The gain of an opamp in the inverting configuration is Rf/Ri (Rf=feedback
They use a voltage divider, a tapped string of resistors to select
from a different angle.
---
I think you are making a mistake many others have made.
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 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?On Thu, 2 Aug 2012 07:46:44 -0700 (PDT), panfilero
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
no, maybe i confuse people by using ground and common interchangeably though. 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.....
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
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
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 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?
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
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
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 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?
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
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
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 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?
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
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
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 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?
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
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
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 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?
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
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
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 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?
It doesn't. An opamp *can* give you the difference between two voltages
Think about a DMM: it has _2_ leads, right? One lead is connected to
As others have said, a voltage divider, between the 2 input leads:
---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
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
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 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?
---On Thu, 2 Aug 2012 22:23:40 -0700 (PDT), panfilero
panfilero@gmail.com> wrote:
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
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
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 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?
---
I meant whether one side your load had to be grounded and the other
high-side driven only, or whether it could be low-side driven.
In any case, here's a couple of ideas on how to do it either way,
without the LM317:
John, yes I've run it and it seems perfect for my application, I actually 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.On Fri, 03 Aug 2012 09:27:27 -0500, John Fields
On Thu, 2 Aug 2012 22:23:40 -0700 (PDT), panfilero
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
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
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 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?
---
I meant whether one side your load had to be grounded and the other
high-side driven only, or whether it could be low-side driven.
In any case, here's a couple of ideas on how to do it either way,
without the LM317:
---
Panfilero,
Did you run the simulation?
Do you need a circuit description?
--
JF
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 output and force 1V voltage across the 4k resistor, and then adjusting the 240k resistor doesn't impact the 1V across the 4k... great circuit, thanksOn Fri, 03 Aug 2012 09:27:27 -0500, John Fields
wrote:
On Thu, 2 Aug 2012 22:23:40 -0700 (PDT), panfilero
wrote:
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
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
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 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?
---
I meant whether one side your load had to be grounded and the other
high-side driven only, or whether it could be low-side driven.
In any case, here's a couple of ideas on how to do it either way,
without the LM317:
---
Panfilero,
Did you run the simulation?
Do you need a circuit description?
--
JF
What's the purpose of the 1k at the output of the opamp going to the transistor?On Fri, 03 Aug 2012 09:27:27 -0500, John Fields
On Thu, 2 Aug 2012 22:23:40 -0700 (PDT), panfilero
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
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
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 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?
---
I meant whether one side your load had to be grounded and the other
high-side driven only, or whether it could be low-side driven.
In any case, here's a couple of ideas on how to do it either way,
without the LM317:
---
Panfilero,
Did you run the simulation?
Do you need a circuit description?
--
JF
---