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555 timer chip tone voltage monitor ???

  • Thread starter twilightzonepinball
  • Start date
T

twilightzonepinball

Jan 1, 1970
0
Hello,

My name is Tom and I am new to this group. Please forgive my
newbieness. I am an Industrial controls tech with limited electronics
knowledge.


I was hoping I could get some help designing a simple tone generator
for a unique application. I have so far built a simple tone generator
using a 555 timer chip and some resistors and caps with a circuit I
found online. I am using pin 5 to control it as a VCO astable.
Everything works great, the input dc voltage is varying the tone, but
I want the frequency to go up with voltage. Kind of like an audio
voltage monitor. This input voltage is about 0-14 volts. The problem
is that the frequency goes down when the voltage goes up. Just the
opposite of what I need.


Can anyone suggest a solution?


Thanks in advance,


-Tom
 
J

John Popelish

Jan 1, 1970
0
twilightzonepinball said:
Yes, I used the one on this website.
http://ourworld.compuserve.com/homepages/Bill_Bowden/page10.htm
The tone generator without the transistor on the output. I am just
using a headphone so not much power needed.

I think you may be able to achieve what you want by just
adding another resistor from your voltage signal to pin 7.
It must not have less than twice the resistance of R1 if you
want to have a tone at zero volts. It will provide the
widest range if the 555 is programmed to produce a pulse,
instead of a nearly square wave, by lowering the value of R2
to something considerably less than R1.
 
J

John Popelish

Jan 1, 1970
0
twilightzonepinball said:
So no need for the inverting opamp? Just another resistor to pin 7
from pin 5? I am using trimmer pots for all the resistors so I can
tweak things.-Tom


No opamp, but the resistor goes between your input signal
and pin 7, instead of pin 5. Just leave pin 5 unconnected.
 
J

John Fields

Jan 1, 1970
0
Hello,

My name is Tom and I am new to this group. Please forgive my
newbieness. I am an Industrial controls tech with limited electronics
knowledge.


I was hoping I could get some help designing a simple tone generator
for a unique application. I have so far built a simple tone generator
using a 555 timer chip and some resistors and caps with a circuit I
found online. I am using pin 5 to control it as a VCO astable.
Everything works great, the input dc voltage is varying the tone, but
I want the frequency to go up with voltage. Kind of like an audio
voltage monitor. This input voltage is about 0-14 volts. The problem
is that the frequency goes down when the voltage goes up. Just the
opposite of what I need.


Can anyone suggest a solution?

---
You can do it by using an inverting opamp to drive the 555 control
voltage pin so that with zero volts into the opamp the control
voltage input is higher than when the input to the opamp is more
positive than 14V.

Can you post a schematic of what you have so far?

(Not here, since this is a text-only newsgroup, but to
alt.binaries.schematics.electronic or to a website somewhere)
 
T

twilightzonepinball

Jan 1, 1970
0
---
You can do it by using an inverting opamp to drive the 555 control
voltage pin so that with zero volts into the opamp the control
voltage input is higher than when the input to the opamp is more
positive than 14V.

Can you post a schematic of what you have so far?

(Not here, since this is a text-only newsgroup, but to
alt.binaries.schematics.electronic or to a website somewhere)
Yes, I used the one on this website.
http://ourworld.compuserve.com/homepages/Bill_Bowden/page10.htm
The tone generator without the transistor on the output. I am just
using a headphone so not much power needed.

Thanks,

-Tom
 
T

twilightzonepinball

Jan 1, 1970
0
twilightzonepinballwrote:

I think you may be able to achieve what you want by just
adding another resistor from your voltage signal to pin 7.
It must not have less than twice the resistance of R1 if you
want to have a tone at zero volts. It will provide the
widest range if the 555 is programmed to produce a pulse,
instead of a nearly square wave, by lowering the value of R2
to something considerably less than R1.

So no need for the inverting opamp? Just another resistor to pin 7
from pin 5? I am using trimmer pots for all the resistors so I can
tweak things.-Tom
 
T

twilightzonepinball

Jan 1, 1970
0
twilightzonepinballwrote:


No opamp, but the resistor goes between your input signal
and pin 7, instead of pin 5. Just leave pin 5 unconnected.- Hide quoted text -

- Show quoted text -

Thanks for the clarification. I will try it and get back.

-Tom
 
T

twilightzonepinball

Jan 1, 1970
0
twilightzonepinballwrote:


No opamp, but the resistor goes between your input signal
and pin 7, instead of pin 5. Just leave pin 5 unconnected.- Hide quoted text -

- Show quoted text -

It works!!!!!!!!!!

You are the man, thanks so much.

-Tom
 
T

twilightzonepinball

Jan 1, 1970
0
twilightzonepinballwrote:


No opamp, but the resistor goes between your input signal
and pin 7, instead of pin 5. Just leave pin 5 unconnected.- Hide quoted text -

- Show quoted text -

My circuit is working great except I need to control the volume from
pin 3. I have a 100uf cap in series with a headphone. I have tried
using a 10k pot wired to ground, but I don't get the control I need.
The volume is either too low or full blast at the top of the pot. I
guess I am using the wrong value pot. How do I choose the correct
value?

Thanks,

-Tom
 
R

Randy Day

Jan 1, 1970
0
twilightzonepinball wrote:

[snip]
My circuit is working great except I need to control the volume from
pin 3. I have a 100uf cap in series with a headphone. I have tried
using a 10k pot wired to ground, but I don't get the control I need.
The volume is either too low or full blast at the top of the pot. I
guess I am using the wrong value pot. How do I choose the correct
value?

You probably need to order a 'linear taper' pot.
It sounds like you have an audio taper one.
 
J

John Popelish

Jan 1, 1970
0
twilightzonepinball said:
My circuit is working great except I need to control the volume from
pin 3. I have a 100uf cap in series with a headphone. I have tried
using a 10k pot wired to ground, but I don't get the control I need.
The volume is either too low or full blast at the top of the pot. I
guess I am using the wrong value pot. How do I choose the correct
value?

If you don't need to be able to adjust the volume all the
way to zero, try disconnecting the ground end of the pot
element. I don't know what ohms your headphones are, but a
series variable resistor (pot with only two connections)
with a total resistance from 10 to 100 times the headphone
resistance would probably work better.
 
M

Michael Black

Jan 1, 1970
0
twilightzonepinball" ([email protected]) said:
My circuit is working great except I need to control the volume from
pin 3. I have a 100uf cap in series with a headphone. I have tried
using a 10k pot wired to ground, but I don't get the control I need.
The volume is either too low or full blast at the top of the pot. I
guess I am using the wrong value pot. How do I choose the correct
value?
The pot is too large in value, in comparison with the headphone.

It's not clear how you have it wired, but to begin let's pretend the
pot is in series with the headphone, one end and the arm being used
and leaving the other end unconnected. That gives you a variable resistor.
You've now created a voltage divider, with the variable resistor as one
of the resistors, and the headphone as the other. When the variable
resistor is the same value as the headphone, it's halving the voltage.
Lower the variable resistor more, and the voltage divider will be
having less and less effect, until the variable resistor is at zero
ohms, and then there's full volume.

But when you start raising the resistance of the variable resistor above
the value of the headphone, it won't take long before the variable resistor
is much larger in value than the headphone, and thus the voltage is divided
by a very large amount.

It's not that it's no longer working, it's that once the variable resistance
goes above a certain level, you just won't notice any more difference as
you increase the resistance. Because you've already got the volume below
a useable level.

You only see basically "loud" and "no volume" because the useable resistance
of the pot is on a very small percentage of its travel.

Use a much lower value pot, like the same value as the impedance of the
headphones or not more than two or three times the impedance of the
headphones, and then you'll see a real volume control. Because then
over all or most of the potentiometer's travel, it will have some effect.

Michael
 
T

twilightzonepinball

Jan 1, 1970
0
The pot is too large in value, in comparison with the headphone.

It's not clear how you have it wired, but to begin let's pretend the
pot is in series with the headphone, one end and the arm being used
and leaving the other end unconnected. That gives you a variable resistor.
You've now created a voltage divider, with the variable resistor as one
of the resistors, and the headphone as the other. When the variable
resistor is the same value as the headphone, it's halving the voltage.
Lower the variable resistor more, and the voltage divider will be
having less and less effect, until the variable resistor is at zero
ohms, and then there's full volume.

But when you start raising the resistance of the variable resistor above
the value of the headphone, it won't take long before the variable resistor
is much larger in value than the headphone, and thus the voltage is divided
by a very large amount.

It's not that it's no longer working, it's that once the variable resistance
goes above a certain level, you just won't notice any more difference as
you increase the resistance. Because you've already got the volume below
a useable level.

You only see basically "loud" and "no volume" because the useable resistance
of the pot is on a very small percentage of its travel.

Use a much lower value pot, like the same value as the impedance of the
headphones or not more than two or three times the impedance of the
headphones, and then you'll see a real volume control. Because then
over all or most of the potentiometer's travel, it will have some effect.

Michael

Does that pose any risk to the output of the chip using a low value
pot?

-Tom
 
T

twilightzonepinball

Jan 1, 1970
0
The pot is too large in value, in comparison with the headphone.

It's not clear how you have it wired, but to begin let's pretend the
pot is in series with the headphone, one end and the arm being used
and leaving the other end unconnected. That gives you a variable resistor.
You've now created a voltage divider, with the variable resistor as one
of the resistors, and the headphone as the other. When the variable
resistor is the same value as the headphone, it's halving the voltage.
Lower the variable resistor more, and the voltage divider will be
having less and less effect, until the variable resistor is at zero
ohms, and then there's full volume.

But when you start raising the resistance of the variable resistor above
the value of the headphone, it won't take long before the variable resistor
is much larger in value than the headphone, and thus the voltage is divided
by a very large amount.

It's not that it's no longer working, it's that once the variable resistance
goes above a certain level, you just won't notice any more difference as
you increase the resistance. Because you've already got the volume below
a useable level.

You only see basically "loud" and "no volume" because the useable resistance
of the pot is on a very small percentage of its travel.

Use a much lower value pot, like the same value as the impedance of the
headphones or not more than two or three times the impedance of the
headphones, and then you'll see a real volume control. Because then
over all or most of the potentiometer's travel, it will have some effect.

Michael

I mean with it wired with one end to ground and the wiper going to the
headphone and the other end to pin 3. I assume that is the correct way
to wire a volume control.

-Tom
 
M

Michael Black

Jan 1, 1970
0
"twilightzonepinball" ([email protected]) writes:

I mean with it wired with one end to ground and the wiper going to the
headphone and the other end to pin 3. I assume that is the correct way
to wire a volume control.
I was going to explain that, but at that point I'd gotten tired.

It's basically the same concept.

MIchael
 
J

John Popelish

Jan 1, 1970
0
twilightzonepinball said:
Does that pose any risk to the output of the chip using a low value
pot?

The worst case is when the volume is turned all the way up,
and the headphone and pot resistor are in parallel across
the output. As long as the pot resistance is several times
the headphone resistance, it does not add much load to that
case. If the chip cannot handle the current drawn by a
direct connection to the headphones, then no pot value will
be safe at full volume.
 
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