Lowering digital potentiometer value.

Thread Starter

Bod

Joined Sep 18, 2016
242
If you saw the end of my previous thread, you might know I am using this schematic:
transistor amplifier.png
To calculate the gain on this it is R3/R4. So for this, it would be 2000/200 = 2. It was suggested I use a 100R potentiometer which is a good idea because I can stop audio clipping with this by adjusting it if the volume is too loud.

The problem is, where I wanted to position my Arduino is out of reach and so using a pot would be difficult. I could run wires to and from the Arduino and place the pot somewhere else but it would be a lot of wire and it couldn't be easily moved around. I then thought - ok - how can I make this wireless? It occurred that I should use an IR remote with a receiver.
The main con is how am I supposed to adjust a pot with an IR pulse? I could use a servo to turn the pot but that's like a plan Z. I did a bit of research on it and found what's called a 'digital potentiometer'. Unfortunately, (looking on RS, Digikey, and Mouser) the lowest value of digital pot is 1K and RS was the only one selling a DIP package; all others were SMD - I need a 100R pot.
I looked around more but I couldn't find any lower than 1K. I also could not find a way to lower the value. It looks like the simple method of adding a resistor in parallel won't work.
What can I do to lower the value of this digital pot? Or is there a better approach to this?

FYI: here's the chip: https://uk.rs-online.com/web/p/digital-potentiometers/9215848/

Bod
 

Alec_t

Joined Sep 17, 2013
10,378
Couldn't you just increase both R3 and R4 by a factor of five? Then R4 would be 1k, the digipot minimum.
Or, just use a 2k digipot for R3?
 

Thread Starter

Bod

Joined Sep 18, 2016
242
Couldn't you just increase both R3 and R4 by a factor of five? Then R4 would be 1k, the digipot minimum.
Or, just use a 2k digipot for R3?
I could. I really was over complicating things (it's getting late for electronics :p)

Thank you!
Bod.
 

Audioguru

Joined Dec 20, 2007
11,251
This transistor is a preamp that you do not need since your input is from your phone that produces plenty of output level. If this transistor with its gain of 10 times is used then the output of this preamp will be severely distorted squarewaves.

You forgot to say where in this transistor circuit you will add a 100 ohms pot and how it will be connected.
Why can't you adjust the loudness (volume) on your phone instead?
 

iimagine

Joined Dec 20, 2010
388
You are just totally confused and lost, and you are not trying to understand or solving problems, instead you just jump from circuit to circuit hoping to find the one that you think is 'working' without further thought of how it worked.

Anyway, you can always control input voltage level by connecting a resister or digi pot in series with the coupling cap. I have mentioned this also in the other thread.
 

Thread Starter

Bod

Joined Sep 18, 2016
242
Do you need the full 0-100 range? And does it need to be linear?

I get 10.
I would prefer to have the full range but it's not necessary. Also, yes it would help if it was linear
(And also - whoops - my bad. It is meant to be 10!)

This transistor is a preamp that you do not need since your input is from your phone that produces plenty of output level. If this transistor with its gain of 10 times is used then the output of this preamp will be severely distorted squarewaves.

You forgot to say where in this transistor circuit you will add a 100 ohms pot and how it will be connected.
Why can't you adjust the loudness (volume) on your phone instead?
It's going in place of R4.
I could adjust my phone volume but I have gotten this idea and it's pretty feasible so I would like to try it.

You are just totally confused and lost, and you are not trying to understand or solving problems, instead you just jump from circuit to circuit hoping to find the one that you think is 'working' without further thought of how it worked.
This is part of the same circuit. I'm not looking for another circuit. I was finding a way to lower the resistance of a digital pot and I was given one. I still have the circuit on my desk, built on the breadboard.
As I said, this is not me asking for another circuit - if I was, I would bump the other thread.

Bod
 

Audioguru

Joined Dec 20, 2007
11,251
You said you want to reduce clipping by adjusting the gain by using a pot for R4.
But R4 is used for DC feedback in addition to AC feedback and reducing its value increases both DC and AC gain, it does not reduce it.

Increased DC gain causes the bottom of the waveform to be clipping.
Increased AC gain causes the top of the waveform to have squashed distortion and the bottom to be clipping.

A volume control is used capacitor-coupled at the input of an amplifier, not inside it affecting DC and AC gain.
 

Attachments

TeeKay6

Joined Apr 20, 2019
393
If you saw the end of my previous thread, you might know I am using this schematic:
View attachment 181736
To calculate the gain on this it is R3/R4. So for this, it would be 2000/200 = 2. It was suggested I use a 100R potentiometer which is a good idea because I can stop audio clipping with this by adjusting it if the volume is too loud.

The problem is, where I wanted to position my Arduino is out of reach and so using a pot would be difficult. I could run wires to and from the Arduino and place the pot somewhere else but it would be a lot of wire and it couldn't be easily moved around. I then thought - ok - how can I make this wireless? It occurred that I should use an IR remote with a receiver.
The main con is how am I supposed to adjust a pot with an IR pulse? I could use a servo to turn the pot but that's like a plan Z. I did a bit of research on it and found what's called a 'digital potentiometer'. Unfortunately, (looking on RS, Digikey, and Mouser) the lowest value of digital pot is 1K and RS was the only one selling a DIP package; all others were SMD - I need a 100R pot.
I looked around more but I couldn't find any lower than 1K. I also could not find a way to lower the value. It looks like the simple method of adding a resistor in parallel won't work.
What can I do to lower the value of this digital pot? Or is there a better approach to this?

FYI: here's the chip: https://uk.rs-online.com/web/p/digital-potentiometers/9215848/

Bod
@Bod
In your #1 post schematic, the base of Q1 is set to about 0.9V via R1 & R2. That means the emitter of Q1 will be about 0.4V to 0.2V. Any input signal (via C1) that drives the base such that the emitter voltage reaches 0V will be clipped. That is, once Q1 is off, you cannot drive it further off. (Note that this analysis is independent of the value of R3 & R4.) If your input signal is that great, then it must be reduced before it drives Q1.
 

Audioguru

Joined Dec 20, 2007
11,251
Reducing the value of R4 increases the AC gain which is already too high.
Reducing the value of R4 also increases the DC gain which causes the bottom of the waveform to have clipping.
 

dl324

Joined Mar 30, 2015
8,954
I would prefer to have the full range but it's not necessary. Also, yes it would help if it was linear
Putting a 110 ohm resistor in parallel with the digital pot would give you a range from 0-99.1 ohms. It isn't linear, but it's monotonic:
upload_2019-7-17_8-27-23.png
An audio taper pot would give more like 10% of the volume at the 50% position.
 

Attachments

Thread Starter

Bod

Joined Sep 18, 2016
242
You said you want to reduce clipping by adjusting the gain by using a pot for R4.
But R4 is used for DC feedback in addition to AC feedback and reducing its value increases both DC and AC gain, it does not reduce it.

Increased DC gain causes the bottom of the waveform to be clipping.
Increased AC gain causes the top of the waveform to have squashed distortion and the bottom to be clipping.

A volume control is used capacitor-coupled at the input of an amplifier, not inside it affecting DC and AC gain.
Reducing the value of R4 increases the AC gain which is already too high.
Reducing the value of R4 also increases the DC gain which causes the bottom of the waveform to have clipping.
Ok so, what I get from this is that do not replace R4, instead, place the pot next to the input cap - as iimagine & TeeKay6 said previously.
If so, what value?

Putting a 110 ohm resistor in parallel with the digital pot would give you a range from 0-99.1 ohms. It isn't linear, but it's monotonic:
View attachment 181857
An audio taper pot would give more like 10% of the volume at the 50% position.
I think that will. There is no reason for me to go that low - or high.
 

TeeKay6

Joined Apr 20, 2019
393
Ok so, what I get from this is that do not replace R4, instead, place the pot next to the input cap - as iimagine & TeeKay6 said previously.
If so, what value?


I think that will. There is no reason for me to go that low - or high.
@Bod
A 1K pot connected across V2, with wiper connected to C1 is likely to work okay...provided V2 is not disturbed by that loading.
 

Audioguru

Joined Dec 20, 2007
11,251
To control audio loudness then use a pot with an "audio taper" not a linear pot.
Since you are using the earphones output of your phone then a 1k or 10k audio pot is fine.

Edit: A digital pot with the AC input from your phone needs a plus and minus power supply unless you bias the input signal at half a positive supply for it.
Also, a digital pot will be linear, not for audio level adjustment.
 
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