Single Op-Amp Current Amplifier
in [Design]
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From: Andy on 8 Sep 2005 10:21 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.
From: John Popelish on 8 Sep 2005 12:11 Andy wrote: > 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.
From: Andy on 8 Sep 2005 13:56 "John Popelish" <jpopelish(a)rica.net> wrote in message news:CvOdnfxiYY_9_73eRVn-sw(a)adelphia.com... > Andy wrote: >> 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. 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.
From: Ol' Duffer on 8 Sep 2005 19:46 In article <432048dc(a)212.67.96.135>, andrewpreece(a)onetel.com says... > 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.
From: Ken Smith on 8 Sep 2005 20:10
In article <MPG.1d8a9732f331f74a989700(a)news.bright.net>, Ol' Duffer <DontSend(a)MeSpam.net> wrote: [...] 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 -- -- kensmith(a)rahul.net forging knowledge |