# opamp inputs

• posted

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

• posted

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?

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

• posted

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?

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.

• posted

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?

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```
• posted

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

• posted

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

• posted

actually i started confusing myself when looking at figure 6.38b on pg 355= in art of electronics.

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

• posted

art of electronics.

follower to cancel out that small current that goes into the adjust pin.

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! :-)

• posted

panfilero presented the following explanation :

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```
• posted

355 in art of electronics.

follower to cancel out that small current that goes into the adjust pin. = =20

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

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?

• posted

mp=20

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no, maybe i confuse people by using ground and common interchangeably thoug= h. my trouble is not understanding how the opamp can make sense out of volt= ages at its input that are referenced to a separate ground... in my mind it= would have to take whatever is at its input and reference it to its own gr= ound and whatever voltage it sees there will have to be within the opamp's = rails. but how does it do that? and does it even have to do that? I think = it must. There some common input voltage that you can't exceed right, and t= his is the voltage from either input pin to ground (the opamps ground)... h= mmmmm.....

• posted

g 355 in art of electronics.

a follower to cancel out that small current that goes into the adjust pin. = =20

cross 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 hi= gher than its rails... then i started thinking.... the inverting input is t= ied to the adjust pin, which is ~1.2V below the output pin... the opamp wil= l try to do whatever it needs to do to make its inputs match (right?) so it= s going to try to make 60V to 1.2V ??? and I'm going to have to run my opam= p off an isolated supply cause the 60V will be with respect to the same gro= und the 317 is on (i can find a high voltage 317)... at this point my brain= started melting

ve to run my opamp off its own isolated ground. i think the circuit will wo= rk, i just don't know how the opamp can tolerate high voltages at its input= without blowing up... if the voltage at the opamp's inputs is referenced t= o a seperate ground, then i don't think the opamp knows what the hell volta= ge is there, it would have to internally reference it to its own ground to = make sense of it... wouldn't it?

g 355 in art of electronics.

a follower to cancel out that small current that goes into the adjust pin. = =20

cross 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 hi= gher than its rails... then i started thinking.... the inverting input is t= ied to the adjust pin, which is ~1.2V below the output pin... the opamp wil= l try to do whatever it needs to do to make its inputs match (right?) so it= s going to try to make 60V to 1.2V ??? and I'm going to have to run my opam= p off an isolated supply cause the 60V will be with respect to the same gro= und the 317 is on (i can find a high voltage 317)... at this point my brain= started melting

ve to run my opamp off its own isolated ground. i think the circuit will wo= rk, i just don't know how the opamp can tolerate high voltages at its input= without blowing up... if the voltage at the opamp's inputs is referenced t= o a seperate ground, then i don't think the opamp knows what the hell volta= ge is there, it would have to internally reference it to its own ground to = make sense of it... wouldn't it?

g 355 in art of electronics.

a follower to cancel out that small current that goes into the adjust pin. = =20

cross 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 hi= gher than its rails... then i started thinking.... the inverting input is t= ied to the adjust pin, which is ~1.2V below the output pin... the opamp wil= l try to do whatever it needs to do to make its inputs match (right?) so it= s going to try to make 60V to 1.2V ??? and I'm going to have to run my opam= p off an isolated supply cause the 60V will be with respect to the same gro= und the 317 is on (i can find a high voltage 317)... at this point my brain= started melting

ve to run my opamp off its own isolated ground. i think the circuit will wo= rk, i just don't know how the opamp can tolerate high voltages at its input= without blowing up... if the voltage at the opamp's inputs is referenced t= o a seperate ground, then i don't think the opamp knows what the hell volta= ge is there, it would have to internally reference it to its own ground to = make sense of it... wouldn't it?

g 355 in art of electronics.

a follower to cancel out that small current that goes into the adjust pin. = =20

cross 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 hi= gher than its rails... then i started thinking.... the inverting input is t= ied to the adjust pin, which is ~1.2V below the output pin... the opamp wil= l try to do whatever it needs to do to make its inputs match (right?) so it= s going to try to make 60V to 1.2V ??? and I'm going to have to run my opam= p off an isolated supply cause the 60V will be with respect to the same gro= und the 317 is on (i can find a high voltage 317)... at this point my brain= started melting

ve to run my opamp off its own isolated ground. i think the circuit will wo= rk, i just don't know how the opamp can tolerate high voltages at its input= without blowing up... if the voltage at the opamp's inputs is referenced t= o a seperate ground, then i don't think the opamp knows what the hell volta= ge is there, it would have to internally reference it to its own ground to = make sense of it... wouldn't it?

g 355 in art of electronics.

a follower to cancel out that small current that goes into the adjust pin. = =20

cross 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 hi= gher than its rails... then i started thinking.... the inverting input is t= ied to the adjust pin, which is ~1.2V below the output pin... the opamp wil= l try to do whatever it needs to do to make its inputs match (right?) so it= s going to try to make 60V to 1.2V ??? and I'm going to have to run my opam= p off an isolated supply cause the 60V will be with respect to the same gro= und the 317 is on (i can find a high voltage 317)... at this point my brain= started melting

ve to run my opamp off its own isolated ground. i think the circuit will wo= rk, i just don't know how the opamp can tolerate high voltages at its input= without blowing up... if the voltage at the opamp's inputs is referenced t= o a seperate ground, then i don't think the opamp knows what the hell volta= ge is there, it would have to internally reference it to its own ground to = make sense of it... wouldn't it?

g 355 in art of electronics.

a follower to cancel out that small current that goes into the adjust pin. = =20

cross 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 hi= gher than its rails... then i started thinking.... the inverting input is t= ied to the adjust pin, which is ~1.2V below the output pin... the opamp wil= l try to do whatever it needs to do to make its inputs match (right?) so it= s going to try to make 60V to 1.2V ??? and I'm going to have to run my opam= p off an isolated supply cause the 60V will be with respect to the same gro= und the 317 is on (i can find a high voltage 317)... at this point my brain= started melting

ve to run my opamp off its own isolated ground. i think the circuit will wo= rk, i just don't know how the opamp can tolerate high voltages at its input= without blowing up... if the voltage at the opamp's inputs is referenced t= o a seperate ground, then i don't think the opamp knows what the hell volta= ge is there, it would have to internally reference it to its own ground to = make sense of it... wouldn't it?

• posted

goes

is

without

and

my trouble is not understanding how the opamp can make sense out of voltages at its input that are referenced to a separate ground... in my mind it would have to take whatever is at its input and reference it to its own ground and whatever voltage it sees there will have to be within the opamp's rails. but how does it do that? and does it even have to do that? I think it must. There some common input voltage that you can't exceed right, and this is the voltage from either input pin to ground (the opamps ground)... hmmmmm.....

It doesn't. An opamp *can* give you the difference between two voltages (differential). Where one of the voltages is the opamp circuit's "ground" (reference, really), it's a special case and some components go away. BTW, an opamp needn't be connected to "ground" at all.

The easiest way to "get" an opamp is to solve the equations for the current/voltage in the input and feedback components. Assume that an opamp has an infinite input impedance, zero output impedance, and infinite differential gain. Infinite input impedance means that the current in the feedback resistor is the same as the current in the input resistor. Infinite gain means that the differential input voltage is zero (the '+' and '-' inputs must be at the same voltage). These assumptions hold remarkably well as long as the output isn't railed.

• posted

voltages at its input that are referenced to a separate ground...

Think about a DMM: it has _2_ leads, right? One lead is connected to the voltage and the other to the voltage's reference. You are giving the DMM a reference for the voltage. A crude DMM could simply have its internal reference/"ground" (the opamp's reference) and the external reference tied together. )Most likely it would be isolated, but let's leave them connected - it's conceptually correct and much simpler.)

but how does it do that? ...

As others have said, a voltage divider, between the 2 input leads:

I know that you know about voltage dividers, I think that you aren't grasping its use between the _2_ inputs.

I(desperately)HTH, Bob

• posted

in art of electronics.

follower to cancel out that small current that goes into the adjust pin.

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

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 without 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 make sense of it... wouldn't it?

```--
high-side driven only, or whether it could be low-side driven.```
• posted

in art of electronics.

follower to cancel out that small current that goes into the adjust pin.

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

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 without 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 make sense of it... wouldn't it?

```--
Panfilero,

Did you run the simulation?```
• posted

pg 355 in art of electronics.

n a follower to cancel out that small current that goes into the adjust pin= . =20

across my full load. but, then i started thinking... ok, if i do this, th= at 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 w= ill 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 op= amp off an isolated supply cause the 60V will be with respect to the same g= round the 317 is on (i can find a high voltage 317)... at this point my bra= in started melting

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 inp= ut without 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 vol= tage is there, it would have to internally reference it to its own ground t= o make sense of it... wouldn't it?

John, yes I've run it and it seems perfect for my application, I actually b= readboarded the 317 current source from the "Art of Electronics" book (with= the voltage follower in it) and I couldn't get that one to behave very wel= l... but the ones in your simulation seem great and seem to hold as I adjus= t the 240k load. I've been trying to go through the circuit and understand= how it works, sure I'd appreciate any description of how it works that you= can give me. seems like an awesome little current source.

thanks again!

• posted

pg 355 in art of electronics.

n a follower to cancel out that small current that goes into the adjust pin= . =20

across my full load. but, then i started thinking... ok, if i do this, th= at 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 w= ill 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 op= amp off an isolated supply cause the 60V will be with respect to the same g= round the 317 is on (i can find a high voltage 317)... at this point my bra= in started melting

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 inp= ut without 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 vol= tage is there, it would have to internally reference it to its own ground t= o make sense of it... wouldn't it?

ok i think i get it, on both of these your essentially setting a voltage at= the non-inverting input of the opamp, that will get duplicated at the outp= ut and force 1V voltage across the 4k resistor, and then adjusting the 240k= resistor doesn't impact the 1V across the 4k... great circuit, thanks

• posted

pg 355 in art of electronics.

n a follower to cancel out that small current that goes into the adjust pin= . =20

across my full load. but, then i started thinking... ok, if i do this, th= at 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 w= ill 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 op= amp off an isolated supply cause the 60V will be with respect to the same g= round the 317 is on (i can find a high voltage 317)... at this point my bra= in started melting

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 inp= ut without 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 vol= tage is there, it would have to internally reference it to its own ground t= o make sense of it... wouldn't it?

What's the purpose of the 1k at the output of the opamp going to the transi= stor?

• posted

breadboarded the 317 current source from the "Art of Electronics" book (with the voltage follower in it) and I couldn't get that one to behave very well... but the ones in your simulation seem great and seem to hold as I adjust the 240k load. I've been trying to go through the circuit and understand how it works, sure I'd appreciate any description of how it works that you can give me. seems like an awesome little current source.

```--
You're welcome, and thanks for the compliment. :-)

Each of the circuits comprises an opamp supply regulator, a reference```

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