Forums

opamp inputs

Started by panfilero August 1, 2012
can anyone tell me how an opamp powered off of 9V, can be used to take meas=
urements of really high voltages, like 100V? Like a DMM.... I realize the o=
pamp is floating and isolated from whatever its measuring, but the inputs o=
f the opamp can't go over the rails of the opamps... right? but you can pro=
be large votlages with DMMs.... what are these inputs referenced to?

much thanx
panfilero wrote:
> can anyone tell me how an opamp powered off of 9V, can be used to take measurements of really high voltages, like 100V? Like a DMM.... I realize the opamp is floating and isolated from whatever its measuring, but the inputs of the opamp can't go over the rails of the opamps... right? but you can probe large votlages with DMMs.... what are these inputs referenced to? > > much thanx
Voltage divider string.. basic ohms law to drop the voltage down to where the op-amp is in useable range. Any thing above that should be protected with some sort of clamps. Auto ranging meters have a front end that can handle the max input voltage and switches the network around to match the metering circuit's range. Other types of DMM that are not auto ranging have some sort of protection from burn out, like a fuse or just blow itself up. Jamie
On Wed, 1 Aug 2012 13:33:57 -0700 (PDT), panfilero <panfilero@gmail.com>
wrote:

>can anyone tell me how an opamp powered off of 9V, can be used to take measurements of really high voltages, like 100V? Like a DMM.... I realize the opamp is floating and isolated from whatever its measuring, but the inputs of the opamp can't go over the rails of the opamps... right? but you can probe large votlages with DMMs.... what are these inputs referenced to? > >much thanx
The gain of an opamp in the inverting configuration is Rf/Ri (Rf=feedback resistor, Ri=input resistor) , so for suitable values of Rf and Ri one can get the output into the range of the opamp (Ri >> Rf). The input side of the input resistor can be at any (reasonable) voltage. Only the output of the opamp has to be inside the power supply voltage. You can also choose any voltage divider you want to scale the input voltage into the range of your amplifier.
On Wed, 1 Aug 2012 13:33:57 -0700 (PDT), panfilero
<panfilero@gmail.com> wrote:

>can anyone tell me how an opamp powered off of 9V, can be used to take measurements of really high voltages, like 100V? Like a DMM.... I realize the opamp is floating and isolated from whatever its measuring, but the inputs of the opamp can't go over the rails of the opamps... right? but you can probe large votlages with DMMs.... what are these inputs referenced to? > >much thanx
They use a voltage divider, a tapped string of resistors to select ranges. The total string is usually 10 megohms. There must be some protection, so that the opamp (or whatever) doesn't explode if you select the lowest voltage range and apply lots of voltage. -- John Larkin Highland Technology, Inc jlarkin at highlandtechnology dot com http://www.highlandtechnology.com Precision electronic instrumentation Picosecond-resolution Digital Delay and Pulse generators Custom laser drivers and controllers Photonics and fiberoptic TTL data links VME thermocouple, LVDT, synchro acquisition and simulation
On Wed, 1 Aug 2012, panfilero wrote:

> can anyone tell me how an opamp powered off of 9V, can be used to take > measurements of really high voltages, like 100V? Like a DMM.... I > realize the opamp is floating and isolated from whatever its measuring, > but the inputs of the opamp can't go over the rails of the opamps... > right? but you can probe large votlages with DMMs.... what are these > inputs referenced to? >
Others have pointed out the voltage divider, but it's worth looking at it from a different angle. The higher the voltage, the less sensitive the meter you need. But you want to measure low voltages, too, which needs a more sensitive meter, which is why an amplifier is added, well that and to buffer the input signal so the meter isn't loading down the circuit. So the amplifier and meter become a fixed meter measuring a very low voltage, and then the voltage divider ahead of it allows for higher voltage readings. It's a much better method than trying to adjust the amplification of the meter, or to change the meter to some other value for each voltage range needed to measure. Michael
you know, i guess really what i'm having a hard time with is... the opamp t=
hat is doing the measuring is referenced to its own ground, and then it goe=
s and takes a differential measurement of something, some high voltage that=
 is referenced to its own ground... does the voltage from the opamps ground=
 to one of its inputs matter? i mean it must be lower than the opamps rails=
 right? but how does the opamp know anything about what its measuring witho=
ut having access to its grounds? i think i'm missing something obvious here=
 and overcomplicating it for myself
actually i started confusing myself when looking at figure  6.38b on pg 355=
 in art of electronics.

http://books.google.com/books?id=3DbkOMDgwFA28C&pg=3DPA355&source=3Dgbs_toc=
_r&cad=3D4#v=3Donepage&q&f=3Dfalse

i want to make a current source using the lm317, and it says throw in a fol=
lower to cancel out that small current that goes into the adjust pin. =20

i want to run... 250uA through a 240k load... which will give me 60V across=
 my full load.  but, then i started thinking... ok, if i do this, that mean=
s the input of my opamp follower will be seeing 60V... which is way higher =
than its rails... then i started thinking.... the inverting input is tied t=
o the adjust pin, which is ~1.2V below the output pin... the opamp will try=
 to do whatever it needs to do to make its inputs match (right?) so its goi=
ng to try to make 60V to 1.2V ??? and I'm going to have to run my opamp off=
 an isolated supply cause the 60V will be with respect to the same ground t=
he 317 is on (i can find a high voltage 317)... at this point my brain star=
ted melting
On Thu, 2 Aug 2012 07:46:44 -0700 (PDT), panfilero
<panfilero@gmail.com> wrote:

>actually i started confusing myself when looking at figure 6.38b on pg 355 in art of electronics. > >http://books.google.com/books?id=bkOMDgwFA28C&pg=PA355&source=gbs_toc_r&cad=4#v=onepage&q&f=false > >i want to make a current source using the lm317, and it says throw in a follower to cancel out that small current that goes into the adjust pin. > >i want to run... 250uA through a 240k load... which will give me 60V across my full load. but, then i started thinking... ok, if i do this, that means the input of my opamp follower will be seeing 60V... which is way higher than its rails... then i started thinking.... the inverting input is tied to the adjust pin, which is ~1.2V below the output pin... the opamp will try to do whatever it needs to do to make its inputs match (right?) so its going to try to make 60V to 1.2V ??? and I'm going to have to run my opamp off an isolated supply cause the 60V will be with respect to the same ground the 317 is on (i can find a high voltage 317)... at this point my brain started melting
--- I know the feeling! :-) Does your load have to be grounded? -- JF
panfilero presented the following explanation :
> you know, i guess really what i'm having a hard time with is... the opamp > that is doing the measuring is referenced to its own ground, and then it goes > and takes a differential measurement of something, some high voltage that is > referenced to its own ground... does the voltage from the opamps ground to > one of its inputs matter? i mean it must be lower than the opamps rails > right? but how does the opamp know anything about what its measuring without > having access to its grounds? i think i'm missing something obvious here and > overcomplicating it for myself
I think you are making a mistake many others have made. There does not have to b a GROUND in any particular circuit. There is generaly a COMMON to which most things are referenced but GROUND, EARTH, is not important to a circuit in a box with its own power supply. In this case the Op Amp should see the Common and the divided down Hi. -- John G
On Thursday, August 2, 2012 3:17:22 PM UTC-5, John Fields wrote:
> On Thu, 2 Aug 2012 07:46:44 -0700 (PDT), panfilero >=20 >=20 >=20 >=20 >=20 > >actually i started confusing myself when looking at figure 6.38b on pg =
355 in art of electronics.
>=20 > > >=20 > >http://books.google.com/books?id=3DbkOMDgwFA28C&pg=3DPA355&source=3Dgbs_=
toc_r&cad=3D4#v=3Donepage&q&f=3Dfalse
>=20 > > >=20 > >i want to make a current source using the lm317, and it says throw in a =
follower to cancel out that small current that goes into the adjust pin. = =20
>=20 > > >=20 > >i want to run... 250uA through a 240k load... which will give me 60V acr=
oss my full load. but, then i started thinking... ok, if i do this, that m= eans the input of my opamp follower will be seeing 60V... which is way high= er than its rails... then i started thinking.... the inverting input is tie= d to the adjust pin, which is ~1.2V below the output pin... the opamp will = try to do whatever it needs to do to make its inputs match (right?) so its = going to try to make 60V to 1.2V ??? and I'm going to have to run my opamp = off an isolated supply cause the 60V will be with respect to the same groun= d the 317 is on (i can find a high voltage 317)... at this point my brain s= tarted melting
>=20 >=20 >=20 > --- >=20 > I know the feeling! :-) >=20 >=20 >=20 > Does your load have to be grounded? >=20 >=20 >=20 > --=20 >=20 > JF
well, my load has to be on the same ground as the current source. I'll have= to run my opamp off its own isolated ground. i think the circuit will work= , i just don't know how the opamp can tolerate high voltages at its input w= ithout blowing up... if the voltage at the opamp's inputs is referenced to = a seperate ground, then i don't think the opamp knows what the hell voltage= is there, it would have to internally reference it to its own ground to ma= ke sense of it... wouldn't it?