Post by thread:[PIC] PWM-> Circuit ?-> 2.4 to 2.6 VDC

Hi: I'm working with a sensor that's specified to have an output between 2.4 and 2.6 VDC (Vcc +5 V) and would like to amplify it's DC output. Since the signal output is on the order of a few millivolts an adjustable voltage reference is needed to keep the op amp swing between the rails. I've been using a couple of fixed resistors and a pot for this but would like to use the PWM output from a 16F88 and some type of circuit that would provide the 2.4 to 2.6 VDC reference signal that would not change (the actual value would be established once during calibration). When the PWM output was a constant +5 volts the circuit output would be +2.6 VDC and when the PWM output was a constant 0 volts the circuit output would be +2.4 VDC. Is there a circuit that does this? Since op amps come in dual and quad configurations one or two would be available for this function. Thanks, Brooke Clarke, N6GCE -- w/Java http://www.PRC68.com w/o Java www.pacificsites.com/~brooke/PRC68COM.shtml http://www.precisionclock.com

> Is there a circuit that does this? Since op amps come in dual > and quad configurations one or two would be available for this > function Sounds like you might be able to use a voltage summer, with rectified and smoothed PWM as a variable voltage http://www.web-ee.com/Schematics/OpAmps/AN-20.pdf If I understand you, the sensor has an output of 2.5 +/- 0.1V. Correct ? And you want to amplify this 0.2V range ? For example, like a Hall Effect sensor with a small output centred about Vcc/2 ? > Since the signal output is on the order of a few millivolts Except I don't quite understand that bit. If the signal is a few mV, what is the 2.5 +/- 0.1V ? Is the signal AC on top of the DC ? Or is the "few mV" the +/- 100mV ? Presently are you feeding the voltage from your 2 resistors and pot into one of the op-amp inputs to shift the ouput ? The other option is to use the F88's ADC directly, with -Vref and +Vref set appropriately. But I'm not entirely clear what you ultimately want to do with the signal - measure it or just amplify

On 25 Dec 2005 at 18:26, Brooke Clarke wrote: > I'm working with a sensor that's specified to have an output between 2.4 > and 2.6 VDC (Vcc +5 V) and would like to amplify it's DC output. > Since the signal output is on the order of a few millivolts an > adjustable voltage reference is needed to keep the op amp swing between > the rails. I've been using a couple of fixed resistors and a pot for > this but would like to use the PWM output from a 16F88 and some type of > circuit that would provide the 2.4 to 2.6 VDC reference signal that > would not change (the actual value would be established once during > calibration). When the PWM output was a constant +5 volts the circuit > output would be +2.6 VDC and when the PWM output was a constant 0 volts > the circuit output would be +2.4 VDC. Hi Brooke, If you use filtered PWM to generate a reference voltage with a range of .2 volts from a possible 5 volts, basically your gain is .2/5, or 1/25, so the ripple will be reduced. However, if your sensor amp is using it, you'll be amplifying the ripple back up again. So if you have your opamp gain set to 500, you'll be amplifying the ripple to 20 times higher than it was originally. So you must filter your PWM reference with a very low frequency knee and be extrememly careful with circuit layout. Is it important to have a DC amp? Why not just have an AC amp with good low frequency response and if you really need to know, read the DC voltage of the sensor with another A/D channel? Cheerful regards, Bob

Hi Jinx and Bob: It is a Hall sensor. The quiescent voltage spec with no magnetic field is about 2.5 +/- 0.1 V. The sensitivity is 5 millivolts per gauss. Since the earth's filed is 0.5 gauss the actual signal will always be less than 2.5 mv (more than 300 times smaller than the offset). In order to feed the 16F88 A/D converter the largest possible earth's field signal a lot of DC amplification is needed but the initial +/- 75 mv offset needs to be backed out first. I do want to read the unamplified sensor output on one A/D input channel and the highly amplified output on another channel. If VRef- and VRef+ are set to 2.4 and 2.6 volts then the gain stage is limited to 100 mv / 2.5 mv or 40 and the large signal output from the sensor can not be used. But if the 10 bit PWM output is filtered (this is not a problem since the value does not change often) then the gain can be more like 2.5 V from zero / 2.5 mv or 1000. This has a lot of advantages. Have Fun & Happy Holidays, Brooke Clarke, N6GCE -- w/Java http://www.PRC68.com w/o Java www.pacificsites.com/~brooke/PRC68COM.shtml http://www.precisionclock.com

Brooke Clarke, N6GCE wrote: >do want to read the unamplified sensor output on one A/D input channel >and the highly amplified output on another channel. If VRef- and VRef+ >are set to 2.4 and 2.6 volts then the gain stage is limited to 100 mv / >2.5 mv or 40 and the large signal output from the sensor can not be used. Although I haven't been following this thread very closely, it appears from above that you want to set VRef+ and VRef- to a 200mV difference. If so, you have a problem. Most of the PICs specify the minimum delta's to be between 1.8 and 3.0V depending on the device and Vdd. Check the A/D converter characteristics table in the data sheet for your device (deltaVref). If I misinterpreted what you were saying in the above sentence - well, never mind! Ken spam_OUTklumiaTakeThisOuTadelphia.net

Brooke Clarke wrote: > I'm working with a sensor that's specified to have an output between 2.4 > and 2.6 VDC (Vcc +5 V) and would like to amplify it's DC output. > Since the signal output is on the order of a few millivolts an > adjustable voltage reference is needed to keep the op amp swing between > the rails. I've been using a couple of fixed resistors and a pot for > this but would like to use the PWM output from a 16F88 and some type of > circuit that would provide the 2.4 to 2.6 VDC reference signal that > would not change (the actual value would be established once during > calibration). When the PWM output was a constant +5 volts the circuit > output would be +2.6 VDC and when the PWM output was a constant 0 volts > the circuit output would be +2.4 VDC. > > Is there a circuit that does this? Yes, this requires 3 resistors and a capacitor at minimum, although an extra pair or two of resistors and capacitors might be a good idea to get better PWM filtering. Think of it as a voltage divider between the PWM output and a 2.5V source. You should be able to come up with resistances for that, or at least ratios. Now realize the bottom resistor of the divider as two separate resistors, one to 5V and the other to GND. Each of these have twice the value of the bottom resistor of the hypothetical divider. Together they are indistiguishable from a single resistor to a 2.5V source. Look up Norton equivalence if you don't understand. > Since op amps come in dual and quad > configurations one or two would be available for this function. You might want one as a buffer, but if you're really clever you can combine the PWM divider and filter with the diff amp directly. ****************************************************************** Embed Inc, Littleton Massachusetts, (978) 742-9014. #1 PIC consultant in 2004 program year. http://www.embedinc.com/products

Hi Jinx: What I'm trying to do has not been done before. Hall effect devices are typically used with strong magnetic fields, i.e. near magnets. The Allegro app notes are all aimed at that application. Yes, the circuit at magnet.htm is using a couple of pots to generate current to buck out the Hall offset. This circuit uses a split (+ and -) power supply but they are only using a positive output, don't know why. But again here the magnetic field driving the Hall effect device is from a nearby magnet, not the earth's magnetic field. I think I have figured out how to do the subject circuit. Step one is to take the PWM output through a simple voltage divider changing the output to zero to 0.2 volts (i.e. divide by 25) and place a very large cap that's good at the PWM output frequency across the bottom resistor, now the output is a DC voltage with some ripple that can be added to 2.4 volts from another voltage divider. Have Fun & Happy Holidays, Brooke Clarke, N6GCE -- w/Java http://www.PRC68.com w/o Java www.pacificsites.com/~brooke/PRC68COM.shtml http://www.precisionclock.com Date: Tue, 27 Dec 2005 11:30:25 +1300 From: Jinx <.....joecolquittKILLspam@spam@clear.net.nz> Subject: Re: To: "Microcontroller discussion list - Public." <piclistKILLspammit.edu> Message-ID: <005e01c60a6b$f924ba30$0100a8c0@ivp2000> Content-Type: text/plain; charset=iso-8859-1 >> It is a Hall sensor. The quiescent voltage spec with no magnetic field >> is about 2.5 ± 0.1 V > > Have you tried specific interest groups ? http://www.magnetometer.org/ http://www.pacificsites.com/~brooke/Sensors.shtml (??!!) http://www.uksmg.org/magnet.htm Is the resistor-based adjustment in that circuit something like what you have now ? 634SS2 datasheet (169kB) http://www.farnell.com/datasheets/34553.pdf Did you see the thread "[EE] Use 405x instead was 4066 help" from the third week of 12/05 ? In it was described the technique of using PWM to make a variable resistor with a 4066, Perhaps you could forget about making a variable V and make a variable R instead

part 1 3491 bytes content-type:text/plain; charset=ISO-8859-1 (decoded quoted-printable) I did not caught the moment when the thread jumped off-list. So I'm reposting it now. ----------------------------------------------------------------------- Brooke Clarke wrote: > If the 0.2 volt range is expanded to 0 to 5 volts then the amplification > is 5 / .2 = 25 (fairly low). > Using an external 12 or 14 bit A/D adds cost and complication. For more > money a MR sensor would work much better, but the idea is to see of a > very low cost Hall sensor will work. Brooke, let's take PIC16F676 worth one dollar or so. If after pre-amplifying your signal 25 times from m volts to, say, 100 mV, you think the signal is still weak for PIC16F676 A/D, you can add a 4-bit R2R ladder to the another input resistors of your Op Amp to adjust the Op Amp's offset, thus allowing bigger amplification. The 4-bit R2R ladder is to be controlled by PIC16F676. This way you can narrow you input range up to 16 times, making possible not 25, but 250 amplification. So your useful signal will be up to 1v. Not that weak signal for 10 bit A/D with 2v reference level. Don't use noisy PWM when measuring mVolts Regards, Mike ----------------------------------------------------------------------- Hi Mike: I've just been surfing Microchip App Note 655 and the R2R ladder looks like a very nice solution. Thanks very much, Brooke ----------------------------------------------------------------------- HI again Mike: I've been trying to figure out the circuit and think I'm back to the original question. For example an R2R ladder driven from the PIC provides a zero to five volt ramp with 5v/(2^n) where n is the number of bits. I can dedicate 5 I/O pins to the ladder so get 0.156 volt steps out of the D/A converter. So the problem is the same as with PWM, that's to say how to get the 0->5 volt range compressed into a 2.4 to 2.6 volt range? Happy New Year, ----------------------------------------------------------------------- Brooke Clarke wrote: > I've been trying to figure out the circuit and think I'm back to the > original question. > For example an R2R ladder driven from the PIC provides a zero to five volt > ramp with 5v/(2^n) where n is the number of bits. > I can dedicate 5 I/O pins to the ladder so get 0.156 volt steps out of the > D/A converter. So the problem is the same as with PWM, that's to say how to > get the 0->5 volt range compressed into a 2.4 to 2.6 volt range? One input of the Op Amp - to the device's output. The second input of Op Amp - to the central point of the voltage divider (2.5V). The bottom resistor of the divider is compound -some resistor and R2R ladder in series (the ladder - on the bottom). Regards, Mike ----------------------------------------------------------------------- Hi Mike: I can't visualize the circuit. Could you send it as a hand drawn schematic (jpg, bmp, tif, etc)? Thanks, Brooke ----------------------------------------------------------------------- Brooke Clarke wrote: > I can't visualize the circuit. I can hardly too now. > Could you send it as a hand drawn schematic > (jpg, bmp, tif, etc)? Yes, definitely with my Epson Perfection 2480 (heck with CADs), after I sobber up a bit tomorrow :-) It's 2:10 am now here. Happy New Year, Mike. ----------------------------------------------------------------------- Tomorrow has come. The "scheme" is attached :-) Regards, Mike part 2 14956 bytes content-type:image/jpeg; name=1.jpg (decode) part 3 35 bytes content-type:text/plain; charset="us-ascii" (decoded 7bit)