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First of all, thanks a lot for everyone's kindness in guiding me on my doubts before this. Would like to ask regarding the following matter: Firstly, the voltage output of my solar panel varies from 4V to 10V (Will not exceed 10V). I intend to shut the PIC down if the output of the solar panel is lower than 4V. Apart from that, I intend to power-up my PIC using the power directly from the solar panel. About reading voltage value using ADC, is there a way where I can scale the read down the values from the solar panel output? Something like: What ever reading value I obtain from the solar panel's output, I make sure it is being divided by 2 before being applied to the ADC input of the PIC using a voltage divider. If yes, do you guys mind to guide me in this simple circuit? I tried doing some simulations to tap the voltage from a source and divide it to obtain a half of through simulations but fail :( Meaning, overall, I intend do the following: By having 2 ADC inputs, a.) one would be an input of a "limited to 5V input". At this point, if the PIC reads a value of lower than 4V, I shut off the PIC. b.) another would be an input of a "voltage from the solar panel divided by two". This will be the one to control my PWM duty ratio. By the way, the duty ratio will be used to control my on/off of a MOSFET and I need approximately 16mA to turn it on. Would like to confirm, about this: Does it mean that if the voltage used to power-up my PIC is 4V, I will obtained an output current of approx. 20mA? Whereas if it is 3V, it will be approx. 15mA? _________________ |
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Circuit to read voltage for ADC - PIC
Started by ●December 14, 2004
Reply by ●December 14, 20042004-12-14
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I assume you will be using a low-drop-out regulator (like a 3940-5) to regulate the output of the solar panel? The 'drop-out' voltage (0.5 volts for the 3940) is how much voltage over the desired one you must produce, for the regulator to work. Never having done what you are trying to do, I'm not sure of the voltage range of the PIC. It does have a 'brown-out' setting, though. --- In , "devonsc" <devonsc@y...> wrote: > > First of all, thanks a lot for everyone's kindness in guiding me on my > doubts before this. Would like to ask regarding the following matter: > > Firstly, the voltage output of my solar panel varies from 4V to 10V > (Will not exceed 10V). I intend to shut the PIC down if the output of > the solar panel is lower than 4V. Apart from that, I intend to > power-up my PIC using the power directly from the solar panel. > > About reading voltage value using ADC, is there a way where I can > scale the read down the values from the solar panel output? Something > like: What ever reading value I obtain from the solar panel's output, > I make sure it is being divided by 2 before being applied to the ADC > input of the PIC using a voltage divider. If yes, do you guys mind to > guide me in this simple circuit? I tried doing some simulations to tap > the voltage from a source and divide it to obtain a half of through > simulations but fail :( > > Meaning, overall, I intend do the following: By having 2 ADC inputs, > a.) one would be an input of a "limited to 5V input". At this point, > if the PIC reads a value of lower than 4V, I shut off the PIC. > > b.) another would be an input of a "voltage from the solar panel > divided by two". This will be the one to control my PWM duty ratio. > > By the way, the duty ratio will be used to control my on/off of a > MOSFET and I need approximately 16mA to turn it on. Would like to > confirm, about this: Does it mean that if the voltage used to power- up > my PIC is 4V, I will obtained an output current of approx. 20mA? > Whereas if it is 3V, it will be approx. 15mA? > _________________ |
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Reply by ●December 14, 20042004-12-14
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Many PICs will operate at 3.3V - I just used a 16LF628A at this voltage quite successfully. A voltage divider of, say, 2 is quite simple: use two 1K ohm resistors in series. Put one end of the string at ground, the input voltage at the other end and the center point will be 1/2 of the input. Want divide by 3? Put the input into a 2K resistor (or 2 1K resistors in series) use the same 1K resistor to ground and measure at the upper end of the 1K, just like before. Now, here's where it gets ugly: The input impedance of the A/D converter is in parallel with the lower resistor, reducing its value by probably 10%. To get a more accurate solution, you may need to increase the lower resistor to 1.1K. You can do this with 2 2.2K resistors in parallel. Of course, the proper way to do this is with an Op Amp because the output impedance of the Op Amp is quite low and the input impedance of the A/D converter won't affect the accuracy of the division. --- In , "Allan Lane" <allan.lane@h...> wrote: > > I assume you will be using a low-drop-out regulator (like a 3940- 5) > to regulate the output of the solar panel? > > The 'drop-out' voltage (0.5 volts for the 3940) is how much voltage > over the desired one you must produce, for the regulator to work. > > Never having done what you are trying to do, I'm not sure of the > voltage range of the PIC. It does have a 'brown-out' setting, though. > > --- In , "devonsc" <devonsc@y...> wrote: > > > > First of all, thanks a lot for everyone's kindness in guiding me on > my > > doubts before this. Would like to ask regarding the following > matter: > > > > Firstly, the voltage output of my solar panel varies from 4V to 10V > > (Will not exceed 10V). I intend to shut the PIC down if the output > of > > the solar panel is lower than 4V. Apart from that, I intend to > > power-up my PIC using the power directly from the solar panel. > > > > About reading voltage value using ADC, is there a way where I can > > scale the read down the values from the solar panel output? > Something > > like: What ever reading value I obtain from the solar panel's > output, > > I make sure it is being divided by 2 before being applied to the ADC > > input of the PIC using a voltage divider. If yes, do you guys mind > to > > guide me in this simple circuit? I tried doing some simulations to > tap > > the voltage from a source and divide it to obtain a half of through > > simulations but fail :( > > > > Meaning, overall, I intend do the following: By having 2 ADC inputs, > > a.) one would be an input of a "limited to 5V input". At this point, > > if the PIC reads a value of lower than 4V, I shut off the PIC. > > > > b.) another would be an input of a "voltage from the solar panel > > divided by two". This will be the one to control my PWM duty ratio. > > > > By the way, the duty ratio will be used to control my on/off of a > > MOSFET and I need approximately 16mA to turn it on. Would like to > > confirm, about this: Does it mean that if the voltage used to power- > up > > my PIC is 4V, I will obtained an output current of approx. 20mA? > > Whereas if it is 3V, it will be approx. 15mA? > > _________________ |
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Reply by ●December 14, 20042004-12-14
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Hi there, thanks for the explanation. Do you mind checking the circuit that I've simulated (as attached). By the way, with reference to the Point A in the circuit, is it a must for me to include a low resistance resistor at Point A before tapping the voltage off the line? If it is a must, do you mind explaining what is the purpose of doing so as I don't seem to find any difference in having the resistor there or not there. Should I just use the voltage divider circuit in my project? As I might face difficulties in supplying my op-amp. Is it true that most of the supply voltage of the op-amp is -/+15V? I will not be able to power it up with my solar panel. Thanks in advance... --- rtstofer <> wrote: > > > Many PICs will operate at 3.3V - I just used a > 16LF628A at this > voltage quite successfully. > > A voltage divider of, say, 2 is quite simple: use > two 1K ohm > resistors in series. Put one end of the string at > ground, the input > voltage at the other end and the center point will > be 1/2 of the > input. > > Want divide by 3? Put the input into a 2K resistor > (or 2 1K > resistors in series) use the same 1K resistor to > ground and measure > at the upper end of the 1K, just like before. > > Now, here's where it gets ugly: The input impedance > of the A/D > converter is in parallel with the lower resistor, > reducing its value > by probably 10%. To get a more accurate solution, > you may need to > increase the lower resistor to 1.1K. You can do > this with 2 2.2K > resistors in parallel. > > Of course, the proper way to do this is with an Op > Amp because the > output impedance of the Op Amp is quite low and the > input impedance > of the A/D converter won't affect the accuracy of > the division. > --- In , "Allan Lane" > <allan.lane@h...> wrote: > > > > I assume you will be using a low-drop-out > regulator (like a 3940- > 5) > > to regulate the output of the solar panel? > > > > The 'drop-out' voltage (0.5 volts for the 3940) is > how much > voltage > > over the desired one you must produce, for the > regulator to work. > > > > Never having done what you are trying to do, I'm > not sure of the > > voltage range of the PIC. It does have a > 'brown-out' setting, > though. > > > > --- In , "devonsc" > <devonsc@y...> wrote: > > > > > > First of all, thanks a lot for everyone's > kindness in guiding me > on > > my > > > doubts before this. Would like to ask regarding > the following > > matter: > > > > > > Firstly, the voltage output of my solar panel > varies from 4V to > 10V > > > (Will not exceed 10V). I intend to shut the PIC > down if the > output > > of > > > the solar panel is lower than 4V. Apart from > that, I intend to > > > power-up my PIC using the power directly from > the solar panel. > > > > > > About reading voltage value using ADC, is there > a way where I can > > > scale the read down the values from the solar > panel output? > > Something > > > like: What ever reading value I obtain from the > solar panel's > > output, > > > I make sure it is being divided by 2 before > being applied to the > ADC > > > input of the PIC using a voltage divider. If > yes, do you guys > mind > > to > > > guide me in this simple circuit? I tried doing > some simulations > to > > tap > > > the voltage from a source and divide it to > obtain a half of > through > > > simulations but fail :( > > > > > > Meaning, overall, I intend do the following: By > having 2 ADC > inputs, > > > a.) one would be an input of a "limited to 5V > input". At this > point, > > > if the PIC reads a value of lower than 4V, I > shut off the PIC. > > > > > > b.) another would be an input of a "voltage from > the solar panel > > > divided by two". This will be the one to control > my PWM duty > ratio. > > > > > > By the way, the duty ratio will be used to > control my on/off of a > > > MOSFET and I need approximately 16mA to turn it > on. Would like to > > > confirm, about this: Does it mean that if the > voltage used to > power- > > up > > > my PIC is 4V, I will obtained an output current > of approx. 20mA? > > > Whereas if it is 3V, it will be approx. 15mA? > > > _________________ > > __________________________________ | |||
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Reply by ●December 14, 20042004-12-14
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At 01:39 PM 12/14/2004, rtstofer wrote: >Want divide by 3? Put the input into a 2K resistor (or 2 1K >resistors in series) use the same 1K resistor to ground and measure >at the upper end of the 1K, just like before. > >Now, here's where it gets ugly: The input impedance of the A/D >converter is in parallel with the lower resistor, reducing its value >by probably 10%. To get a more accurate solution, you may need to >increase the lower resistor to 1.1K. You can do this with 2 2.2K >resistors in parallel. The statement about the a/d input impedance is off by a large factor. Microchip specs say that source impedance must be less than 10K for 8 bit a/d or less than 2.5K for 10 bit a/d for 1/2 LSB error. Your source impedance is considered to be the voltage divider resistors in parallel: 3:1 voltage divider with 2K feeding 1K is a source impedance of about 667 Ohms - well within the allowable range. Note 1: putting a cap in parallel with the lower resistor is always a good idea - it both filters the input as well as providing a low impedance source for that brief instant when the s/h switch operates inside the a/d converter. Its not absolutely necessary but I do include them on all my low speed a/d inputs. Note 2: Increasing source impedance above the recommended values can NOT be compensated for by manipulating the divider ratio. The problem is that input impedance related errors are caused by leakage current which can be either positive or negative and which can (will) change depending on the input voltage. I used to be able to cheat the input impedance specs on the older 16c71 & 16c73 / 74 chips - the leakage on those parts was extremely low and source impedance errors were dominated by the s/h charge current. That was easily fixed by adding a filter cap to each a/d input. But all the newer flash parts with the smaller process geometries have much higher leakage currents than the old OTP parts and you can't cheat that way any more. Its actually quite a step backwards when trying to design a very low cost product that requires extensive filtering on the a/d inputs. I'm currently re-designing one product that originally used a high impedance 3 stage RC filter on the a/d inputs - now I have to add an opamp buffer to deal with much higher a/d leakage currents (16c71 vs 16f676). The added a/d resolution makes it worthwhile, though. dwayne -- Dwayne Reid <> Trinity Electronics Systems Ltd Edmonton, AB, CANADA (780) 489-3199 voice (780) 487-6397 fax Celebrating 20 years of Engineering Innovation (1984 - 2004) .-. .-. .-. .-. .-. .-. .-. .-. .-. .- `-' `-' `-' `-' `-' `-' `-' `-' `-' Do NOT send unsolicited commercial email to this email address. This message neither grants consent to receive unsolicited commercial email nor is intended to solicit commercial email. |
Reply by ●December 15, 20042004-12-15
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--- In , Devon Lee <devonsc@y...> wrote: > Hi there, thanks for the explanation. Do you mind > checking the circuit that I've simulated (as > attached). Your sketch didn't stay attached. No way to check it. > > By the way, with reference to the Point A in the > circuit, is it a must for me to include a low > resistance resistor at Point A before tapping the > voltage off the line? If it is a must, do you mind > explaining what is the purpose of doing so as I don't > seem to find any difference in having the resistor > there or not there. > > Should I just use the voltage divider circuit in my > project? As I might face difficulties in supplying my > op-amp. Is it true that most of the supply voltage of > the op-amp is -/+15V? I will not be able to power it > up with my solar panel. A voltage divider will probably work out fine. You may have to do some kind of calibration but that shouldn't be a problem. The TI TLV2474 Op Amp is a quad unit that will operate from the same supply as the PIC, either 3.3V or 5V. Dual supplies are not required. There are variations: check www.ti.com. While you are there, get "Op Amps For Everyone" and check out chapter 4. The book is free. Specifically, chapter 4 deals with scaling and offset of input signals. For example, I was once interested in battery voltages between 20 and 28 volts. It was possible to change this to a 0..5V input with a single op amp and a few resistors. This made maximum use of the A/D converter because the batteries would cease to function if they dropped below 20V anyway. Kind of like your solar panel; below a certain level is meaningless and it is better to offset the signal and rescale for enhanced accuracy of the important range of voltages. |
