Can a single supply op amp be used for a summer?

[John Fields]

On Thu, 8 Oct 2009 03:56:06 -0700 (PDT), Andy <andrewkgentile@gmail.com>
wrote:

JF,
I found a good document from TI explaining single ended op amp design,
so I will continue from there. I really do appreciate you effort,
but unfortunately there is some data being lost in translation here.
This is apparent by your equation defining a voltage divider. A
divider consisting of two resistors will create a center voltage of:
Vout = (E1-Vin)*R2/(R1+R2).
---

Only when Vout is referred to Vin.

In the case where Vout is referred to 0V (your case, I believe) and Vin
varies, then Vin must be added to Vout.

For example, more clearly annotating the circuit:

E1
|
[R1]
|
+--->E2
|
[R2]
|
E3

and solving:


(E1 - E3) * R2
E2 = ----------------
R1 + R2


for various values of E3 results in the following table with R1 set to
4, R2 set to 1, and E2 referred to E3

E1 E3 E2
VOLTS VOLTS VOLTS
------|--------|--------
5 0 1.0
5 1 0.8
5 2 0.6
5 3 0.4
5 4 0.2
5 5 0.0


That's nonsensical for your application, which I believe refers the
output of the divider to ground instead of to Vin.

Such being the case, we can redraw the circuit to include the ground
reference,:


E1>----+
|
[R1]
|
+------>E2
|
[R2]
|
E3>----+

GND>---------->0V

rewrite the equation in order to include the ground reference,:

(E1 - E3) * R2
E2 = ---------------- + E3
R1 + R2

and wind up with:


E1 E3 E2
VOLTS VOLTS VOLTS
------|--------|--------
5 0 1.0
5 1 1.8
5 2 2.6
5 3 3.4
5 4 4.2
5 5 5.0

Which is what you want, yes?

 

[George Herold]

On Oct 8, 6:56 am, Andy <andrewkgent...@gmail.com> wrote:

On Oct 7, 8:52 pm, John Fields <jfie...@austininstruments.com> wrote:





On Wed, 7 Oct 2009 15:06:18 -0700 (PDT), Andy <andrewkgent...@gmail.com
wrote:

I appreciate your taking the time to write this up, but I'm going to
have to simply believe you that this circuit is workable with a single
ended op amp.

---
Bad idea.

You need to believe that everything I, or anyone else say(s) is suspect
until you confirm that it's true for yourself.
---

I cannot make heads or tails out of what you sent.  

---
Sorry about that.

It's an LTspice circuit list containing all the data you need to
simulate the circuit.

The simulator's fantastic, and free, and you can download it from:

http://www.linear.com/designtools/software/
---

But thank you for your effort.

---
My pleasure.

For a voltage divider configured to properly do what you want:

      E1
      |
     [R1]
      |
      +--->Vout
      |
     [R2]
      |
     Vin

Vout will vary as a function of Vin like this:

             (E1 - Vin) * R2
     Vout = ----------------- + Vin    
                 R1 + R2

and not like:

    Vout = (5-Vin)/5

JF

JF,
I found a good document from TI explaining single ended op amp design,
so I will continue from there.   I really do appreciate you effort,
but unfortunately there is some data being lost in translation here.
This is apparent by your equation defining a voltage divider.  A
divider consisting of two resistors will create a center voltage of:
Vout = (E1-Vin)*R2/(R1+R2).  Your equation shown above as Vout = (E1-
Vin)*R2/(R1+R2) + Vin is the description of another circuit.  This is
the circuit which I cannot see, but it clearly consists of more than
two resistors in series.  Anyway, if you are interested, google
"single supply op amp design Mancini" and you will find the paper from
TI.  It shows diagrams of several typical op amp circuits along with
the equations for Vout = f(Vin).- Hide quoted text -

- Show quoted text -

Hi Andy, I don't want to be critical, but I would forget reading app
notes for a bit and sit down with Kirchoffs rules for voltages and
currents in circuits with batteries and resistors. If you can't
generate John F's equation... Well....

Once you do that build the circuit.

George H.

 

[gearhead]

On 8 oct, 07:56, Andy <andrewkgent...@gmail.com> wrote:
SNIP

I found a good document from TI explaining single ended op amp design,
so I will continue from there.   I really do appreciate you effort,
but unfortunately there is some data being lost in translation here.
This is apparent by your equation defining a voltage divider.  A
divider consisting of two resistors will create a center voltage of:
Vout = (E1-Vin)*R2/(R1+R2).  Your equation shown above as Vout = (E1-
Vin)*R2/(R1+R2) + Vin is the description of another circuit.  This is
the circuit which I cannot see, but it clearly consists of more than
two resistors in series.  Anyway, if you are interested, google
"single supply op amp design Mancini" and you will find the paper from
TI.  It shows diagrams of several typical op amp circuits along with
the equations for Vout = f(Vin).- Ocultar texto de la cita -

- Mostrar texto de la cita -

Andy,
You have been abstracting this and not seeing the forest for
the trees, as the saying goes. Your best bet now is to take some
resistors, a multimeter, and a power supply that you can vary the
voltage output of.
Once you get your fingers in a breadboard with some resistors on it
and your eyes on a meter readout, it will bring you back to reality.
Build the circuit. Vary the input, test the output. Even if you
don´t have a variable power supply, you can vary the voltage somehow.
Put some 1.5 volt cells in series, or put some diodes in series with
the output of a six volt battery. Whatever. I just suggested a
couple of kluges that will give you a low impedance voltage source.
Attach your power supply´s positive output to the bottom of the
resistor string. Then measure the output voltage with respect to
ground of your power supply.
Michael

 

[Michael A. Terrell]

Andy wrote:

All,
I think these forums are an excellent source of information. But it
should also be quite apparent that communications in these forums is
often imperfect, and due to this a lot of assumptions are made. We
don't have the privilege of knowing one another with respect to
background and/or skill level, and any assumptions made about users in
the forum are almost certainly incorrect. These forums will be most
valuable if we can focus on the questions being asked, not on what we
perceive to be the skill level of the asker.
AG

This is not a 'Forum'. It is a Newsgroup, and part of Usenet. You
are accessing it through Google's crappy HTML interface.


--
The movie 'Deliverance' isn't a documentary!

 

[John Fields]

On Sat, 10 Oct 2009 11:00:28 -0700 (PDT), Andy
<andrewkgentile@gmail.com> wrote:

On Oct 9, 10:18 am, gearhead <nos...@billburg.com> wrote:
On 8 oct, 07:56, Andy <andrewkgent...@gmail.com> wrote:
SNIP

I found a good document from TI explainingsingleended op amp design,
so I will continue from there.   I really do appreciate you effort,
but unfortunately there is some data being lost in translation here.
This is apparent by your equation defining a voltage divider.  A
divider consisting of two resistors will create a center voltage of:
Vout = (E1-Vin)*R2/(R1+R2).  Your equation shown above as Vout = (E1-
Vin)*R2/(R1+R2) + Vin is the description of another circuit.  This is
the circuit which I cannot see, but it clearly consists of more than
two resistors in series.  Anyway, if you are interested, google
"singlesupplyop amp design Mancini" and you will find the paper from
TI.  It shows diagrams of several typical op amp circuits along with
the equations for Vout = f(Vin).- Ocultar texto de la cita -

- Mostrar texto de la cita -

Andy,
          You have been abstracting this and not seeing the forest for
the trees, as the saying goes.  Your best bet now is to take some
resistors, a multimeter, and a powersupplythat you can vary the
voltage output of.
Once you get your fingers in a breadboard with some resistors on it
and your eyes on a meter readout, it will bring you back to reality.
     Build the circuit.  Vary the input, test the output.  Even if you
don´t have a variable powersupply, you can vary the voltage somehow.
Put some 1.5 volt cells in series, or put some diodes in series with
the output of a six volt battery.  Whatever.  I just suggested a
couple of kluges that will give you a low impedance voltagesource.
     Attach your powersupply´s positive output to the bottom of the
resistor string.  Then measure the output voltage with respect to
ground of your powersupply.
Michael

All,
I think these forums are an excellent source of information. But it
should also be quite apparent that communications in these forums is
often imperfect, and due to this a lot of assumptions are made. We
don't have the privilege of knowing one another with respect to
background and/or skill level, and any assumptions made about users in
the forum are almost certainly incorrect. These forums will be most
valuable if we can focus on the questions being asked, not on what we
perceive to be the skill level of the asker.

---
That's not true, since the answers must be constructed in a manner the
querant can understand.

Because these are technical fora we often assume at least a basic level
of technical competence in the querant when we answer their questions.
Sometimes, however, as in your case, we find that that assumption was
unjustified and have to get more and more basic.

Unfortunately, instead of digging in and finding out why you're having
trouble or asking for help with what you specifically don't understand
about what was explained to you, you try to set yourself up as a judge
and excuse yourself by saying that the trouble isn't with you, it's
because the presenter isn't presenting the material properly, or the
presenter doesn't understand the circuit, or the math, or some other
nonsensical crap.

I, for one, went to a good deal of trouble to try to help you out and
was extremely polite in the bargain, since this is
sci.electronics.basics, (where there are no stupid questions) while you
decided to play your little "I know what I'm doing and you don't" games.

Are you still stuck, or has what we've all been telling you, so far,
sunk in?

JF

 

[Andy]

On Oct 9, 10:18 am, gearhead <nos...@billburg.com> wrote:

On 8 oct, 07:56, Andy <andrewkgent...@gmail.com> wrote:
SNIP

I found a good document from TI explainingsingleended op amp design,
so I will continue from there.   I really do appreciate you effort,
but unfortunately there is some data being lost in translation here.
This is apparent by your equation defining a voltage divider.  A
divider consisting of two resistors will create a center voltage of:
Vout = (E1-Vin)*R2/(R1+R2).  Your equation shown above as Vout = (E1-
Vin)*R2/(R1+R2) + Vin is the description of another circuit.  This is
the circuit which I cannot see, but it clearly consists of more than
two resistors in series.  Anyway, if you are interested, google
"singlesupplyop amp design Mancini" and you will find the paper from
TI.  It shows diagrams of several typical op amp circuits along with
the equations for Vout = f(Vin).- Ocultar texto de la cita -

- Mostrar texto de la cita -

Andy,
          You have been abstracting this and not seeing the forest for
the trees, as the saying goes.  Your best bet now is to take some
resistors, a multimeter, and a powersupplythat you can vary the
voltage output of.
Once you get your fingers in a breadboard with some resistors on it
and your eyes on a meter readout, it will bring you back to reality.
     Build the circuit.  Vary the input, test the output.  Even if you
don´t have a variable powersupply, you can vary the voltage somehow.
Put some 1.5 volt cells in series, or put some diodes in series with
the output of a six volt battery.  Whatever.  I just suggested a
couple of kluges that will give you a low impedance voltagesource.
     Attach your powersupply´s positive output to the bottom of the
resistor string.  Then measure the output voltage with respect to
ground of your powersupply.
Michael

All,
I think these forums are an excellent source of information. But it
should also be quite apparent that communications in these forums is
often imperfect, and due to this a lot of assumptions are made. We
don't have the privilege of knowing one another with respect to
background and/or skill level, and any assumptions made about users in
the forum are almost certainly incorrect. These forums will be most
valuable if we can focus on the questions being asked, not on what we
perceive to be the skill level of the asker.
AG

 

[Jon Slaughter]

"Andrew" <andrewkgentile@gmail.com> wrote in message
news:7beb4b0b-700a-439c-a282-3ac92aa11ecf@u16g2000pru.googlegroups.com...

Hello,
I'm working on a circuit where I want to shift/scale a 0-5V input to
1-5V. I am using single supply op amps. Can a summing circuit be
built using single supply op amps?

thanks

Yes, use a summing op amp configuration. But you inputs are the scaled
versions of the two signals using voltage dividers.


so

Vin ---R1+------R3---*
|
R2
|
GND

Vcc ---R4+------R6---*
|
R5
|
GND

(use fixed width font to view)

*'s are connected to the op amp in the summer configuration


R1 and R2 form a voltage didivder scaling down Vin. So a 4:5 ratio.

R4 and R5 scale down Vcc to 1V so a 1:5 ratio.

R3 and R6 are your "weights" for the op amp summer.

So proceede as normal.

The idea is simple. You simply add the divided signals together. 0.8*Vin +
0.2*Vcc. To do scaling you use voltage dividers(an op amp confugred as an
amplifier is an "active" voltage divider).

You must choose the resistors appropriately for this to work. Not just the
ratio's but the relative magnitudes.

 

[Andy]

On Oct 10, 12:53 pm, John Fields <jfie...@austininstruments.com>
wrote:

On Sat, 10 Oct 2009 11:00:28 -0700 (PDT), Andy



andrewkgent...@gmail.com> wrote:
On Oct 9, 10:18 am, gearhead <nos...@billburg.com> wrote:
On 8 oct, 07:56, Andy <andrewkgent...@gmail.com> wrote:
SNIP

I found a good document from TI explainingsingleended op amp design,
so I will continue from there.   I really do appreciate you effort,
but unfortunately there is some data being lost in translation here.
This is apparent by your equation defining a voltage divider.  A
divider consisting of two resistors will create a center voltage of:
Vout = (E1-Vin)*R2/(R1+R2).  Your equation shown above as Vout = (E1-
Vin)*R2/(R1+R2) + Vin is the description of another circuit.  This is
the circuit which I cannot see, but it clearly consists of more than
two resistors in series.  Anyway, if you are interested, google
"singlesupplyop amp design Mancini" and you will find the paper from
TI.  It shows diagrams of several typical op amp circuits along with
the equations for Vout = f(Vin).- Ocultar texto de la cita -

- Mostrar texto de la cita -

Andy,
          You have been abstracting this and not seeing the forest for
the trees, as the saying goes.  Your best bet now is to take some
resistors, a multimeter, and a powersupplythat you can vary the
voltage output of.
Once you get your fingers in a breadboard with some resistors on it
and your eyes on a meter readout, it will bring you back to reality.
     Build the circuit.  Vary the input, test the output.  Even if you
don´t have a variable powersupply, you can vary the voltage somehow.
Put some 1.5 volt cells in series, or put some diodes in series with
the output of a six volt battery.  Whatever.  I just suggested a
couple of kluges that will give you a low impedance voltagesource.
     Attach your powersupply´s positive output to the bottom of the
resistor string.  Then measure the output voltage with respect to
ground of your powersupply.
Michael

All,
I think these forums are an excellent source of information.  But it
should also be quite apparent that communications in these forums is
often imperfect, and due to this a lot of assumptions are made.  We
don't have the privilege of knowing one another with respect to
background and/or skill level, and any assumptions made about users in
the forum are almost certainly incorrect.  These forums will be most
valuable if we can focus on the questions being asked, not on what we
perceive to be the skill level of the asker.

---
That's not true, since the answers must be constructed in a manner the
querant can understand.

Because these are technical fora we often assume at least a basic level
of technical competence in the querant when we answer their questions.
Sometimes, however, as in your case, we find that that assumption was
unjustified and have to get more and more basic.

Unfortunately, instead of digging in and finding out why you're having
trouble or asking for help with what you specifically don't understand
about what was explained to you, you try to set yourself up as a judge
and excuse yourself by saying that the trouble isn't with you, it's
because the presenter isn't presenting the material properly, or the
presenter doesn't understand the circuit, or the math, or some other
nonsensical crap.

I, for one, went to a good deal of trouble to try to help you out and
was extremely polite in the bargain, since this is
sci.electronics.basics, (where there are no stupid questions) while you
decided to play your little "I know what I'm doing and you don't" games.

Are you still stuck, or has what we've all been telling you, so far,
sunk in?

JF

JF,
I sincerely apologize if I offended you, or anyone for that matter.
It is not in my interest to offend. However, the tone of your
response is exactly what I was referring to in my previous point.
This kind of misunderstanding is typical in these groups. And by the
way, I did make a point of personally thanking your for your efforts.
But this thread...read it. It's like we are all speaking different
languages. Please excuse me, but I cannot seem to make a statement
which comes across as I mean it to. I will not participate further.