Single Op-Amp Current Amplifier

[Andy]

I've been revisiting my op-amp theory, and note that there are cicuits for
voltage-voltage amplifiers (non-inverting ), current-voltage amplifiers (
see-saw or inverting amplifier ), but that it's a little more obscure with
voltage-current and current-current amplifiers.

I've managed to derive the last two configurations but the load is floating
in both cases. Of course I understand that, for instance, a voltage follower
provides current gain,
but it is not what I would call a voltage-current amplifier, the reason
being that if you were to change the value of the output load, the output
current would change too ( as the o/p voltage remains constant ).

A real voltage-current amplifier would give a gain of a constant value of
Amps/Volt, even if the output load varied in resistance. Same with a
current-current amplifier, it would have a constant gain of whatever
Amps/Amp even if the load varied i.e it would apparently have a current
source output.

So, has anyone come across either of these last two mythical beasts, with,
like I say, a non-floating load ( i.e. one end referenced to ground ).

thanks,

Andy.

 

[John Popelish]

I've been revisiting my op-amp theory, and note that there are cicuits for
voltage-voltage amplifiers (non-inverting ), current-voltage amplifiers (
see-saw or inverting amplifier ), but that it's a little more obscure with
voltage-current and current-current amplifiers.

I've managed to derive the last two configurations but the load is floating
in both cases. Of course I understand that, for instance, a voltage follower
provides current gain,
but it is not what I would call a voltage-current amplifier, the reason
being that if you were to change the value of the output load, the output
current would change too ( as the o/p voltage remains constant ).

A real voltage-current amplifier would give a gain of a constant value of
Amps/Volt, even if the output load varied in resistance. Same with a
current-current amplifier, it would have a constant gain of whatever
Amps/Amp even if the load varied i.e it would apparently have a current
source output.

So, has anyone come across either of these last two mythical beasts, with,
like I say, a non-floating load ( i.e. one end referenced to ground ).

thanks,

Andy.

The ideal current input amplifier has zero ohms input resistance, so
it can pass whatever input current the source offers. The virtual
ground of an inverting amplifier is a pretty fair approximation of
such a beast.

A current output amplifier, ideally has infinite output resistance,
and has feedback based on the current through the load, not the
voltage across it. This feedback is often taken from the voltage drop
across a low value resistor in series with the load current.

If you want a current to current amplifier, you may need to use two
opamps, one to sum the input and feedback currents, and one to amplify
the small drop across the current sense resistor and produce a current
for the input current summing junction.

 

[Andy]

The ideal current input amplifier has zero ohms input resistance, so it
can pass whatever input current the source offers. The virtual ground of
an inverting amplifier is a pretty fair approximation of such a beast.

A current output amplifier, ideally has infinite output resistance, and
has feedback based on the current through the load, not the voltage across
it. This feedback is often taken from the voltage drop across a low value
resistor in series with the load current.

If you want a current to current amplifier, you may need to use two
opamps, one to sum the input and feedback currents, and one to amplify the
small drop across the current sense resistor and produce a current for the
input current summing junction.

Yes, that idea ocurred to me though I haven't drawn it out. I just found it
a little improbable that these circuits hadn't been invented years ago
by some bright spark, and thought that they weren't shown in my text
books because they were little used. Like I say, you can make
current input and current output opamp circuits, but my intellect has
only succeeded by allowing both ends of the load resistor to be
floating, which must severely limit the utility of these circuits.

Perhaps it is something to do with the voltage amplifier nature of
op-amps, after all with high input impedance and low input impedance
they're the opposite of what you want in a current amplifier.

thanks,

Andy.

 

[Ol' Duffer]

So, has anyone come across either of these last two mythical beasts, with,
like I say, a non-floating load ( i.e. one end referenced to ground ).

Have you heard of the Howland Current Pump? It is a 5-resistor
configuration that drives current into a grounded load. The
input is a floating voltage, so one side of it could be grounded
if necessary.

Vin1---RRRR--+---RRRR--+---Iout
| |
| |\ R
| | \ R
+---|+ \ |
| >-+
+---|- / |
| | / |
| |/ |
| |
Vin2---RRRR--+---RRRR--+

From memory - I hope I got it right.

The equations tend to be messy since a portion of load current
also flows in the feedback, but you can make the output sense
resistor significantly lower in value than the feedback resistors
and pre-calculate out the error.

Also, its input would ideally connect to a zero impedance source,
but if the source happened to be a sense resistor of significantly
lower value than the feedback resistors, it seems that it also
could be pre-calculated out.

 

[Ken Smith]

[...]
Labels added:


R1 Ra

Vin1---RRRR--+---RRRR--+---Iout
| |
| |\ R Rb
| | \ R
+---|+ \ |
| >-+
+---|- / |
| | / |
| |/ |
| |
Vin2---RRRR--+---RRRR--+
R2 R3

The equations tend to be messy since a portion of load current
also flows in the feedback,

If you think of Ra and Rb as one resistor (r4) with a tab at the Iout, the
math is a lot less messy.

For infinite impedance:

R1 = R2
R3 = R4 = K*R1

Iout = Vin * K / Ra

 

[Andy]

[...]
Labels added:


R1 Ra

Vin1---RRRR--+---RRRR--+---Iout
| |
| |\ R Rb
| | \ R
+---|+ \ |
| >-+
+---|- / |
| | / |
| |/ |
| |
Vin2---RRRR--+---RRRR--+
R2 R3

The equations tend to be messy since a portion of load current
also flows in the feedback,

If you think of Ra and Rb as one resistor (r4) with a tab at the Iout, the
math is a lot less messy.

For infinite impedance:

R1 = R2
R3 = R4 = K*R1

Iout = Vin * K / Ra

Thanks for that, yes, I did consider the Howland Current pump, but
didn't give it too much attention as it apparently has a voltage input,
however, rereading my own question, I see that I asked about
voltage-current and current-current amps, so the Howland would
qualify as a voltage-current amp with a load referenced to zero.
I shall look at it again.

That only leaves me with a proper single op-amp current-current
amp to find!

Andy

 

[Fred Bloggs]

I just found it
a little improbable that these circuits hadn't been invented years ago
by some bright spark, and thought that they weren't shown in my text
books because they were little used.

You are talking about a transconductance amplifier, voltage-to-current,
they are all smoke and mirrors, and a current-to-current is left as an
exercise for the student.

 

[Ken Smith]

[...]
That only leaves me with a proper single op-amp current-current
amp to find![/QUOTE]

 

[John Woodgate]

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

That only leaves me with a proper single op-amp current-current amp to
find!

Ground V1in and leave out R2. Current-current!

 

[[email protected]]

Andy wrote:
(snip)

That only leaves me with a proper single op-amp current-current
amp to find!

It doesn't have a ground connected load (but a virtul ground connected
load), but an inverting amplifier with no input resistor and the load
as the feedbback resistor is a current to current configuration.

 

[Andy]

Andy wrote:
(snip)

It doesn't have a ground connected load (but a virtul ground connected
load), but an inverting amplifier with no input resistor and the load
as the feedbback resistor is a current to current configuration.

Ha, I spotted that one! In fact I drew a few block diagrams of a two input
port/two output port amplifier
and fiddled around with the feedback until I got a
configuration that was current output, and when I translated it back to an
op-amp I found I had what you suggest. The trouble with all my
'breakthroughs', is that they always have a snag, and it is the floating
load in this
case. Doubtless a useful circuit in special
circumstances though.

I will look at the suggestions regarding the Howland current souce and
modifying it, that might work as
another poster suggested.

thanks All,

Andy.