Autoranging Digital Voltmeter Help needed

In summary: The datasheet says that the Vref voltage "multiplies" the input voltage by 2. This means that the full-scale voltage on the display will be four times the Vref voltage.
  • #1
kjnhu
14
0
I am building an autoranging digital voltmeter as the subject line suggests and am having a problem with the ICL7107 chip that I am using. Looking at figure 15 on: http://www.intersil.com/data/fn/fn3082.pdf
specifically RefHI and RefLO, with a 2V max input difference between the pos and neg inputs. I do not necessarily have the negative input at 0 volts, and therefore would like assistance on how to use the Vref inputs. Do I need to set Vref=1V greater than the low input, or should it be 1V with repect to ground? Also why does the figure show a potentiometer rather than simply using resistors? Also should I just short the COM pin to my low input? I am unsure of how it actually functions after reading the cryptic explanation on the data sheet. Thank you for your assistance.
Kevin Hurley
 
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  • #2
Hi kjnhu,

You're in luck -- I'm a designer at Intersil. While I didn't work on the specific chip you're using, I'm sure I can help you with it.

The REF HI and REF LO pins are used to specify the voltage range for the ADC. An input voltage equal to REF LO will be converted to zero on the display, while an input voltage equal to REF HI will be converted to full-scale on the display.

The potentiometer is your scale-adjusting input. By varying the value of REF HI, you're varying the range of voltages shown on the display.

You can leave COMMON floating, in most cases, and you'll be just fine. I assume you're presenting single-ended signals to the device.

- Warren
 
  • #3
chroot said:
You're in luck -- I'm a designer at Intersil. While I didn't work on the specific chip you're using, I'm sure I can help you with it.
Now is that cool or what? PF rocks!
 
  • #4
chroot said:
The REF HI and REF LO pins are used to specify the voltage range for the ADC. An input voltage equal to REF LO will be converted to zero on the display, while an input voltage equal to REF HI will be converted to full-scale on the display.

The potentiometer is your scale-adjusting input. By varying the value of REF HI, you're varying the range of voltages shown on the display.

- Warren

Thank you a million! I have emailed Intersil support about five times and not received anywhere near the quality of response that you provided me with. So since INHI and INLO can be floating with respect to the power supply for the chip, if I connect RefLo to my INLO pin, and then connect a 2V source between REFLO and REFHI I would be getting a chip that reads with a maximum of a 2V scale? Also if I am correct, would this be the easiest way to do this?

You can leave COMMON floating, in most cases, and you'll be just fine. I assume you're presenting single-ended signals to the device.
One more thing, what do you mean "single-ended signals"? The signals I am presenting are voltages scaled to 2V max difference between INHI and INLO, with INLO always being the lower of the two.
Thank you again for your help, it is very much appreciated!
Kevin
 
  • #5
"Single Ended Signals" are contrasted with "differential" signals. A single-ended signal is a single v(t) that is measured with respect to an overall steady reference voltage, usually ground. A differential voltage is a + and - wire connection, where the common-mode voltage (the average of the two voltages) does not matter, and the difference voltage is what carries the information. If you are grounding your INLO and RefLo signals, and you are connecting your input - wire to ground, then you are treating the input signal as a single-ended signal. Just be sure that the input signal always falls in the 0V to 2V range of your ADC reference range.
 
  • #6
Thank you berkeman, that clears that part of my question up. My input would then be a differential input, so are you saying I need to connect COM to the average of the two inputs? Thank you again,
Kevin
 
  • #7
kjnhu said:
Thank you berkeman, that clears that part of my question up. My input would then be a differential input, so are you saying I need to connect COM to the average of the two inputs? Thank you again,
Kevin
It would be better if chroot answers your questions the next time he checks in with PF, but my impression from scanning the datasheet is that you should not make any connections between COM and your input signals. Just let the input circuitry of the ADC take care of doing the differential-to-single-ended conversion for you. Also, it looks like there is a 2x multiplication from the Vref voltage to the full scale input voltage range, so keep that in mind as you design your circuit. The datasheet has some pretty good application info in it, so be sure to read it carefully.
 
  • #8
kjnhu said:
Thank you a million! I have emailed Intersil support about five times and not received anywhere near the quality of response that you provided me with. So since INHI and INLO can be floating with respect to the power supply for the chip, if I connect RefLo to my INLO pin, and then connect a 2V source between REFLO and REFHI I would be getting a chip that reads with a maximum of a 2V scale? Also if I am correct, would this be the easiest way to do this?

Have you consulted some of the app notes referenced in the article? These seem very useful:

http://www.intersil.com/data/an/an023.pdf
http://www.intersil.com/data/an/an046.pdf

From page 10 of the datasheet, section "Reference Voltage," you can see that VREF should be half of the desired full-scale. If you want a 2V input to be full-scale, then VREF should be 1V. While the datasheet does not seem to be explicit about it, I'm sure VREF = REF HI - REF LO.

Depending upon how you're generating the reference voltage, you might want to tie REF LO to either ground or COMMON.

You can tie IN LO to COMMON to eliminate a little bit of noise, but it's also not necessary. Just make sure you observe the limitations on the value of VIN as provided in the same paragraph.

- Warren
 
  • #9
I posted this on anther forum but it seems less active, so ill post it again here, hopefully that isn't violating any rules, if it is please let me know, but I really need to know what part numbers these components could be so that I can build this project. Could someone please recommend part values for the following components, or tell me if they do not exist:
1. differential op-amp that can source enough current to turn on a BJT and an LED, and also outputs a voltage near the +-5V supply inputs
2. a BJT that can withstand up to 2V across it
3. An LED with a reverse breakdown voltage of greater than 5V (i have looked at radioshack but it doesn't show the reverse breakdown)
4. An PNP Transistor that can withstand very large voltages but turns on with below 5V difference (Vth).
5. The connectors that you see on a voltmeter to plug the testing leads into.
Thank you for any help you can be.
Kevin
 
  • #10
-1- Linear Technology or TI should have what you are looking for. You can use their selector guides at their websites to find good candidate parts, and then use Findchips.com or Digikey.com to see if they are easily available. The best opamps I've used lately are the National Semiconductor LMV321 and the Burr-Brown OPA343.

-2- 2V is tiny for BJT BVCEO. What voltage and current specs are you really talking about? What topology are you using for the BJT? Would a vanilla TO-92 transistor like the 2N3904 (NPN) and 2N3906 (PNP) work for you?

-3- Put a diode in series with the LED if you need a larger reverse standoff voltage.

-4- How high a voltage? Same question about the 2N3906 -- will it work? What are the current and voltage specs that you need?

-5- Do you mean female banana plugs? You can get those anywhere, including Digikey.com or your local Radio Shack.
 
  • #11
Thank you for the help berkeman.
1. I may be misreading the datasheets for the op-amps/comparitors I have been looking at, but can they only source about 50-60mA current? If this is the case, do I need to use a pullup resistor or something similar in order to source the current I need to keep and LED and BJT on? I changed my design specification to use an op-amp or comparitor that outputs +5 or 0 since it seems more are available that do that, however I am still stuck with the current sourcing I believe.

2. I am using the BJT between my 2 inputs to decide which path to use for scaling for the autoranging component. For example if I get +5V signal indicating that the input is on the order of 10^2 that certain BJT will turn on and let the input pass through it to scale it down to 2V through a resistor network. It was my error to type it needs to withstand 2V, it should actually be able to withstand at least 500Vdc, so in this case it appears 2N3904 can not work for that high of voltages. Also, will using a BJT effect the signal I am passing? The signal will be almost 0 current.

3. Thank you very much, I didnt even think about this.

4. I need an enhanced MOSFET that can withstand up to 500V drain to source, but turns on with a difference of no more than probably 4V max between gate and source (pnp)

5. Yes that is what I am looking for, thank you, I couldn't remember what to search for.
 
  • #12
-1- Opamps are not meant to drive any significant power. You need to buffer the opamp outputs with transistors to get current or voltage gain.

-2- 500V, yikes. I sure hope you know what you're doing working with voltages that high. Please be careful not to hurt yourself. Also, you probably shouldn't be passing your input signal through a BJT and then measuring it DVM-wise. BJTs are not good choices for series switches where you want low/consistent voltage drop. A relay or FET would be a better choice. Also, 500V transistors are extremely specialized, and don't belong in a DVM. Consider dividing down the input voltage with resistors before doing any measurements on higher voltages (or use an external HV probe).

-4- Same comment about 500V transistors.
 
  • #13
1. Hmmm I am not sure what you mean by "buffer the opamp outputs with transistors to get current or voltage gain". How exactly would I connect a transistor to source the current but maintain the 5V signal? I am assuming for this I would use a BJT to provide current gain? But I am not sure how I would implement this if I wanted to keep the constant 5V.

2. As you can probably tell, I am not exactly an expert, but I suppose that can be expected since I have only completed 1.5 semesters of school and have only learned transistors/diodes/anything remotely advanced on my own time as I have not covered them in school yet. But, you don't need to worry about me, I won't actually be dealing with voltages that high, it is merely a proof of concept thing for an English class actually. I adjusted my design to scale down the input to only a max of 2V where the input to the BJT would have been. I think I will replace the BJT with a relay, do you have any part numbers you would recommend that can be turned on with a 5V signal and pass a maximum of a 2V signal without distorting it?

4. I think I could implement this part with relays also, but I would like to use some MOSFET's because I enjoy learning new things and would like to try them out, so that being said, my design in theory calls for MOSFETs that can withstand up to 1000V actually, but I would actually never put more than probably 30V across any one, so I think I will design it in that manner but show that I could have replaced them with relays, so I think the transistor you suggested earlier, 2N3906 would work, should I encounter any other issues with this transistor, or would you still recommend it for this application? Thank you once again berkeman you are an enormous help!
Kevin
 
  • #14
-1- To buffer a pullup current, connect the opamp as a follower, but with the output into an NPN base, the - input connected to the NPN emitter, the NPN collector to the + supply, and the emitter connected to the load which is connected to ground. The opamp follower configuration ensures that the input voltage on its + input is also present at the NPN emitter (top of the load), and you get a beta current gain through the NPN transistor. You can use a Darlington for beta^2 current gain if necessary, and as long as you have the extra voltage headroom. Keep in mind what the input-to-output voltage transfer characteristic is for this circuit, so if you need +5V followed onto your load, the opamp will need higher than a 5V supply (quiz question -- why?).

-2- Look up 5V relays at Digikey.com or Mouser.com. Keep in mind that they take a fair amount of coil current, though, and generally you will use a dedicated coil drive interface IC for that. Also, on the 500V-->2V divider issue, keep in mind that you will lose accuracy the farther down you divide your input signal, and any errors introduced in the measurement circuit on the 2V end will be multiplied by 250.

-4- The 1000V MOSFET is unrealistic for a DVM application. High voltage MOSFETs and BJTs are used for power switching applications, and come in giant packages with lots of heat sinking. The DVM paradigm for measuring high voltages is to use a high voltage probe that plugs into the banana jacks of the DVM. The high voltage probes have precision dividers in them.
 
  • #15
1. In this case, could I simply connect my op-amp that is operating at its rails to generate +-8V or so depending on if the +input or the -input is higher and connect it's output to the base of my transistor, connect 5V to the collector and connect my load then ground to the emitter, generating 5V at my load or 0V depending on if my op-amp is outputting a + or - voltage? I don't see why I need another op-amp connected as a follower. Would my above plan work to generate enough current? And I believe the answer to the quiz question is that it needs greater than 5V to turn on the transistor because as it starts to turn on there will be 5V at the emitter so no current will flow from the base to the emitter, shutting the transistor off again.

2. I looked up some relays, the most applicable one I saw would probably be the FTR-H1CA005V at mouser. To drive it, could I do the following: Have my 5V source at the top of the coil, a diode in parallel w/ the coil to prevent spikes, at the other side of the coil/diode connect a BC337 transistor, with the emitter connected to ground and the base connected to the 5V signal coming off of the transistor from part 1, connected to a 33 ohm resistor before connecting to the base in order to source .15A to keep the transistor on when 5V is applied to base?

I think I am finally nearing a more solid design through your help, thanks once again!
 
  • #16
One more thing, I haven't really covered anything in classes regarding voltage regulators, but would like to use the LM317 or LM117 in my circuit. I want to use it in two separate applications, one as a floating source generating 5V over a signal called "b", which is my low input to the DVM, and another application where it generates 5V w/ respect to ground. The way I understand it, the LM317 generates 1.25V across R1, giving a current of 1.25/R1. What I don't understand is how to calculate Iadj to use in the formula for Vout. Also, to make this a floating source, do I connect my 'b' voltage to the spot where ground is? Thanks again,
Kevin
 
  • #17
-1- Yeah, you can do it like that. I thought you wanted 5V at the load, which is why I suggested the follower connection for the opamp and transistor.

-2- Sounds fine.
 
  • #18
kjnhu said:
One more thing, I haven't really covered anything in classes regarding voltage regulators, but would like to use the LM317 or LM117 in my circuit. I want to use it in two separate applications, one as a floating source generating 5V over a signal called "b", which is my low input to the DVM, and another application where it generates 5V w/ respect to ground. The way I understand it, the LM317 generates 1.25V across R1, giving a current of 1.25/R1. What I don't understand is how to calculate Iadj to use in the formula for Vout. Also, to make this a floating source, do I connect my 'b' voltage to the spot where ground is? Thanks again,
Kevin
As the datasheet says, Iadj (the current out of the ADJ terminal) is nominally 100uA. It adds an error term to the output voltage, which is the total of the voltage across R1 and R2. And yes, the LM317 output voltage is just with respect to the "ground" shown in the datasheet, so you can float the LM317 on top of your b input instead if you want. You still need to put the decoupling capacitors on the input and output, though, as shown in the datasheet.
 
  • #19
Thank you once again berkeman, you have been an immense help. One more question (for now :D ) - I need a MOSFET like IRF150 with a maximum of Vth of 4V, but that one seems kinda pricy, as in $2-3 for each FET, do you know of a cheaper alternative?
 
  • #20
For jellybean MOSFETs I use the BSS98 and BSS110 (n- and p-channel) in TO-92 packages. They're under $1, but I don't know if they have the specs that you need. I've used some nice dual MOSFETs recently -- the small FDG6301N/6302P FETs and the medium-size ALD1101 FETs. The matching isn't all that great, but otherwise they're pretty good parts. A little more expensive, though. Do you really need the 75A rating of the IRF150? That's a serious DVM!
 
  • #21
Hopefully this is one of the last questions I have to beg off of you, but I am unsure about using the LM317. I do not see the purpose of the Vin pin. If for example I want to make a 5V source, would it work for R1 to be 330 ohms and R2 to be 1000 ohms, would I get 5V out regardless of the input to Vin? Also, can I use the LM317 to generate -5V w/ respect to ground if I connect Vout to ground and the pin usually connected to ground I would connect to the input to my IC that needs -5V? Thanks again,
Kevin
 
  • #22
Vin is the input pin -- that's where the power comes from to supply the output voltage rail. Look at the equivalent circuit for the LM317 in its datasheet. Also, keep in mind that there is a minimum allowed input-to-output voltage drop in order for the 317 (or any voltage reulator) to stay in regulation. For most non-low-dropout regulators, the minimum input-to-output drop is something like 2-3V.

If you want to make a negative rail, use the LM117 instead of the LM317. All of these voltage regulators have stability considerations, so trying to run them upside-down from their intended purpose can cause stability issues.

Good luck! -Mike-
 
  • #23
I don't think I see the figure you are referring to, do you happen to have the link for the datasheet you are using? I am using one from Nat'l Semi and either do not have that figure, or am misinterpreting what you are saying. I want to power my DVM from a 9V battery, but would like the LM317 to output 5V. Could I simply input the 9V at Vin, and set R2=1k, R1=330 ohms to get the 5V output? Thank you once again!
 
  • #24
Hmmm. The LM317 datasheet in my old National Semiconductor Linear Databook has a "schematic" on page 1-35. It's not a simplified block diagram, but it does show some pretty descriptive details. It's right after the Application Hints section and before the Typical Applications schematics.

BTW, if you only need 5V, just use a 7805 voltage regulator instead of the LM317. The 317 is for non-standard positive voltages, and the 117 is for non-standard negative voltages. For +/-5V and +/-12V, just use the appropriate 78xx and 79xx regulators.
 

1. What is an autoranging digital voltmeter?

An autoranging digital voltmeter is a measuring instrument used to determine the voltage of an electrical circuit. It automatically adjusts its range to display the most accurate reading, making it easier to use than a manual ranging voltmeter.

2. How does an autoranging digital voltmeter work?

An autoranging digital voltmeter works by sending a low current through the circuit being measured and then using a digital display to show the voltage. It continually adjusts its range until it displays the most accurate reading, making it ideal for both low and high voltage measurements.

3. What are the benefits of using an autoranging digital voltmeter?

There are several benefits to using an autoranging digital voltmeter. It is more accurate and easier to use than a manual ranging voltmeter, as it automatically adjusts its range. It is also versatile, as it can measure both low and high voltages. Additionally, it is more compact and portable than traditional analog voltmeters.

4. How do I use an autoranging digital voltmeter?

To use an autoranging digital voltmeter, first ensure that it is set to the correct voltage range for your measurement. Then, connect the voltmeter probes to the circuit being measured, making sure to match the positive and negative terminals. The voltmeter will automatically adjust its range and display the voltage reading on its digital display.

5. Are autoranging digital voltmeters safe to use?

Yes, autoranging digital voltmeters are safe to use as long as they are used correctly. Make sure to follow all safety precautions and instructions when using the voltmeter. Also, avoid measuring voltages higher than the voltmeter's maximum range to prevent damage to the instrument or potential harm to the user.

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