power supply can produce constant 26V at continuous
load and the signal is sinusoid. This is not the case with unregulated power
supply and music signals. We have to take into account how much energy is
stored in power supply capacitors and how much will the transformer voltages
sag under these conditions and that music signal has much lower effective power
comparing to instantaneous power.

Gain value
----------

Using inverted topology since we want to reduce common mode distortion in the
input stage.
For this power amplifier we are using non-inverting topology for simplicity
reasons. If you would like to have less distortion then LM1875 should be used
in inverting configuration.

The equivalent gain circuit resistance needs to stay below 600ohms. This is so
because all noise measurements in data-sheet were done with 600ohms or 0ohms.

Using low feedback gain is preferred for several reasons:

* there is more loop gain available to reduce the distortion
* reduced outout noues
* reduced output noise
* lower offset at output

Nominal gain is:

.. math::
G=-Rf/Rg
G=-Rf/Rg+1
Using E24 series of resistors:

+-----------+-----------+---------+
| Rf [Ohm] | Rg [kOhm] | G [V/V] |
+-----------+-----------+---------+
| 510 | 7.5 | -14.7 |
| 510 | 7.5 | 15.7 |
+-----------+-----------+---------+
| *510* | *8.2* | *-16.0* |
| *510* | *8.2* | *17.0* |
+-----------+-----------+---------+
| 510 | 9.1 | -17.8 |
| 510 | 9.1 | 18.8 |
+-----------+-----------+---------+
| 510 | 10.0 | -19.6 |
| 510 | 10.0 | 20.6 |
+-----------+-----------+---------+
| 510 | 11.0 | -21.5 |
| 510 | 11.0 | 22.5 |
+-----------+-----------+---------+

Using E48 series of resistors:

+-----------+-----------+---------+
| Rf [Ohm] | Rg [kOhm] | G [V/V] |
+-----------+-----------+---------+
| 511 | 7.50 | -14.7 |
| 511 | 7.50 | 15.7 |
+-----------+-----------+---------+
| 511 | 7.87 | -15.4 |
| 511 | 7.87 | 16.4 |
+-----------+-----------+---------+
| *511* | *8.25* | *-16.1* |
| *511* | *8.25* | *17.1* |
+-----------+-----------+---------+
| 511 | 8.66 | -16.9 |
| 511 | 8.66 | 17.9 |
+-----------+-----------+---------+
| 511 | 9.09 | -17.8 |
| 511 | 9.09 | 18.8 |
+-----------+-----------+---------+
| 511 | 9.53 | -18.6 |
| 511 | 9.53 | 19.6 |
+-----------+-----------+---------+
| 511 | 10.00 | -19.6 |
| 511 | 10.00 | 20.6 |
+-----------+-----------+---------+
| 511 | 10.50 | -20.5 |
| 511 | 10.50 | 21.5 |
+-----------+-----------+---------+
| 511 | 11.00 | -21.5 |
| 511 | 11.00 | 22.5 |
+-----------+-----------+---------+
| 499 | 7.50 | -15.0 |
| 499 | 7.50 | 16.0 |
+-----------+-----------+---------+

Chosen values for E24 series:
Expand All @@ -303,109 +242,6 @@ Chosen values for E48 series:
* Rf = 7.5kOhm
* Rg = 499 Ohm

Chosen values when using parallel E24 series (two resistor):

* Rf = 15kOhm
* Rg = 1kOhm

Chosen values when using parallel E48 series (two resistor):

* Rf = 15kOhm
* Rg = 1kOhm


Gain errors when using parallel solution
````````````````````````````````````````

Nominal absolute gain is:

.. math::
G=Rf/Rg
Where ``Rf`` is the resistor towards output and ``Rg`` is the resistor towards
signal source. We are using absolute gain here since it's more natural to work
with positive numbers. The resistor tolerance is 0.1%. Maximum value for gain
due to resistor tolerances in this case is:

.. math::
G(max)=Rf(max)/Rg(min)
G(max)=(Rf*(1+pp))/(Rg*(1-pp))=G*(1+pp)/(1-pp)
Minimum gain is:

.. math::
G(min)=Rf(min)/Rg(max)
G(min)=(Rf*(1-pp))/(Rg*(1+pp))=G*(1-pp)/(1+pp)
Maximum voltage difference by resistor tolerances can be calculated by:

.. math::
Uin=Uout(max)/G
Urdiff(max)=G(max)*Uin-G(min)*Uin=Uin*(G(max)-G(min))
Urdiff(max)=(Uout(max)/G)*(G(max)-G(min))
This approximates to:

.. math::
Udiff(max)=Uout(max)*4*pp
For 0.1% the pp is 0.001, so if ``pp=0.001`` and ``uout(max) = 30V``, we get:

.. math::
Urdiff(max) = 120mV
Maximum voltage difference due to different open loop gains can be calculated,
too:

.. math::
Eadiff(max)=uout(max)/A(min)
Typical open loop gain in the data-sheet is 115dB. Minimum open loop gain is
90dB. This calculates to the difference of input voltage, 90dB is approx.
30.000:

.. math::
Eadiff(max)=30/30000=1mV
This calculates to:

.. math::
Uadiff(max)=Eadiff(max)*g=30mV
Total max difference voltage is sum of voltages created from resistor
tolerances and a voltage from open loop gain deficiency:

.. math::
Udiff(max)=Urdiff(max)+Uadiff(max)=120+30=150mV
For this part of circuit there is no advantage of using multiple resistors
(parallel or series) to get the desired resistance but lower the tolerance.
The reason the tolerances do not decrease when using multiple resistors is
because of the involved manufacturing process. Using multiple resistors is
OK only in situation when wanting bigger power dissipation ability or to get
a specific non E24 resistance.

The equivalent resistance of the loop gain circuitry must be below 600ohms.

The LM1875 shall be in differential connection. The lower arm of the gain loop
circuitry shall use ~500ohm resistor. Using 470uF we get 0.68Hz lower corner
frequency. Also, the signal is applied to inverting input. See Bob Cordell
super gain clone ``.ppt``.

Frequency compensation
----------------------

Expand All @@ -417,7 +253,7 @@ The LM1875 is modeled in the following way:
* ``Fp3``, pole which probably originates from input or intermediate stages.
* ``Fp4 Hz``, pole which probably originates from input or intermediate stages.
* ``Rops``, open loop output stage impedance. The OPS open loop impedance is
unusually low because the LM3886 uses output inclusive Miller compensation
unusually low because the LM1875 uses output inclusive Miller compensation
which can be observed on the equivalent schematic in the data-sheet.

+-----------+-----------+-----------+-----------+-----------+-----------+-----------+
Expand Down Expand Up @@ -489,9 +325,9 @@ For LM1875 we would get:
Rf = 7.5kOhm
fp2 = 1.6e6 Hz
fp2 = 1.5e6 Hz
Cl=1/(2*pi*Rf*fp2)=13.3pF
Cl=1/(2*pi*Rf*fp2)=14.1pF
Outcome:

Expand Down Expand Up @@ -572,7 +408,7 @@ For LM1875 we get:

.. math::
Cf=Cl+Csi+Cadd=13.3+0.8+2pF=16.1pF
Cf=Cl+Csi+Cadd=14.1+0.8+2pF=16.9pF
Since the closest, standard values of capacitors are 15pF and 18pF, we choose
the 15pF as the final value for `Cl` capacitor:
Expand Down Expand Up @@ -613,44 +449,6 @@ powers of 20W-30W @ 8 Ohm.
* On case chassis there should be a safety ground screw just near at the input
220V socket.


Parallel chip solution
----------------------

Transformer specification for LM1875 amplifier is the following:

* ``S=200VA``, power rating.
* ``Usn1=20Veff``, first secondary nominal voltage.
* ``Usn2=20Veff``, second secondary nominal voltage.
* ``k=5%``, regulation.

Secondary internal resistance is:

.. math::
Usu=Usn1*(1+(k/100))
Isn=S/(Usn1+Usn2)
Ri=(Usn1-Usu)/Isn
Using values from above we get:

.. math::
Usu=20*(1+(5/100))=21Veff
Isn=5Aeff
Ri=200mOhm
The power supply section is using two banks of 10mF capacitors with 0.22Ohm
resistor in series between them. This arrangement gives time constant about
100ms when going from unloaded to full load state.

Single chip solution
--------------------

Transformer specification for LM1875 amplifier is the following:

* ``S=80VA``, power rating.
Expand Down

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