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
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
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.
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 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
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.
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
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! :-) Does your load have to be grounded?
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
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?
mp=20
t goes=20
at is=20
to=20
=20
thout=20
e and=20
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.....
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?
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.
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
On Thu, 2 Aug 2012 22:23:40 -0700 (PDT), panfilero wrote:
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 meltingrun 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:
Version 4 SHEET 1 952 680 WIRE -112 16 -848 16 WIRE 192 16 96 16 WIRE 336 16 192 16 WIRE 464 16 336 16 WIRE 720 16 464 16 WIRE -112 64 -112 16 WIRE 720 64 720 16 WIRE -848 96 -848 16 WIRE -768 96 -848 96 WIRE -624 96 -688 96 WIRE -528 96 -624 96 WIRE -384 96 -528 96 WIRE 192 96 192 16 WIRE 336 96 336 16 WIRE -528 128 -528 96 WIRE 720 160 720 144 WIRE 720 160 400 160 WIRE -112 192 -112 144 WIRE 720 192 720 160 WIRE -384 208 -384 96 WIRE 464 208 464 16 WIRE -416 224 -448 224 WIRE 400 224 400 160 WIRE 432 224 400 224 WIRE -288 240 -352 240 WIRE -176 240 -208 240 WIRE 544 240 496 240 WIRE 656 240 624 240 WIRE -528 256 -528 208 WIRE -416 256 -528 256 WIRE 336 256 336 176 WIRE 432 256 336 256 WIRE -448 320 -448 224 WIRE -112 320 -112 288 WIRE -112 320 -448 320 WIRE 336 320 336 256 WIRE -848 352 -848 96 WIRE -624 352 -624 96 WIRE -528 352 -528 256 WIRE -112 352 -112 320 WIRE 96 416 96 16 WIRE 720 416 720 288 WIRE 192 432 192 176 WIRE 336 432 336 400 WIRE 336 432 192 432 WIRE 464 432 464 272 WIRE 464 432 336 432 WIRE 192 448 192 432 WIRE -848 480 -848 432 WIRE -624 480 -624 416 WIRE -624 480 -848 480 WIRE -528 480 -528 432 WIRE -528 480 -624 480 WIRE -384 480 -384 272 WIRE -384 480 -528 480 WIRE -112 480 -112 432 WIRE -112 480 -384 480 WIRE 96 544 96 496 WIRE 192 544 192 512 WIRE 192 544 96 544 WIRE 720 544 720 496 WIRE 720 544 192 544 WIRE -848 560 -848 480 WIRE 96 624 96 544 FLAG -848 560 0 FLAG 96 624 0 SYMBOL res -128 48 R0 SYMATTR InstName R1 SYMATTR Value 240k SYMBOL npn -176 192 R0 SYMATTR InstName Q1 SYMATTR Value 2N3904 SYMBOL res -128 336 R0 SYMATTR InstName R2 SYMATTR Value 4k SYMBOL Opamps\\LT1001 -384 176 R0 SYMATTR InstName U1 SYMBOL res -544 112 R0 SYMATTR InstName R3 SYMATTR Value 44000 SYMBOL res -544 336 R0 SYMATTR InstName R4 SYMATTR Value 4k SYMBOL voltage -848 336 R0 WINDOW 123 0 0 Left 2 WINDOW 39 0 0 Left 2 SYMATTR InstName V2 SYMATTR Value 70 SYMBOL res -192 224 R90 WINDOW 0 0 56 VBottom 2 WINDOW 3 32 56 VTop 2 SYMATTR InstName R5 SYMATTR Value 1000 SYMBOL res 704 400 R0 SYMATTR InstName R6 SYMATTR Value 240k SYMBOL res 704 48 R0 SYMATTR InstName R7 SYMATTR Value 4k SYMBOL Opamps\\LT1001 464 176 R0 SYMATTR InstName U2 SYMBOL voltage 96 400 R0 WINDOW 123 0 0 Left 2 WINDOW 39 0 0 Left 2 SYMATTR InstName V4 SYMATTR Value 70 SYMBOL res 640 224 R90 WINDOW 0 0 56 VBottom 2 WINDOW 3 32 56 VTop 2 SYMATTR InstName R10 SYMATTR Value 1000 SYMBOL pnp 656 288 M180 SYMATTR InstName Q2 SYMATTR Value 2N3906 SYMBOL zener 208 512 R180 WINDOW 0 -38 33 Left 2 WINDOW 3 -88 -3 Left 2 SYMATTR InstName D1 SYMATTR Value 1N5371B SYMBOL res 176 80 R0 WINDOW 3 36 68 Left 2 SYMATTR InstName R8 SYMATTR Value 1000 SYMBOL res 352 416 R180 WINDOW 0 -41 72 Left 2 WINDOW 3 -38 36 Left 2 SYMATTR InstName R9 SYMATTR Value 9k SYMBOL res 320 80 R0 SYMATTR InstName R11 SYMATTR Value 1k SYMBOL res -672 80 R90 WINDOW 0 0 56 VBottom 2 WINDOW 3 32 56 VTop 2 SYMATTR InstName R12 SYMATTR Value 1000 SYMBOL zener -608 416 R180 WINDOW 0 54 33 Left 2 WINDOW 3 31 -3 Left 2 SYMATTR InstName D2 SYMATTR Value BZX84C12L TEXT -714 520 Left 2 !.tran .1 uic
-- JF
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 meltingrun 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? Do you need a circuit description?
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!
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
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?
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 generator and a current driver.
In the circuit with the load tied high, I've assumed a raw supply of about 70V in order to keep 60 volts across the load, with 10 volts of headroom.
R12 and D2 comprise a 12V Zener shunt regulator which is used to generate a 12 volt supply for the opamp and the reference generator, the R3/R4 voltage divider.
I arbitrarily chose 1 volt to be the output from the current sensing resistor, R2, which means that in order to drop 1 volt with 250µA through it, it has to be 4k ohms.
Then, in order for the reference to be equal to 1 volt, I chose R4 to be equal to R2 and calculated the value of R3 to drop 11 volts with
250µA through it, which comes out to 44k.Now, with the reference set at 1 volt and connected to the + input of the opamp, the the opamp will do whatever is required to drive the - input to 1 volt.
With Q1 in the loop, what will happen is that if the signal from R2 tries to go more positive than the reference, the opamp output will decrease, lowering the voltage into the base of Q1, which will increase its resistance and lower the current into R2 until its output voltage is equal to the reference voltage.
The opposite happens if the voltage from R2 falls below 1 volt, the result being that since current in a series circuit is the same throughout the circuit, regulating the current through R2 to 250µA will cause the current through the load to be regulated as well.
In the circuit with the load tied low, we'll need to get 60V into it, so a PNP with its emitter tied to 70V and dropping 10 volts across the C/E junction will work.
In order to drive the base properly we need to get it up to about a diode drop more negative than the emitter voltage, so to do that we can use 70V for the opamp + supply and 60v for the - supply, putting
10 across the opamp's supply terminals.R8 and D1 comprise a 60 volt Zener shunt regulator, which supplies the
60 volts needed for the opamp - supply and also drives the reference generator, R11/R9.R7 is the current sensing resistor, and with 250µA through it it'll drop 1 volt, placing 69 volts on the opamp's - input.
Now, since R11 is connected to 70 volts and R9 is connected to 60 volts, we have a voltage divider with 10 volts across it and, since we need 69 volts for the reference, we need the voltage divider to drop 1 volt across one resistor and 9 volts across the other.
Arbitrarily making R11 equal to 1k and and calculating R9 to drop 9V will accomplish that.
Then, since the opamp must do whatever it has to in order to make the voltage on its - input equal the voltage on its + input, it'll drive Q2 in the direction required to make sure that the current through R7 (and the load) is 250µA.
Be aware that the sim is just a rough sketch used to illustrate an idea, and is in no way a finished design.
-- JF
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