Differential to single ended

[Rene Tschaggelar]

I should convert a differential signal to a single ended one.
This is usually done with an OpAmp in a subtracting setup.
For speed reasons I only have a current feedback amp.
That should work, shouldn't it ? With the same 4 resistors ?

I somehow lost a paper from { analog ? linear ? national ? }
named "me and the current feedback amp" or similar. And these
parts are too tiny and too fast to just give them a try on
a breadboard.

Rene

 

[Mac]

I should convert a differential signal to a single ended one.
This is usually done with an OpAmp in a subtracting setup.
For speed reasons I only have a current feedback amp.
That should work, shouldn't it ? With the same 4 resistors ?

There are some pretty fast voltage feedback amps around nowadays, too.

I somehow lost a paper from { analog ? linear ? national ? }
named "me and the current feedback amp" or similar. And these
parts are too tiny and too fast to just give them a try on
a breadboard.

Rene

I've never done what you are trying to do. But I have used current
feedback op-amps before and I have read their datasheets.

The thing to watch out for with a current feedback amp is that the
non-inverting input is NOT high impedance. Also, you don't have as much
flexibility in setting the feedback resistor. Whatever value or values the
datasheet recommends is what you should use.

So, if you can live with the impedance issue, and the feedback resistor
value is OK, it should work. Be sure to check the common-mode input range
on the op-amp, too.

You can probably breadboard it on perfboard. If the amp is sensitive to
capacitive loading, you may want to put a series resistor immediately
after the output, and before the load. Also, make the feedback path as
simple and direct as possible.

Good luck.

Mac

 

[Ben Bradley]

I should convert a differential signal to a single ended one.
This is usually done with an OpAmp in a subtracting setup.
For speed reasons I only have a current feedback amp.
That should work, shouldn't it ? With the same 4 resistors ?

I somehow lost a paper from { analog ? linear ? national ? }
named "me and the current feedback amp" or similar. And these
parts are too tiny and too fast to just give them a try on
a breadboard.

I dunno if this is what you lost (it's nothing like the title you
gave), but it does address the issue of differential inputs using CFB
amps:

http://focus.ti.com/docs/apps/catalog/resources/appnoteabstract.jhtml?abstractName=sboa081

 

[Mac]

There are some pretty fast voltage feedback amps around nowadays, too.


I've never done what you are trying to do. But I have used current
feedback op-amps before and I have read their datasheets.

The thing to watch out for with a current feedback amp is that the
non-inverting input is NOT high impedance. Also, you don't have as much
flexibility in setting the feedback resistor. Whatever value or values the
datasheet recommends is what you should use.

OOPS! I meant to say that the inverting input is NOT high impedance. The
non-inverting input typically is high-impedance.

[snip]

Mac

 

[John Popelish]

I should convert a differential signal to a single ended one.
This is usually done with an OpAmp in a subtracting setup.
For speed reasons I only have a current feedback amp.
That should work, shouldn't it ? With the same 4 resistors ?

I somehow lost a paper from { analog ? linear ? national ? }
named "me and the current feedback amp" or similar. And these
parts are too tiny and too fast to just give them a try on
a breadboard.

I have little experience with current feedback opamps, but I think the
two amp subtractor will allow a high impedance input with the resistor
network values selected to be optimum for the inverting low impedance
and amplifier speed. Then you can terminate the high impedance inputs
as you desire. See second example on page 16 of:
http://www.national.com/an/AN/AN-31.pdf

I haven't browsed through them, but there are several current feedback
opamp application notes about half way down on:
http://www.national.com/appinfo/amps/0,2175,967,00.html

 

[Rene Tschaggelar]

OOPS! I meant to say that the inverting input is NOT high impedance. The
non-inverting input typically is high-impedance.

I'll have to either do a simulation or an experiment to figure out
the useability for the intended purpose.

Rene

 

[Rene Tschaggelar]

I dunno if this is what you lost (it's nothing like the title you
gave), but it does address the issue of differential inputs using CFB
amps:

http://focus.ti.com/docs/apps/catalog/resources/appnoteabstract.jhtml?abstractName=sboa081

Thanks I'll have a look at it.
My title was fictious, meaning I lost the title
of the document.

Rene

 

[Rene Tschaggelar]

I have little experience with current feedback opamps, but I think the
two amp subtractor will allow a high impedance input with the resistor
network values selected to be optimum for the inverting low impedance
and amplifier speed. Then you can terminate the high impedance inputs
as you desire. See second example on page 16 of:
http://www.national.com/an/AN/AN-31.pdf

I haven't browsed through them, but there are several current feedback
opamp application notes about half way down on:
http://www.national.com/appinfo/amps/0,2175,967,00.html

Thanks,
The later reference was about like the one I remember.
I'll have a look at the documents.

Rene

 

[Rene Tschaggelar]

I should convert a differential signal to a single ended one.
This is usually done with an OpAmp in a subtracting setup.
For speed reasons I only have a current feedback amp.
That should work, shouldn't it ? With the same 4 resistors ?

I somehow lost a paper from { analog ? linear ? national ? }
named "me and the current feedback amp" or similar. And these
parts are too tiny and too fast to just give them a try on
a breadboard.

Thanks all for the suggestions.
Beside doing some reading, I guess I'll give it a try.
I happen to have two amplifiers both in the MSOP8 case.
One is a voltage feedback amp, the THS4275, the other is
a current feedback amp, the THS3001.
While the VFA has a rather low slewrate of 1V/ns causing
a limited full power bandwidth, the CFA is absolutely stunning,
its slewrate is supply dependent, but somewhat higher.
A bit more than 1V swing would be appreciated.
The noise voltages are 3nV/rtHz for the VFA and 1.6nV/rtHz
for the CFA.
And the CFA tends to be less prone to oscillations.

Rene

 

[John Popelish]

Thanks,
The later reference was about like the one I remember.
I'll have a look at the documents.

If you come across specific info that relates to using current
feedback opamps as subtractors, please come back and educate me.
Thanks.

 

[Bill Sloman]

I should convert a differential signal to a single ended one.
This is usually done with an OpAmp in a subtracting setup.
For speed reasons I only have a current feedback amp.
That should work, shouldn't it ? With the same 4 resistors ?

I somehow lost a paper from { analog ? linear ? national ? }
named "me and the current feedback amp" or similar. And these
parts are too tiny and too fast to just give them a try on
a breadboard.

I've done this - some 15 years ago - with Comlinear parts, and the
National Semiconductor equivalents, which used somewhat higher
resistance feedback resitor values, which saved a useful amount of
current.

Both parts are long obsolete, so I can't usefully give you part
numbers, even if I could remember them.

In the second go-around, we used a transmission line transformer (
wound with a twisted pair of 36 swg enamelled transformer wire) to
push up the common mode rejection at high frequencies. Nobody was more
surprised than I was when the damn things worked straight off the
drawing board, but I managed to keep a straight face.

I think I might have schematics at home in Nijmegen - e-mail me if you
want more detail.

 

[John Woodgate]

I read in sci.electronics.design that Bill Sloman <[email protected]>

Bill Sloman, Melbourne, Australia (but back in Nijmegen on Monday)

I wouldn't bother, Bill. The weather is b. awful here, and we keep
swapping the rain and fog back and forth with the Netherlands.

Marginally happy new year, though. (;-)

 

[Rene Tschaggelar]

Rene Tschaggelar wrote:



If you come across specific info that relates to using current
feedback opamps as subtractors, please come back and educate me.
Thanks.

I found an article amongst the material mentioned that was
convcerned about subtractors. As I understand it, the same
4 resistors do the subtract with a fine detail though.
The '+' input is highimpedance and the '-' input tries to follow
the '+' input, but the '-' input is a low impedance current input
Whatever current appears there due to different voltages or so,
is amplified and appears at the output. So in a closed loop
such as a subtractor, the input current at the '-' input is kept
at zero, statically. Again the feedback resistor makes the
speed, so it should be rather low, as suggested in the datasheet,
in the few hundred ohms range.

Actually :
AN-597, OA-07 of NationalSemi mentione something about the subject
OA-25 has a model of the CFA.

Rene

 

[Fred Bloggs]

Thanks I'll have a look at it.
My title was fictious, meaning I lost the title
of the document.

Rene

That TI app note is their typically low-quality stuff, and the Fig 6 for
the differential amplifier shows the wrong equation, and why not, this
is in keeping with the incoherence of the rest of the note, using
undefined variables and loop equations that make no sense. The "author"
must be a cooperative student on training assignment- confused individual.
The CFB differential amplifier is exactly the same as in the voltage
feedback case, the only difference is that you are constrained in the
choice of feedback resistor from OUT to IN(-) to a small range of
recommended values -selected for performance and stability purposes. I
suppose that some confusion can arise because the IN(-) is actually a
buffered output, BUT, since VOUT is a very large transimpedance gain
function of the IN(-) output current, a negative feedback configuration
using voltage-to-current converting resistors, will drive that IN(-)
current to zero, AND, since the voltage at IN(-) is a unity gain
buffered version of the voltage at IN(+), you have zero net differential
input voltage. This means that for frequencies where the transimpedance
gain can be taken as infinitely large, the circuit equations are
identical to those of the ideal opamp:no input currents, zero input
differential voltage, and zero output impedance.

 

[Mac]

I'll have to either do a simulation or an experiment to figure out
the useability for the intended purpose.

Rene

Actually, I think it will work. The input impedance thing is largely moot
because of feedback. The only catch is that the feedback resistor has to
be as specified in the datasheet. This will probably force other resistors
to particular values, depending on the gain you are looking for.

If you are looking for unity-gain, then all four resistors will be the
same as the feedback one. The input impedance for differential signals
will be 2R. The input impedance for common-mode signals will be R. If that
is acceptable, then I don't think you have any problem.

Mac


If you
want, draw the circuit, post it somewhere (you can post in here in
ascii-art) and I'll be happy to look at it.

 

[Fred Bloggs]

Fred Bloggs wrote:
[..SNIP..]

If wideband differential to single-ended conversion is required with
minimal input loading, then you have this option in CFB- trim the R's in-:
Please view in a fixed-width font such as Courier.




R R R R
+---/\/\--+------/\/\---+---/\/\--+------/\/\--+
| | | | |
--- | | | |
COM | |\ | | |\ |
+--|-\ | +--|-\ |
| >-------+ | >------+-> 2x(V2-V1)
+--|+/ +--|+/
| |/ | |/
| |
| |
V1 >--------+ |
|
|
V2 >--------------------------------+

 

[Spehro Pefhany]

Fred Bloggs wrote:
[..SNIP..]

If wideband differential to single-ended conversion is required with
minimal input loading, then you have this option in CFB- trim the R's in-:
Please view in a fixed-width font such as Courier.




R R R R
+---/\/\--+------/\/\---+---/\/\--+------/\/\--+
| | | | |
--- | | | |
COM | |\ | | |\ |
+--|-\ | +--|-\ |
| >-------+ | >------+-> 2x(V2-V1)
+--|+/ +--|+/
| |/ | |/
| |
| |
V1 >--------+ |
|
|
V2 >--------------------------------+

Why would you use that nastly asymmetrical configuration instead of
the below?

Vo = (Vinv- Vnon-inv) * 2 * Rf/Rg
V(inv) |\|
o----------|+\
| >--+ _R_
+-|-/ | +-|___|-+
| |/| | | |
| | | |
| ___ | _R_ | |
+-|___|-+-|___|----| |
| Rf | |\| |
.-. +--|-\ | Vo
| | | >-+---o
Rg | | +--|+/
'-' Rf | |/|
| ___ R__ |
+-|___|-+-|___|----+
| | |
| |\| | .-.
+-|-\ | | |
| >--+ | | R
o---------|+/ '-'
V(non-inv) |/| |
|
===
GND


Best regards,
Spehro Pefhany

 

[John Woodgate]

I read in sci.electronics.design that Spehro Pefhany <speffSNIP@interlog

Why would you use that nastly asymmetrical configuration instead of
the below?

Because the simpler circuit is quite satisfactory, AIUI.

 

[Fred Bloggs]

Fred Bloggs wrote:
[..SNIP..]

If wideband differential to single-ended conversion is required with
minimal input loading, then you have this option in CFB- trim the R's in-:
Please view in a fixed-width font such as Courier.




R R R R
+---/\/\--+------/\/\---+---/\/\--+------/\/\--+
| | | | |
--- | | | |
COM | |\ | | |\ |
+--|-\ | +--|-\ |
| >-------+ | >------+-> 2x(V2-V1)
+--|+/ +--|+/
| |/ | |/
| |
| |
V1 >--------+ |
|
|
V2 >--------------------------------+

Why would you use that nastly asymmetrical configuration instead of
the below?

Vo = (Vinv- Vnon-inv) * 2 * Rf/Rg
V(inv) |\|
o----------|+\
| >--+ _R_
+-|-/ | +-|___|-+
| |/| | | |
| | | |
| ___ | _R_ | |
+-|___|-+-|___|----| |
| Rf | |\| |
.-. +--|-\ | Vo
| | | >-+---o
Rg | | +--|+/
'-' Rf | |/|
| ___ R__ |
+-|___|-+-|___|----+
| | |
| |\| | .-.
+-|-\ | | |
| >--+ | | R
o---------|+/ '-'
V(non-inv) |/| |
|
===
GND


Best regards,
Spehro Pefhany

Actually there is a very elegant way to make gain adjust symmetrical for
that circuit-and my apologies to JP who already pointed this ckt out on
p16 of National AN-31 OpAmp circuit collection:

Please view in a fixed-width font such as Courier.


Rg
_
/|
+----------/\/\ --------+
| / |
| |
R | R R | R
+---/\/\--+------/\/\---+---/\/\--+------/\/\--+
| | | | |
--- | | | |
COM | |\ | | |\ |
+--|-\ | +--|-\ |
| >-------+ | >------+-> Gx(V2-V1)
+--|+/ +--|+/
| |/ | |/
| |
| |
V1 >--------+ |
|
|
V2 >--------------------------------+

 

[Rene Tschaggelar]

Fred Bloggs wrote:
[..SNIP..]

If wideband differential to single-ended conversion is required with
minimal input loading, then you have this option in CFB- trim the R's in-:
Please view in a fixed-width font such as Courier.




R R R R
+---/\/\--+------/\/\---+---/\/\--+------/\/\--+
| | | | |
--- | | | |
COM | |\ | | |\ |
+--|-\ | +--|-\ |
| >-------+ | >------+-> 2x(V2-V1)
+--|+/ +--|+/
| |/ | |/
| |
| |
V1 >--------+ |
|
|
V2 >--------------------------------+

Why would you use that nastly asymmetrical configuration instead of
the below?

Vo = (Vinv- Vnon-inv) * 2 * Rf/Rg
V(inv) |\|
o----------|+\
| >--+ _R_
+-|-/ | +-|___|-+
| |/| | | |
| | | |
| ___ | _R_ | |
+-|___|-+-|___|----| |
| Rf | |\| |
.-. +--|-\ | Vo
| | | >-+---o
Rg | | +--|+/
'-' Rf | |/|
| ___ R__ |
+-|___|-+-|___|----+
| | |
| |\| | .-.
+-|-\ | | |
| >--+ | | R
o---------|+/ '-'
V(non-inv) |/| |
|
===
GND

Thanks both,

I wasn't specific about the application.
The application is an output amplifier for a 300MHz clocked DDS,
with an output signal up to 150MHz or a bit less. It has
differential outputs, great for differential input mixers,
but less usefull for something else.
So the only choice to go single ended is the 4 resistor
subtractor with a damn fast amp such as the THS3001 CFA with
a BW of 420MHz(@gain=1), slewrate 6.5V/ns and 100mA output
current.
There are very few parts available and building my own is
not very rewarding either.

Rene