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

[Andrew]

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

 

[Rich Grise]

On Tue, 06 Oct 2009 11:54:42 -0700, whit3rd wrote:

On Oct 6, 11:16 am, Andrew <andrewkgent...@gmail.com> wrote:

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?

The question is unclear; a summing node with resistors can do this
without any amplifiers. To add a 'single supply' op amp would presumably
either be required

because it's a homework question.

 

[whit3rd]

On Oct 6, 11:16 am, Andrew <andrewkgent...@gmail.com> wrote:

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?

The question is unclear; a summing node with resistors can do this
without any amplifiers. To add a 'single supply' op amp would
presumably either be required because the output range (to +5V)
is at the power supply positive power rail, OR because the
input range (to GND) is at the negative power rail? Or, both?

And, do you need the op amp to keep the input signal unloaded,
or to drive the output into a load?

 

[Tim Wescott]

On Tue, 06 Oct 2009 12:28:33 -0700, Andy wrote:

On Oct 6, 12:54 pm, whit3rd <whit...@gmail.com> wrote:
On Oct 6, 11:16 am, Andrew <andrewkgent...@gmail.com> wrote:

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?

The question is unclear;  a summing node with resistors can do this
without any amplifiers.  To add a 'single supply' op amp would
presumably either be required because the output range (to +5V) is at
the power supply positive power rail, OR because the input range (to
GND) is at the negative power rail?  Or, both?

And, do you need the op amp to keep the input signal unloaded, or to
drive the output into a load?

Let me clarify... I have a 0 to 5V source and I want to convert this to
a 1 to 5V source. The linear mapping of this would be: Vout = 0.8*Vin
+1. Given typical dual supply op amps, this can be done with a summer
and an inverter. However, because a summing op amp has an inverting
gain, the single supply op amp wouldn't be able to do this. So, I was
wondering if there was a way of making that conversion, as stated
earlier, using single supply op amps.

thanks

You can do this with a pair of resistors and no op amps, as long as you
don't care about loading the supply and having an output with a non-zero
source impedance.

--
www.wescottdesign.com

 

[Andy]

On Oct 6, 12:54 pm, whit3rd <whit...@gmail.com> wrote:

On Oct 6, 11:16 am, Andrew <andrewkgent...@gmail.com> wrote:

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?

The question is unclear;  a summing node with resistors can do this
without any amplifiers.  To add a 'single supply' op amp would
presumably either be required because the output range (to +5V)
is at the power supply positive power rail, OR because the
input range (to GND) is at the negative power rail?  Or, both?

And, do you need the op amp to keep the input signal unloaded,
or to drive the output into a load?

Let me clarify... I have a 0 to 5V source and I want to convert this
to a 1 to 5V source. The linear mapping of this would be:
Vout = 0.8*Vin +1. Given typical dual supply op amps, this can be
done with a summer and an inverter. However, because a summing op amp
has an inverting gain, the single supply op amp wouldn't be able to do
this. So, I was wondering if there was a way of making that
conversion, as stated earlier, using single supply op amps.

thanks

 

[George Herold]

On Oct 6, 3:28 pm, Andy <andrewkgent...@gmail.com> wrote:

On Oct 6, 12:54 pm, whit3rd <whit...@gmail.com> wrote:

On Oct 6, 11:16 am, Andrew <andrewkgent...@gmail.com> wrote:

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?

The question is unclear;  a summing node with resistors can do this
without any amplifiers.  To add a 'single supply' op amp would
presumably either be required because the output range (to +5V)
is at the power supply positive power rail, OR because the
input range (to GND) is at the negative power rail?  Or, both?

And, do you need the op amp to keep the input signal unloaded,
or to drive the output into a load?

Let me clarify... I have a 0 to 5V source and I want to convert this
to a 1 to 5V source.  The linear mapping of this would be:
Vout = 0.8*Vin +1.  Given typical dual supply op amps, this can be
done with a summer and an inverter.  However, because a summing op amp
has an inverting gain, the single supply op amp wouldn't be able to do
this.  So, I was wondering if there was a way of making that
conversion, as stated earlier, using single supply op amps.

thanks

You can 'sorta' do this if you don't mind a bit of error. This brute
force approach would be to hang a small resistor to ground from the
non-inverting input. (Let's use 1 k ohm.) Then 'sum' through two
bigger resistors. From a one volt supply put 'say' 10k to the
noninverting input and from the varibale voltage put 8k ohm to the non-
inverting input. Then have a gain of ten in the opamp. The result is
'sorta' what you want. You can make it more 'accurate' by throwing
away more signal and adding more gain... but it looks like a losing
proposition to me.

If you allow for more opamps you might be able to make the resistor
current sources look more ideal. Does a Howland current source
(Gack!) work from a single supply?

George H.

 

[whit3rd]

On Oct 6, 12:28 pm, Andy <andrewkgent...@gmail.com> wrote:

On Oct 6, 12:54 pm, whit3rd <whit...@gmail.com> wrote:

The question is unclear; ...
And, do you need the op amp to keep the input signal unloaded,
or to drive the output into a load?

Let me clarify... I have a 0 to 5V source and I want to convert this
to a 1 to 5V source.

So, a 1k resistor from input to output, and a 4k pullup resistor from
output
to +5V, does it.
Add a voltage-follower at the input end, or a voltage-follower at the
output end, or both, if they're needed. It isn't clear whether they
ARE needed.

 

[George Herold]

On Oct 6, 5:17 pm, whit3rd <whit...@gmail.com> wrote:

On Oct 6, 12:28 pm, Andy <andrewkgent...@gmail.com> wrote:

On Oct 6, 12:54 pm, whit3rd <whit...@gmail.com> wrote:
The question is unclear; ...
And, do you need the op amp to keep the input signal unloaded,
or to drive the output into a load?

Let me clarify... I have a 0 to 5V source and I want to convert this
to a 1 to 5V source.

So, a 1k resistor from input to output, and a 4k pullup resistor from
output
to +5V, does it.
Add a voltage-follower at the input end, or a voltage-follower at the
output end, or both, if they're needed.  It isn't clear whether they
ARE needed.

Excellent, Thanks

George H.

 

[Andy]

On Oct 6, 3:17 pm, whit3rd <whit...@gmail.com> wrote:

On Oct 6, 12:28 pm, Andy <andrewkgent...@gmail.com> wrote:

On Oct 6, 12:54 pm, whit3rd <whit...@gmail.com> wrote:
The question is unclear; ...
And, do you need the op amp to keep the input signal unloaded,
or to drive the output into a load?

Let me clarify... I have a 0 to 5V source and I want to convert this
to a 1 to 5V source.

So, a 1k resistor from input to output, and a 4k pullup resistor from
output
to +5V, does it.
Add a voltage-follower at the input end, or a voltage-follower at the
output end, or both, if they're needed.  It isn't clear whether they
ARE needed.

I'm trying to draw this out. Please clarify. I assume we are talking
about an op amp.

"1k resistor from input to output" --- is this the inverting or non-
inverting input, the feedback loop?

" and a 4k pullup resistor from output to +5V, does it"

So the circuit input goes to the non-inverting input?

thanks

 

[John Fields]

On Wed, 7 Oct 2009 06:37:38 -0700 (PDT), Andy <andrewkgentile@gmail.com>
wrote:

On Oct 6, 3:17 pm, whit3rd <whit...@gmail.com> wrote:
On Oct 6, 12:28 pm, Andy <andrewkgent...@gmail.com> wrote:

On Oct 6, 12:54 pm, whit3rd <whit...@gmail.com> wrote:
The question is unclear; ...
And, do you need the op amp to keep the input signal unloaded,
or to drive the output into a load?

Let me clarify... I have a 0 to 5V source and I want to convert this
to a 1 to 5V source.

So, a 1k resistor from input to output, and a 4k pullup resistor from
output
to +5V, does it.
Add a voltage-follower at the input end, or a voltage-follower at the
output end, or both, if they're needed.  It isn't clear whether they
ARE needed.

I'm trying to draw this out. Please clarify. I assume we are talking
about an op amp.

"1k resistor from input to output" --- is this the inverting or non-
inverting input, the feedback loop?

" and a 4k pullup resistor from output to +5V, does it"

So the circuit input goes to the non-inverting input?

---
What he meant was: (View in Courier

.. +5V
.. |
.. [4k]R1
.. |
.. +---Vout
.. |
.. [1k]R2
.. |
.. Vin

R1 doesn't have to be 4K and R2 doesn't have to be 1k, but their ratio
has to be 4:1.

If you needed a lower output impedance you could also:

.. +5V
.. |
.. [4R1]
.. |
.. +-------|+\
.. | | >--+--->Vout
.. [R1] +--|-/ |
.. | | |
.. Vin +--------+


or, if you needed a higher input impedance:

.. +5V
.. |
.. [4R1]
.. |
.. +-->Vout
.. |
.. [R1]
..Vin---------|+\ |
.. | >--+
.. +--|-/ |
.. | |
.. +--------+

or, if you needed both:

.. +5V
.. |
.. [4R1]
.. |
.. +-------|+\
.. | | >--+--->Vout
.. [R1] +--|-/ |
..Vin---------|+\ | | |
.. | >--+ +--------+
.. +--|-/ |
.. | |
.. +--------+


JF

 

[Andy]

On Oct 7, 10:53 am, John Fields <jfie...@austininstruments.com> wrote:

On Wed, 7 Oct 2009 06:37:38 -0700 (PDT), Andy <andrewkgent...@gmail.com
wrote:



On Oct 6, 3:17 pm, whit3rd <whit...@gmail.com> wrote:
On Oct 6, 12:28 pm, Andy <andrewkgent...@gmail.com> wrote:

On Oct 6, 12:54 pm, whit3rd <whit...@gmail.com> wrote:
The question is unclear; ...
And, do you need the op amp to keep the input signal unloaded,
or to drive the output into a load?

Let me clarify... I have a 0 to 5V source and I want to convert this
to a 1 to 5V source.

So, a 1k resistor from input to output, and a 4k pullup resistor from
output
to +5V, does it.
Add a voltage-follower at the input end, or a voltage-follower at the
output end, or both, if they're needed.  It isn't clear whether they
ARE needed.

I'm trying to draw this out.  Please clarify.  I assume we are talking
about an op amp.

"1k resistor from input to output" --- is this the inverting or non-
inverting input, the feedback loop?

" and a 4k pullup resistor from output to +5V, does it"

So the circuit input goes to the non-inverting input?

---
What he meant was: (View in Courier

.     +5V
.      |
.    [4k]R1
.      |
.      +---Vout
.      |
.    [1k]R2
.      |
.     Vin

R1 doesn't have to be 4K and R2 doesn't have to be 1k, but their ratio
has to be 4:1.

If you needed a lower output impedance you could also:

.     +5V
.      |
.    [4R1]
.      |
.      +-------|+\  
.      |       |  >--+--->Vout
.     [R1]  +--|-/   |
.      |    |        |
.     Vin   +--------+

or, if you needed a higher input impedance:

.                 +5V
.                  |
.                [4R1]
.                  |
.                  +-->Vout      
.                  |                      
.                 [R1]                  
.Vin---------|+\   |                  
.            |  >--+                
.         +--|-/   |
.         |        |
.         +--------+

or, if you needed both:

.                 +5V
.                  |
.                [4R1]
.                  |
.                  +-------|+\  
.                  |       |  >--+--->Vout
.                 [R1]  +--|-/   |
.Vin---------|+\   |    |        |
.            |  >--+    +--------+
.         +--|-/   |
.         |        |
.         +--------+

JF

I get it now. Nice and simple. This will work well.

 

[John Fields]

On Wed, 7 Oct 2009 12:34:40 -0700 (PDT), Andy <andrewkgentile@gmail.com>
wrote:

On Oct 7, 10:53 am, John Fields <jfie...@austininstruments.com> wrote:

.                 +5V
.                  |
.                [4R1]
.                  |
.                  +-------|+\  
.                  |       |  >--+--->Vout
.                 [R1]  +--|-/   |
.Vin---------|+\   |    |        |
.            |  >--+    +--------+
.         +--|-/   |
.         |        |
.         +--------+

JF

Given that resistor combination, Vout = (5-Vin)/5

Vin Vx
0 1
1 0.8
2 0.6
3 0.4
4 0.2
5 0

This doesn't work.

---
E1
.                 +5V Non-inverting input
.                  | input
.                [4R1] /
. Non-inverting    | E2 /
. input       +-------|+\  
. \        |       |  >--+--->Vout
.         \      [R1]  +--|-/   |
.Vin---------|+\   |    |        |
.            |  >--+    +--------+
.         +--|-/   |
.         |        |
.         +--------+


Version 4
SHEET 1 880 680
WIRE 144 16 -208 16
WIRE 288 16 144 16
WIRE 144 64 144 16
WIRE 368 96 224 96
WIRE 288 128 288 16
WIRE 224 144 224 96
WIRE 256 144 224 144
WIRE 368 160 368 96
WIRE 368 160 320 160
WIRE 144 176 144 144
WIRE 256 176 144 176
WIRE 144 208 144 176
WIRE -208 288 -208 16
WIRE 64 288 -208 288
WIRE 144 320 144 288
WIRE 144 320 0 320
WIRE 64 352 64 288
WIRE 0 368 0 320
WIRE 32 368 0 368
WIRE 144 384 144 320
WIRE 144 384 96 384
WIRE 32 400 -80 400
WIRE -208 432 -208 288
WIRE -80 432 -80 400
WIRE -208 544 -208 512
WIRE -80 544 -80 512
WIRE -80 544 -208 544
WIRE 64 544 64 416
WIRE 64 544 -80 544
WIRE 288 544 288 192
WIRE 288 544 64 544
WIRE -208 608 -208 544
FLAG -208 608 0
SYMBOL Opamps\\LT1218 288 96 R0
SYMATTR InstName U1
SYMBOL res 128 48 R0
SYMATTR InstName R1
SYMATTR Value 4k
SYMBOL res 128 192 R0
SYMATTR InstName R2
SYMATTR Value 1k
SYMBOL Opamps\\LT1218 64 320 R0
SYMATTR InstName U2
SYMBOL voltage -80 416 R0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
SYMATTR InstName V1
SYMATTR Value PULSE(0 5 0 1)
SYMBOL voltage -208 416 R0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
SYMATTR InstName V2
SYMATTR Value 5
TEXT -192 576 Left 0 !.tran 2 uic

 

[Andy]

On Oct 7, 10:53 am, John Fields <jfie...@austininstruments.com> wrote:

On Wed, 7 Oct 2009 06:37:38 -0700 (PDT), Andy <andrewkgent...@gmail.com
wrote:



On Oct 6, 3:17 pm, whit3rd <whit...@gmail.com> wrote:
On Oct 6, 12:28 pm, Andy <andrewkgent...@gmail.com> wrote:

On Oct 6, 12:54 pm, whit3rd <whit...@gmail.com> wrote:
The question is unclear; ...
And, do you need the op amp to keep the input signal unloaded,
or to drive the output into a load?

Let me clarify... I have a 0 to 5V source and I want to convert this
to a 1 to 5V source.

So, a 1k resistor from input to output, and a 4k pullup resistor from
output
to +5V, does it.
Add a voltage-follower at the input end, or a voltage-follower at the
output end, or both, if they're needed.  It isn't clear whether they
ARE needed.

I'm trying to draw this out.  Please clarify.  I assume we are talking
about an op amp.

"1k resistor from input to output" --- is this the inverting or non-
inverting input, the feedback loop?

" and a 4k pullup resistor from output to +5V, does it"

So the circuit input goes to the non-inverting input?

---
What he meant was: (View in Courier

.     +5V
.      |
.    [4k]R1
.      |
.      +---Vout
.      |
.    [1k]R2
.      |
.     Vin

R1 doesn't have to be 4K and R2 doesn't have to be 1k, but their ratio
has to be 4:1.

If you needed a lower output impedance you could also:

.     +5V
.      |
.    [4R1]
.      |
.      +-------|+\  
.      |       |  >--+--->Vout
.     [R1]  +--|-/   |
.      |    |        |
.     Vin   +--------+

or, if you needed a higher input impedance:

.                 +5V
.                  |
.                [4R1]
.                  |
.                  +-->Vout      
.                  |                      
.                 [R1]                  
.Vin---------|+\   |                  
.            |  >--+                
.         +--|-/   |
.         |        |
.         +--------+

or, if you needed both:

.                 +5V
.                  |
.                [4R1]
.                  |
.                  +-------|+\  
.                  |       |  >--+--->Vout
.                 [R1]  +--|-/   |
.Vin---------|+\   |    |        |
.            |  >--+    +--------+
.         +--|-/   |
.         |        |
.         +--------+

JF

Given that resistor combination, Vout = (5-Vin)/5

Vin Vx
0 1
1 0.8
2 0.6
3 0.4
4 0.2
5 0

This doesn't work.

 

[Jamie]

Andy wrote:

On Oct 7, 3:32 pm, John Fields <jfie...@austininstruments.com> wrote:

On Wed, 7 Oct 2009 12:34:40 -0700 (PDT), Andy <andrewkgent...@gmail.com
wrote:






On Oct 7, 10:53 am, John Fields <jfie...@austininstruments.com> wrote:

. +5V
. |
. [4R1]
. |
. +-------|+\
. | | >--+--->Vout
. [R1] +--|-/ |
.Vin---------|+\ | | |
. | >--+ +--------+
. +--|-/ |
. | |
. +--------+

JF

Given that resistor combination, Vout = (5-Vin)/5

Vin Vx
0 1
1 0.8
2 0.6
3 0.4
4 0.2
5 0

This doesn't work.

---
E1
. +5V Non-inverting input
. | input
. [4R1] /
. Non-inverting | E2 /
. input +-------|+\
. \ | | >--+--->Vout
. \ [R1] +--|-/ |
.Vin---------|+\ | | |
. | >--+ +--------+
. +--|-/ |
. | |
. +--------+

Version 4
SHEET 1 880 680
WIRE 144 16 -208 16
WIRE 288 16 144 16
WIRE 144 64 144 16
WIRE 368 96 224 96
WIRE 288 128 288 16
WIRE 224 144 224 96
WIRE 256 144 224 144
WIRE 368 160 368 96
WIRE 368 160 320 160
WIRE 144 176 144 144
WIRE 256 176 144 176
WIRE 144 208 144 176
WIRE -208 288 -208 16
WIRE 64 288 -208 288
WIRE 144 320 144 288
WIRE 144 320 0 320
WIRE 64 352 64 288
WIRE 0 368 0 320
WIRE 32 368 0 368
WIRE 144 384 144 320
WIRE 144 384 96 384
WIRE 32 400 -80 400
WIRE -208 432 -208 288
WIRE -80 432 -80 400
WIRE -208 544 -208 512
WIRE -80 544 -80 512
WIRE -80 544 -208 544
WIRE 64 544 64 416
WIRE 64 544 -80 544
WIRE 288 544 288 192
WIRE 288 544 64 544
WIRE -208 608 -208 544
FLAG -208 608 0
SYMBOL Opamps\\LT1218 288 96 R0
SYMATTR InstName U1
SYMBOL res 128 48 R0
SYMATTR InstName R1
SYMATTR Value 4k
SYMBOL res 128 192 R0
SYMATTR InstName R2
SYMATTR Value 1k
SYMBOL Opamps\\LT1218 64 320 R0
SYMATTR InstName U2
SYMBOL voltage -80 416 R0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
SYMATTR InstName V1
SYMATTR Value PULSE(0 5 0 1)
SYMBOL voltage -208 416 R0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
SYMATTR InstName V2
SYMATTR Value 5
TEXT -192 576 Left 0 !.tran 2 uic- Hide quoted text -

- Show quoted text -


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. I cannot make heads or tails out of what you sent. But
thank you for your effort.

You need LTSPICE .. It's free..

you import that and it'll show you and simulate for you.

 

[Andy]

On Oct 7, 3:32 pm, John Fields <jfie...@austininstruments.com> wrote:

On Wed, 7 Oct 2009 12:34:40 -0700 (PDT), Andy <andrewkgent...@gmail.com
wrote:





On Oct 7, 10:53 am, John Fields <jfie...@austininstruments.com> wrote:
.                 +5V
.                  |
.                [4R1]
.                  |
.                  +-------|+\  
.                  |       |  >--+--->Vout
.                 [R1]  +--|-/   |
.Vin---------|+\   |    |        |
.            |  >--+    +--------+
.         +--|-/   |
.         |        |
.         +--------+

JF

Given that resistor combination, Vout = (5-Vin)/5

Vin    Vx
0       1
1       0.8
2      0.6
3     0.4
4      0.2
5       0

This doesn't work.

---
                    E1
 .                 +5V       Non-inverting input
 .                  |        input
 .                [4R1]     /
 . Non-inverting    |   E2 /
 .      input       +-------|+\  
 .         \        |       |  >--+--->Vout
 .          \      [R1]  +--|-/   |
 .Vin---------|+\   |    |        |
 .            |  >--+    +--------+
 .         +--|-/   |
 .         |        |
 .         +--------+

Version 4
SHEET 1 880 680
WIRE 144 16 -208 16
WIRE 288 16 144 16
WIRE 144 64 144 16
WIRE 368 96 224 96
WIRE 288 128 288 16
WIRE 224 144 224 96
WIRE 256 144 224 144
WIRE 368 160 368 96
WIRE 368 160 320 160
WIRE 144 176 144 144
WIRE 256 176 144 176
WIRE 144 208 144 176
WIRE -208 288 -208 16
WIRE 64 288 -208 288
WIRE 144 320 144 288
WIRE 144 320 0 320
WIRE 64 352 64 288
WIRE 0 368 0 320
WIRE 32 368 0 368
WIRE 144 384 144 320
WIRE 144 384 96 384
WIRE 32 400 -80 400
WIRE -208 432 -208 288
WIRE -80 432 -80 400
WIRE -208 544 -208 512
WIRE -80 544 -80 512
WIRE -80 544 -208 544
WIRE 64 544 64 416
WIRE 64 544 -80 544
WIRE 288 544 288 192
WIRE 288 544 64 544
WIRE -208 608 -208 544
FLAG -208 608 0
SYMBOL Opamps\\LT1218 288 96 R0
SYMATTR InstName U1
SYMBOL res 128 48 R0
SYMATTR InstName R1
SYMATTR Value 4k
SYMBOL res 128 192 R0
SYMATTR InstName R2
SYMATTR Value 1k
SYMBOL Opamps\\LT1218 64 320 R0
SYMATTR InstName U2
SYMBOL voltage -80 416 R0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
SYMATTR InstName V1
SYMATTR Value PULSE(0 5 0 1)
SYMBOL voltage -208 416 R0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
SYMATTR InstName V2
SYMATTR Value 5
TEXT -192 576 Left 0 !.tran 2 uic- Hide quoted text -

- Show quoted text -

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. I cannot make heads or tails out of what you sent. But
thank you for your effort.

 

[Dan Coby]

Andy wrote:

On Oct 7, 10:53 am, John Fields <jfie...@austininstruments.com> wrote:
On Wed, 7 Oct 2009 06:37:38 -0700 (PDT), Andy <andrewkgent...@gmail.com
wrote:



On Oct 6, 3:17 pm, whit3rd <whit...@gmail.com> wrote:
On Oct 6, 12:28 pm, Andy <andrewkgent...@gmail.com> wrote:
On Oct 6, 12:54 pm, whit3rd <whit...@gmail.com> wrote:
The question is unclear; ...
And, do you need the op amp to keep the input signal unloaded,
or to drive the output into a load?
Let me clarify... I have a 0 to 5V source and I want to convert this
to a 1 to 5V source.
So, a 1k resistor from input to output, and a 4k pullup resistor from
output
to +5V, does it.
Add a voltage-follower at the input end, or a voltage-follower at the
output end, or both, if they're needed. It isn't clear whether they
ARE needed.
I'm trying to draw this out. Please clarify. I assume we are talking
about an op amp.
"1k resistor from input to output" --- is this the inverting or non-
inverting input, the feedback loop?
" and a 4k pullup resistor from output to +5V, does it"
So the circuit input goes to the non-inverting input?
---
What he meant was: (View in Courier

. +5V
. |
. [4k]R1
. |
. +---Vout
. |
. [1k]R2
. |
. Vin

R1 doesn't have to be 4K and R2 doesn't have to be 1k, but their ratio
has to be 4:1.

snip

JF

Given that resistor combination, Vout = (5-Vin)/5

Vin Vx
0 1
1 0.8
2 0.6
3 0.4
4 0.2
5 0

This doesn't work.

You might want to check your equation.

 

[John Fields]

On Wed, 7 Oct 2009 15:06:18 -0700 (PDT), Andy <andrewkgentile@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

 

[Andy]

On Oct 7, 5:11 pm, Jamie
<jamie_ka1lpa_not_valid_after_ka1l...@charter.net> wrote:

Andy wrote:
On Oct 7, 3:32 pm, John Fields <jfie...@austininstruments.com> wrote:

On Wed, 7 Oct 2009 12:34:40 -0700 (PDT), Andy <andrewkgent...@gmail.com
wrote:

On Oct 7, 10:53 am, John Fields <jfie...@austininstruments.com> wrote:

.                 +5V
.                  |
.                [4R1]
.                  |
.                  +-------|+\  
.                  |       |  >--+--->Vout
.                 [R1]  +--|-/   |
.Vin---------|+\   |    |        |
.            |  >--+    +--------+
.         +--|-/   |
.         |        |
.         +--------+

JF

Given that resistor combination, Vout = (5-Vin)/5

Vin    Vx
0       1
1       0.8
2      0.6
3     0.4
4      0.2
5       0

This doesn't work.

---
                   E1
.                 +5V       Non-inverting input
.                  |        input
.                [4R1]     /
. Non-inverting    |   E2 /
.      input       +-------|+\  
.         \        |       |  >--+--->Vout
.          \      [R1]  +--|-/   |
.Vin---------|+\   |    |        |
.            |  >--+    +--------+
.         +--|-/   |
.         |        |
.         +--------+

Version 4
SHEET 1 880 680
WIRE 144 16 -208 16
WIRE 288 16 144 16
WIRE 144 64 144 16
WIRE 368 96 224 96
WIRE 288 128 288 16
WIRE 224 144 224 96
WIRE 256 144 224 144
WIRE 368 160 368 96
WIRE 368 160 320 160
WIRE 144 176 144 144
WIRE 256 176 144 176
WIRE 144 208 144 176
WIRE -208 288 -208 16
WIRE 64 288 -208 288
WIRE 144 320 144 288
WIRE 144 320 0 320
WIRE 64 352 64 288
WIRE 0 368 0 320
WIRE 32 368 0 368
WIRE 144 384 144 320
WIRE 144 384 96 384
WIRE 32 400 -80 400
WIRE -208 432 -208 288
WIRE -80 432 -80 400
WIRE -208 544 -208 512
WIRE -80 544 -80 512
WIRE -80 544 -208 544
WIRE 64 544 64 416
WIRE 64 544 -80 544
WIRE 288 544 288 192
WIRE 288 544 64 544
WIRE -208 608 -208 544
FLAG -208 608 0
SYMBOL Opamps\\LT1218 288 96 R0
SYMATTR InstName U1
SYMBOL res 128 48 R0
SYMATTR InstName R1
SYMATTR Value 4k
SYMBOL res 128 192 R0
SYMATTR InstName R2
SYMATTR Value 1k
SYMBOL Opamps\\LT1218 64 320 R0
SYMATTR InstName U2
SYMBOL voltage -80 416 R0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
SYMATTR InstName V1
SYMATTR Value PULSE(0 5 0 1)
SYMBOL voltage -208 416 R0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
SYMATTR InstName V2
SYMATTR Value 5
TEXT -192 576 Left 0 !.tran 2 uic- Hide quoted text -

- Show quoted text -

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.  I cannot make heads or tails out of what you sent.  But
thank you for your effort.

You need LTSPICE .. It's free..
   you import that and it'll show you and simulate for you.

I've got PSPICE. But I can't imagine such a simple circuit would
require simulation. It's probably two or three lines of algebra. But
I'll get LTSPICE if it will get me to an answer.

 

[Andy]

On Oct 7, 7:58 pm, Andy <andrewkgent...@gmail.com> wrote:

On Oct 7, 5:11 pm, Jamie



jamie_ka1lpa_not_valid_after_ka1l...@charter.net> wrote:
Andy wrote:
On Oct 7, 3:32 pm, John Fields <jfie...@austininstruments.com> wrote:

On Wed, 7 Oct 2009 12:34:40 -0700 (PDT), Andy <andrewkgent...@gmail.com
wrote:

On Oct 7, 10:53 am, John Fields <jfie...@austininstruments.com> wrote:

.                 +5V
.                  |
.                [4R1]
.                  |
.                  +-------|+\  
.                  |       |  >--+--->Vout
.                 [R1]  +--|-/   |
.Vin---------|+\   |    |        |
.            |  >--+    +--------+
.         +--|-/   |
.         |        |
.         +--------+

JF

Given that resistor combination, Vout = (5-Vin)/5

Vin    Vx
0       1
1       0.8
2      0.6
3     0.4
4      0.2
5       0

This doesn't work.

---
                   E1
.                 +5V       Non-inverting input
.                  |        input
.                [4R1]     /
. Non-inverting    |   E2 /
.      input       +-------|+\  
.         \        |       |  >--+--->Vout
.          \      [R1]  +--|-/   |
.Vin---------|+\   |    |        |
.            |  >--+    +--------+
.         +--|-/   |
.         |        |
.         +--------+

Version 4
SHEET 1 880 680
WIRE 144 16 -208 16
WIRE 288 16 144 16
WIRE 144 64 144 16
WIRE 368 96 224 96
WIRE 288 128 288 16
WIRE 224 144 224 96
WIRE 256 144 224 144
WIRE 368 160 368 96
WIRE 368 160 320 160
WIRE 144 176 144 144
WIRE 256 176 144 176
WIRE 144 208 144 176
WIRE -208 288 -208 16
WIRE 64 288 -208 288
WIRE 144 320 144 288
WIRE 144 320 0 320
WIRE 64 352 64 288
WIRE 0 368 0 320
WIRE 32 368 0 368
WIRE 144 384 144 320
WIRE 144 384 96 384
WIRE 32 400 -80 400
WIRE -208 432 -208 288
WIRE -80 432 -80 400
WIRE -208 544 -208 512
WIRE -80 544 -80 512
WIRE -80 544 -208 544
WIRE 64 544 64 416
WIRE 64 544 -80 544
WIRE 288 544 288 192
WIRE 288 544 64 544
WIRE -208 608 -208 544
FLAG -208 608 0
SYMBOL Opamps\\LT1218 288 96 R0
SYMATTR InstName U1
SYMBOL res 128 48 R0
SYMATTR InstName R1
SYMATTR Value 4k
SYMBOL res 128 192 R0
SYMATTR InstName R2
SYMATTR Value 1k
SYMBOL Opamps\\LT1218 64 320 R0
SYMATTR InstName U2
SYMBOL voltage -80 416 R0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
SYMATTR InstName V1
SYMATTR Value PULSE(0 5 0 1)
SYMBOL voltage -208 416 R0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
SYMATTR InstName V2
SYMATTR Value 5
TEXT -192 576 Left 0 !.tran 2 uic- Hide quoted text -

- Show quoted text -

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.  I cannot make heads or tails out of what you sent.  But
thank you for your effort.

You need LTSPICE .. It's free..
   you import that and it'll show you and simulate for you.

I've got PSPICE.  But I can't imagine such a simple circuit would
require simulation.  It's probably two or three lines of algebra.  But
I'll get LTSPICE if it will get me to an answer.

I've got PSPICE. But I can't imagine such a simple circuit would
require simulation. It's probably two or three lines of algebra. But
I'll get LTSPICE if it will get me to an answer.

 

[Andy]

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).