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Arduino Forum DS1804 (100k) produces a lot of noise
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  • noise
  • DS1804
Related

DS1804 (100k) produces a lot of noise

Ricodora
Ricodora over 1 year ago

Hello people!
The volume goes up and down correctly. The lower the volume, the more noise it produces.

DS1804 has 100 positions:

In positions [80,99] it does not produce noise (I do not hear it)
[70-80) -> produces a little noise
[60-70) -> more noise
[50-60) -> even more noise
...
[10-20) -> a lot of distortion

The component is original purchased from "Mouser Electronic".

Component Detail anda Data Sheet

I use a constant 5v power supply, I don't use USB-C

I have 2 units of DS1804, they both do the same thing.

<code>

#include <DS1804.h>

#define increment 14
#define select 33
#define up_down 27

// DS1804( byte CSpin, byte INCpin, byte UDpin, unsigned long maxResistance )
DS1804 digipot = DS1804( select, increment, up_down, DS1804_HUNDRED );

void setup() {
Serial.begin(9600);
digipot.setToZero();
}

void loop() {
if ( Serial.available() > 0 ) {
int position = Serial.parseInt();
if (position > 0) {
digipot.setWiperPosition(position);
}
}
}

</code>

image

Any ideas? Thank you

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Top Replies

  • michaelkellett
    michaelkellett over 1 year ago in reply to Ricodora +4
    The XC9503 is expensive because it has an analogue signal range of +/- 8V while operating froma single rail 5V supply. That's why it works. You can use the DS1804 but you will need input capacitors…
  • michaelkellett
    michaelkellett over 1 year ago +2
    There are problems with your circuit. The audio signal going in will usually swing both positive and negative with respect to GND. But the data sheet for the 1804 says: H - High-Terminal Potentiometer…
  • dougw
    dougw over 1 year ago +1
    Is it white noise or a tone?
Parents
  • michaelkellett
    0 michaelkellett over 1 year ago

    There are problems with your circuit. The audio signal going in will usually swing both positive and negative with respect to GND.

    But the data sheet for the 1804 says:

    H - High-Terminal Potentiometer. This is the high terminal of the potentiometer. It is not required that this terminal be connected to a potential greater than the L-terminal. Voltage applied to the H-terminal cannot exceed the power-supply voltage, VCC, or go below ground.
    L - Low-Terminal Potentiometer. This is the low terminal of the potentiometer. It is not required that this terminal be connected to a potential less than the H-terminal. Voltage applied to the L-terminal cannot exceed the power-supply voltage, VCC, or go below ground.
    W - Wiper of the Potentiometer. This pin is the wiper of the potentiometer. Its position on the resistor array is controlled by the three-terminal control port. Voltage applied to the wiper cannot exceed the power-supply voltage, VCC, or go below ground.

    The next problem is that you are driving headphones directly from the 1804 - but headphones are usually quite a low impedance. You are using the 100K version of the 1804 so the load impedance should not be less than about 500k ohms - a headphone amplifier might be OK for that but not most headphones.

    Could you let us know the source of the audio signal (and spec of the device) and the kind of headphones you are driving. I could then suggest a circuit that might work !

    It's very likely to need a buffer amplifier.

    MK

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  • Ricodora
    0 Ricodora over 1 year ago in reply to michaelkellett

    Hello mk,

    I understand... Since the signal that I put to 'H' takes negative values (less than ground) the component cannot function correctly. This component is not valid for me. Learning is fascinating but it is costing me money and headaches.

    This circuit works well with the X9C503 (AliExpress, thanks to your help) there is no noise or distortion. When I went to buy the original they didn't have the DIP version, so I bought this DS1804 (and I was wrong)

    As the source of the audio signal I am using the PC and the mobile phone, but I want to connect any common device (mp3, tablet, music equipment...). With a 3.5mm or RCA connector

    I use headphones and/or speakers only to test that it works correctly, but instead I use an MSGEQ7.

    What am I looking for? I connect the input line to two 22k resistors and send it to the MSGEQ7 and paint the 7 frequencies with LEDs. Sometimes the music volume is too high and all the LEDs are painted (saturation) and the music is not painted correctly. I use the potentiometer in this situation to lower the volume so that the music is drawn well.

    This is my invention

    I decided to use the X9C503 out of fear, I don't have enough knowledge. My first option was M62429 which lowers the volume and also amplifies, but I was afraid of breaking the MSGEQ7 when amplifying the signal (because sometimes the device or the song has low volume and very few LEDs are painted)

    With this experience I have to look for another potentiometer that does not have this limitation or characteristic.

    Thank you!

    image

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  • michaelkellett
    0 michaelkellett over 1 year ago in reply to Ricodora

    The XC9503 is expensive because it has an analogue signal range of +/- 8V while operating froma single rail 5V supply.

    That's why it works.

    You can use the DS1804 but you will need input capacitors and a buffer amplifier:

    image

    R3 and R4 make a bias voltage of 3V, a bit over half the 5V supply.

    C1 and C2 stop that bias voltage reaching the inputs or the inputs drawing DC current from the bias.

    R5 connects the bias voltage to the H pin and the input caps.

    Now for +/1V on the input pin you'll get about  2- 4 V on the H pin.

    C3 stops noise on the supply getting into the H pin.

    The L pin is grounded for AC but open ciruit for DC by C4. The voltage here will be a few mV less than on H because of Q1 base current and leakage in C4.

    Q1 and Q2 act as and NPN transistor wired as an emitter follower with very low base current.  You need the composite transistor because a single transistor will need too much base current which will cause a lrge voltage drop (more than 1V) across the 50K pot in the DS1804.

    C5 decouples the DC on Q1 emitter from the output pin.

    R1 and R2 and R5 and the 50k pot resistance reduce the gain at maximum to a bit less than 1:

    The input impedance at H is 50k//100k = 33k. For either input gain = 33/(33+10) = 0.767 or approx -3dB.

    Q1 and Q2 can be any (almost) small signal audio type transistor. You are looking for Hfe > 200 at Ic = 5mA for Q1 and at a Ic = 100uA for Q2.

    BC182L or BC184L are good fr NPN, BC212L or BC214L are good for PNP (in this application - not for everything).

    TP3 is the output.

    You may have trouble getting this to work without access to a scope.

    But now you know why the X9C503 is worth the money !

    (But the ciruit above has a much lower output impedance, although it won't drive headphones.and will struggle to deliver more than +/- 2V on the output into a 2k load)

    MK

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  • michaelkellett
    0 michaelkellett over 1 year ago in reply to Ricodora

    The XC9503 is expensive because it has an analogue signal range of +/- 8V while operating froma single rail 5V supply.

    That's why it works.

    You can use the DS1804 but you will need input capacitors and a buffer amplifier:

    image

    R3 and R4 make a bias voltage of 3V, a bit over half the 5V supply.

    C1 and C2 stop that bias voltage reaching the inputs or the inputs drawing DC current from the bias.

    R5 connects the bias voltage to the H pin and the input caps.

    Now for +/1V on the input pin you'll get about  2- 4 V on the H pin.

    C3 stops noise on the supply getting into the H pin.

    The L pin is grounded for AC but open ciruit for DC by C4. The voltage here will be a few mV less than on H because of Q1 base current and leakage in C4.

    Q1 and Q2 act as and NPN transistor wired as an emitter follower with very low base current.  You need the composite transistor because a single transistor will need too much base current which will cause a lrge voltage drop (more than 1V) across the 50K pot in the DS1804.

    C5 decouples the DC on Q1 emitter from the output pin.

    R1 and R2 and R5 and the 50k pot resistance reduce the gain at maximum to a bit less than 1:

    The input impedance at H is 50k//100k = 33k. For either input gain = 33/(33+10) = 0.767 or approx -3dB.

    Q1 and Q2 can be any (almost) small signal audio type transistor. You are looking for Hfe > 200 at Ic = 5mA for Q1 and at a Ic = 100uA for Q2.

    BC182L or BC184L are good fr NPN, BC212L or BC214L are good for PNP (in this application - not for everything).

    TP3 is the output.

    You may have trouble getting this to work without access to a scope.

    But now you know why the X9C503 is worth the money !

    (But the ciruit above has a much lower output impedance, although it won't drive headphones.and will struggle to deliver more than +/- 2V on the output into a 2k load)

    MK

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