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Op-Amp selection for inverting source signal - Bench PSU build

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chopnhack

Apr 28, 2014
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Hello all, I am looking for some assistance in choosing an op-amp to invert an incoming signal. Specifically, I am trying to use a panel meter to read negative voltage.

I hooked up the panel meter to a supply and then tried to use the meter to read the value of a battery in reverse and it was not able to do so. Apparently this type of panel meter can only read positive voltage.

I came across an old thread where someone was trying to use an op-amp to invert the voltage from negative to positive to feed to his meter, but there was much discussion about his op-amp being the wrong choice for use. I am open to suggestion!

After some reading, I learned that the important characteristics would be Avol, Voh and Vol - essentially determining how much overhead the op-amp requires to make its comparision and Vol being how noisy the circuit is? Lower values of Voh,ol are better, as this would mean that as it inverts my incoming negative voltage, it would only lose the difference between the signal and the overhead at the peaks of the rail to rail comparison. (If I have that correct)

I looked at TI's OPA2197 (two opamps since I have two meters that I want to be able to use as negative channels). I am not sure if this is available yet, as it claims new, but a 25μV offset seems to be a very reasonable amount of loss.

Have I made a decent selection? If I have not understood the mechanic behind choosing an op amp, can you please explain?

Thanks in advance!

Edit - I will add OPA2170 to my list as well.
 

BobK

Jan 5, 2010
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Not necessary. Just switch the leads on the meter. Put the positive lead to ground and the negative lead to V-.

Bob
 

chopnhack

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Not necessary. Just switch the leads on the meter. Put the positive lead to ground and the negative lead to V-.

Bob
I did try this, but it did not work - I like simple solutions too ;-)

hooked up the panel meter to a supply and then tried to use the meter to read the value of a battery in reverse and it was not able to do so. Apparently this type of panel meter can only read positive voltage.
 

Bluejets

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Thought about a supply isolator but the combo volt/current arrangement sort of throws a spanner in the works.
Saw a write up on that somewhere if you want I can look it up.
cheers...Jorgo
 

chopnhack

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Take a look at this thread about the same problem.....

http://www.eevblog.com/forum/beginners/digital-panel-meters-measuring-negative-voltages/

Apparently the current sensor line may create a short on the supply.

Thanks Blue! Apparently, it's all been thought of before!!

So reading the negative voltage using an op-amp will work! That's good to know :)

I think I see the problem... the voltmeter uses one leg of the supply which is then used in the current sensing portion of the ammeter. Thanks for pointing this out.
I may drop this line of thinking - since I have come to using lm317 as my v.regulator, there are so many options. I just found a great Nuts and Volts article that is very similar to what I was looking to do!! I was genuinely shocked to see it. I will upload my rough idea sketch of what I was trying to accomplish for comparison.
Thanks all.

bench%20psu_zpsjwbyu0np.jpg
 

chopnhack

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If you go with seperate meters for the current and voltage it might overcome the problem.
The other article I referred to was how to run an LCD volt or current meter which is designed to run from an isolated supply off a common supply.

http://www.thebackshed.com/Windmill/articles/PanelMeter.asp
That is a good point, I was just trying to use the two meters I already had. But if I can find small panel meters just for current, its worth another look. As for isolated, I had always intended to run these off of their own transformers.

Edit: Found meter

I found a meter - please have a look at the schematic (third photo) - they break the positive connection between the battery and the load to splice the meter inline. Being a current meter, wouldn't this work just as well on the negative side?

The reason I ask this is because with a negative voltage supply, Common is more positive than the negative rail, so the meter would be spliced into the common? Do you think that would work?

Thanks again.
 
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hevans1944

Hop - AC8NS
Jun 21, 2012
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This is a very old problem, "easily" solved by using an isolated power supply to power the meter, as in the Nuts And Volts article... which is quite good BTW.

I haven't used panel meters that also have a current function "built in" to the meter, but it would not be a big deal to add an internal shunt to the same basic LED (or LCD) meter and power both meters from the same low-voltage DC supply. The important thing to note is all of these inexpensive digital panel meters measure a DC voltage that is positive with respect to their power supply negative rail. That means the external voltage which you are trying to measure must be positive with respect to the meter power supply negative rail.

For the dual volt/current meter, the current shunt produces a positive voltage with respect to the meter power supply negative when it is inserted in the low side of a load that is powered from a supply that is positive with respect to the low side of the load. That conveniently means the power supply voltage for the load can also be measured by connecting the other meter input directly to the high side of the load. This arrangement corresponds to the circuit diagram on the left in your link to the current meter.

If you power this current meter (or a dual volt/amp panel meter) with an isolated power supply, you can use the circuit on the right, which (in this example) places the current-measuring shunt in the positive lead of the supply voltage to the load. Notice there is no "common" in the illustrated example. Internally, one terminal of the internal shunt is still connected to POWER-BLACK as well as to IN-BLACK and the other terminal is connected to IN-WHITE. The only difference is the shunt terminals are not referenced to an external ground or other common.

For this meter, with an isolated power supply, you just need to make sure that IN-WHITE is positive with respect to IN-BLACK. So, yeah, place it anywhere you want... high side (-Vreg) or low side (COM)... in your power supply. Just make sure you have the polarity correct. I would try this with a 9 V battery first, just to make sure you have the polarity correct. Then add the isolated power supply to replace the battery. Placing the meter in series with the -Vreg output is better than trying to interrupt the -Vreg common to place the meter there, IMO.

I highly recommend the isolated power supply approach. Using an op-amp requires precision matched resistors to get an inverting gain of -1 and much more complexity than needed.

I like the idea of salvaging isolated DC-to-DC converters from old Ethernet cards, as suggested by the link in @Bluejets post #7. I shudder to think of how many of those we threw away when converting from coax to CAT5 cables late in the previous Century. Argh! No, wait! I never throw anything away... surely I can find a pair of those somewhere in my pile of ancient goodies. Hmmm. Might be "cheaper" to buy new ones, like this little gem from Digi-Key.
 

chopnhack

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Nuts And Volts article... which is quite good BTW.
Yes, I liked that most of what I was thinking about was covered and more! It does look like a good unit that will service my needs for - well for as long as I can tell. I was thinking of doing a fixed 5v and 3v for μcu projects - trivial, small transformer and two v-regs, but very useful.

but it would not be a big deal to add an internal shunt to the same basic LED (or LCD) meter and power both meters from the same low-voltage DC supply.
I do not understand the purpose of the shunt. Can you explain? I re-read your explanation a number of times, but I am a little at a loss... what I did gather is that so long as I am feeding this device the voltage to be metered in the correct polarity, the meter should work. My experiment of hooking up the supply to a 9v battery and then measuring a AA battery in reverse, did not get a reading, however, I realize that this is not quite the same thing as having a negative supply.

I highly recommend the isolated power supply approach. Using an op-amp requires precision matched resistors to get an inverting gain of -1 and much more complexity than needed.
The isolated power supply was always a given. In the Nuts and Volts article, the author regulates the voltage output to zero by means of a precision zener - he uses it to bleed off the ref. voltage of the v.reg!! Good thinking to get a linear supply to zero voltage.

Tomorrow I will try to test the unit again with a negative supply.

Thanks!
 

BobK

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My experiment of hooking up the supply to a 9v battery and then measuring a AA battery in reverse, did not get a reading, however, I realize that this is not quite the same thing as having a negative supply.
To do an experiment on how it would work on dual supplies, you would do the following:

Put 2 AA batteries in series by connecting the negative of one to the positive of the other.

Now connect the meter with the negative input to the center and the positive to the free positive end of the battery. This is measuring the positive supply. Next, connect the positive lead of the meter to the center and the negative lead to the free negative terminal of the battery. This is measuring the neative supply.

Bob
 

hevans1944

Hop - AC8NS
Jun 21, 2012
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Yes, I liked that most of what I was thinking about was covered and more! It does look like a good unit that will service my needs for - well for as long as I can tell. I was thinking of doing a fixed 5v and 3v for μcu projects - trivial, small transformer and two v-regs, but very useful. ...
A bench power supply just makes everything easier for the hobbyist and professional alike. Meters, separate enabling switches for each output voltage, and 5-way binding posts really help to move things along while prototyping, testing, or troubleshooting projects. Sure, you can grab a wall-wart and cobble up connections and use a couple of multi-meters to measure voltage and current... I did that several years ago when my wife allowed me to use her sewing table as an electronics bench... but it is far from a satisfactory solution compared to having all the power you need in one location, monitored and measured with dedicated meters. We could do an entire thread here on how to select (or design) a bench power supply... and we probably already have

... I do not understand the purpose of the shunt. Can you explain? I re-read your explanation a number of times, but I am a little at a loss... what I did gather is that so long as I am feeding this device the voltage to be metered in the correct polarity, the meter should work. ...
The shunt is a small resistance connected across the meter input terminals and placed in series with the external load and its power supply. The purpose of the shunt is to create a small voltage (generally in the millivolt range so as not to significantly subtract from the external load's power supply voltage) that is interpreted as the load current. So, a 0.01 Ω shunt will have a "sensitivity" of 10 mV per ampere and a 20 A load current would produce a 200 mV voltage drop across the shunt. On most panel meters with a "200 mV" full-scale range of 199 or 1999 counts, this would be displayed as 19.9 A or 19.99 A.

Yes, as long as you are feeding a voltage panel meter with a voltage of the correct polarity, and an amplitude less than the full-scale input of the meter, it will display the correct value. In actuality, all digital panel meters are voltage panel meters, typically with a full-scale range of 200 mV. Placing two digital panel meters in the same package, and placing a shunt across the input terminals of one of them, converts the meter with a shunt into an ammeter.

The meter doesn't care if the input you apply to it comes from a negative power supply output referenced to power supply common, or from a positive power supply output referenced to power supply common, where the power supply common is external to the meter and the meter is powered from an isolated power supply.

In the first situation (negative supply referenced to power supply common), you connect the positive input of the meter to the power supply common and the negative input of the meter to the negative supply output. In the second situation (positive supply referenced to power supply common), you connect the positive input of the meter to the positive supply output and the negative input of the meter to the power supply common. This is precisely what @BobK described in his post #11. The reason it works is because neither of the meter inputs is connected to an external common until you make it so.

... My experiment of hooking up the supply to a 9v battery and then measuring a AA battery in reverse, did not get a reading, however, I realize that this is not quite the same thing as having a negative supply. ...
Why would you expect the meter to measure the voltage of an AA battery in reverse? The meter has two input leads, a plus lead and a minus lead, and will only read properly when the plus lead is positive with respect to the minus lead. So it will measure the voltage of an AA dry-cell if the + terminal on the cell is connected to the plus lead and the opposite - terminal on the cell is connected to the minus lead. It will not measure anything if you reverse the connections because then you would be applying a negative voltage to the plus lead with respect to the minus lead.

... The isolated power supply was always a given. In the Nuts and Volts article, the author regulates the voltage output to zero by means of a precision zener - he uses it to bleed off the ref. voltage of the v.reg!! Good thinking to get a linear supply to zero voltage. ...
I agree that the zener is a clever solution to getting the LM317 to regulate down to zero output voltage, but it doesn't "bleed off" anything. It simply biases the ADJ terminal to -1.25 V when the voltage adjustment potentiometers are at minimum resistance. This effectively negates the +1.25 V that the regulator establishes between OUT and ADJ, which the regulator does by controlling the conductance between IN and OUT. Note that you must not exceed the 40 V maximum voltage between IN and OUT, which if ADJ were connected to common would be VIN - 1.25 V. Biasing ADJ to -1.25 V effectively subtracts the zener voltage from the normally +1.25 V minimum output. Clever.

... Tomorrow I will try to test the unit again with a negative supply.

Thanks!


I think you are missing the point! The panel meter has two input leads, the first of which is also the negative terminal of the isolated power supply, whether that power supply be a battery or an isolated line-powered supply. The other, second, input lead will only accept and digitize a voltage which is positive with respect to the first lead.

So, you take a AA dry-cell and connect its negative terminal to the first input of the panel meter that is connected to the negative power terminal. You connect the positive terminal of the AA dry-cell to the other, second, input lead of the panel meter. Voila! The panel meter measures the 1.5 V potential of the dry-cell, assuming you have predetermined the maximum input voltage for the panel meter to be more than 1.5 V. Notice in making these connections there is not a ground anywhere to be seen.

There is a positive input (the second wire mentioned above), a negative input (the first wire mentioned above), a power supply negative wire (that is also internally connected to, and in common with, the negative input wire), and a power supply positive wire. Some meters may provide separate connections for the negative input wire and the negative power supply wire, but internally they are connected together. So, some meters will have just three wires: positive input, negative input connected to power supply negative, and power supply positive.

Higher-quality DC digital panel meters will accept bi-polar inputs and digitize both positive and negative inputs with a polarity indicator in the display. Many of these panel meters use a standard integrated circuit (ICL7106 for LCD displays or ICL7107 for LED displays) that accepts a positive 200 mV input and produces an LCD or LED output of 199 or 1999 counts for 199 mV or 199.9 mV input.

Most of the Asian imports DO NOT DIGITIZE inputs that are negative with respect to their meter circuit common. Perhaps this is a patent issue, or it may be just a design decision based on other undisclosed factors. In either case it is not a problem if your input does not change polarity and you power the meter from either batteries or an isolated power supply... as laboriously described above. OTOH, if you need a bi-polar DPM, and if you purchase one of those Maxim or Intersil chips to roll your own bi-polar DPM, the IC alone will cost seven or eight US dollars. Add to that the cost of the displays, a circuit board, and glue components and the cost to build is approaching the high side of ten dollars. That doesn't include assembly cost, marketing, and a package to mount everything in. So which would you choose: a $50 DPM that measures bi-polar inputs and displays a polarity sign, or a $2 DPM that only measures mono-polar inputs? You pays your money and you makes your choices.

If you desire to digitize a signal that is negative with respect to some power supply common, you must think of this as digitizing a common signal that is positive with respect to your negative signal referenced to power supply common. This is exactly what you would do with a digital voltmeter that could only measure positive voltages between its input probe and its common probe: connect the input probe to power supply common and connect the common probe to the negative source. So, for a panel meter that can only measure positive voltages with respect to its negative power supply terminal, you isolate the power supply, connect the positive meter input to your circuit common, and connect the negative meter input terminal to the negative voltage source you are trying to measure.

A combination volt/amp meter is just two panel meters in the same case, but with a shunt resistor wired between the positive meter input and the negative meter input. If the shunt resistor is 0.1 Ω, then one ampere will produce 0.1 V across the shunt, and if the meter is 200 mV full-scale (199 mV or 199.9 mV) it will read 1.00 or 1.000.

So how did your test with a "negative supply" turn out? Are you all clear on how to measure voltage with a digital panel meter now?:D

Hop
 

chopnhack

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Now connect the meter with the negative input to the center and the positive to the free positive end of the battery. This is measuring the positive supply. Next, connect the positive lead of the meter to the center and the negative lead to the free negative terminal of the battery. This is measuring the negative supply.

LOL - it's negative in relation to the "higher" battery which is measured as a positive 1.5v. The center connection between batteries is common to both and the bottom is negative in relation to the first battery. I think I understand. I used two 9volts again and the panel meter measured the voltage, I would just have to remember that the voltage reading is negative in relation.
 

chopnhack

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The shunt is a small resistance connected across the meter input terminals and placed in series with the external load and its power supply. The purpose of the shunt is to create a small voltage (generally in the millivolt range so as not to significantly subtract from the external load's power supply voltage) that is interpreted as the load current. So, a 0.01 Ω shunt will have a "sensitivity" of 10 mV per ampere and a 20 A load current would produce a 200 mV voltage drop across the shunt. On most panel meters with a "200 mV" full-scale range of 199 or 1999 counts, this would be displayed as 19.9 A or 19.99 A.

Yes, as long as you are feeding a voltage panel meter with a voltage of the correct polarity, and an amplitude less than the full-scale input of the meter, it will display the correct value. In actuality, all digital panel meters are voltage panel meters, typically with a full-scale range of 200 mV. Placing two digital panel meters in the same package, and placing a shunt across the input terminals of one of them, converts the meter with a shunt into an ammeter.

Thanks Hop! This makes a lot more sense now. I was wondering how this small unit was measuring the high amperage without burning out. There is a built in shunt, its a short loop of a thicker copper wire. As long as the polarity is connected correctly, it will give a reading.

So how did your test with a "negative supply" turn out? Are you all clear on how to measure voltage with a digital panel meter now?:D

Quite well, thanks! I think its a lot clearer, since these meters should work, I just need to build a BOM, do some layout, organize space and get some other materials lined up. Very exciting and I think I will come in far under what a simple dual supply would cost!!
 

hevans1944

Hop - AC8NS
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One of the nicest dual power supplies I have ever used was the HP (now Agilent) 6227B. They were a bit pricey, in keeping with Hewlett-Packard's design philosophy of not building or selling junk, but they were also virtually bulletproof. And now sometimes available on eBay for much less than their original cost.

Except for the brand name and spiffy case with dual D'Arsonval meters (which I prefer to digital panel meters for power supplies), you should be able to put together an equivalent power supply for less than $200 to $300, depending on how much you have to pay for the transformer. Plus, you gain the not insignificant advantage of being able to maintain it yourself.

I really like HP power supplies, but some models can be a real PITA to maintain. I would recommend obtaining a service manual and going over the schematics and parts list in some detail to see if you can maintain anything purchased as used equipment, regardless of brand name. I would do this even if by some miracle you found some "new old stock" in its original packaging. I own three Tektronix oscilloscopes with genuine cathode ray tubes, but if something ever fails (like the internal fan did on one of them) it is virtually impossible to obtain a cost-effective repair, or even replacement parts for many of the Tektronix custom ICs and other components.

Hop
 

chopnhack

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That is a sweet unit, but at over $400 delivered, way outside my budget - LOL. I can only imagine what it went for new!!

I am not sure on the transformer yet, I really have to spend some time and come up with an anticipated total energy consumption list and then decide how much headroom I want on the main transformer. The smaller transformers should be very inexpensive. I have some free transformers from microwaves that I could rewrap the secondary! Just kidding, they are probably horrible choices for a power supply - I am sure the core makeup affects their intended usage. But, I have seen some 160va between $40-80. I am looking at some that are surplus as well.

Maintaining by myself is a must! Good news is that I don't have to mess with any op amps :) Incidentally, I actually have some of those intersil 7106's in my junkbox :rolleyes:
 

hevans1944

Hop - AC8NS
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... Incidentally, I actually have some of those intersil 7106's in my junkbox :rolleyes:
I don't remember when these first became available, but we were playing with something similar back in the 1970s before digital panel meters became so inexpensive it wasn't worth the effort. I almost SIMPed when I saw a two-and-a-half digit Asian DPM for less than two bucks on eBay yesterday. Free shipping from China!

I don't know anything about rewinding microwave oven transformers (MOTs) but it does seem to be a popular activity on the web. Make sure you first hacksaw off the high-voltage winding, leaving just the primary. With the high-voltage secondary removed, there should be lots of room to add a few dozen turns of wire for a couple of low-voltage secondary windings. Here's a web link.

I cannot imagine what the downside to re-purposing the MOT core and primary would be. Most of these puppies are good for 700 to 1200 watt microwave ovens, so figure maybe half-again that much power to make up for magnetron losses and maybe a hundred watts for the magnetron filament. If someone here has any horror stories about using a re-purposed MOT for low-voltage AC, this would be a good time to bring that up.

Sadly, I have thrown away several microwave ovens over the past thirty years or so without bothering to salvage anything from any of them. At work, someone was trying to use a re-purposed microwave oven to process (sinter?) some exotic rare-earth mixture for making infrared-transparent windows for sensors and lasers, but I didn't get involved with that and didn't want to get involved with that, considering they had no idea how microwaves were generated or propagated. They managed to burn out about a dozen magnetrons by the time I retired several years later.

Hop
 

chopnhack

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I almost SIMPed
LOL, that actually took me a sec. :D:p
Yes, things of this nature are certainly cheaper than they used to be due to the massive amounts of manufacturing in China. Check out these SPDT, cheaper than any SPST I could find! I am going to order two lots, they look very useful for this project and for some others.

Thanks for the MOV links, I will study them. If its cheaper to rewind the secondary, that might make for an interesting experience and reduce a major expense!

Sadly, I have thrown away several microwave ovens over the past thirty years or so without bothering to salvage anything from any of them.
I didn't know about it until I came across a web link showing what could be salvaged out of them - I was in awe, there are literally hundreds of dollars worth of parts that have good uses.
 

roughshawd

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The key word here is shunt. Try different shunts to maintain the signal in a capacitance, then drain the circuit with triacs... Heh heh .. don't try this at work!!!
 
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