Using a Pot to vary resistance to ground?? HELP!! - Electronic Circuits Projects Diagrams Free

kenyan06 · 13th May 2007, 09:37 AM

Hi guys.

I know this might sound like a stupid question but how do i use a Pot to vary the resistance to ground. The reason i ask is because i am doing a project at university where the output from an infrared transistor goes first to an atmega88 's ADC an then to ground through a resistor (normally 10 k ) but i want to use a POT to vary the amount of voltage getting to the atmega88 (changing the sensitivity). Could someone please explain how i am meant to hook the pots up to achieve this.

Cheers

ericgibbs · 13th May 2007, 10:11 AM

hi,

The 10K0 pot has three terminals, one of the outer terminals goes to the voltage output of your IR device, the other outer terminal of the pot goes to 0Volt. The centre terminal of the pot goes to the atmega88.

If you find that when you turn the pot 'clockwise' to increase the output it goes down, just swop over the two outer terminals of the pot.

Is that clear?

kenyan06 · 13th May 2007, 10:35 AM

Just to make sure i understood correctly. Is this what you ment?

thank's a lot for your help

ericgibbs · 13th May 2007, 11:44 AM

hi,

This way.

kenyan06 · 13th May 2007, 01:29 PM

im still not 100% convinced because if i do that its just going to divide the voltage i get from the sensors and reduce it with increasing resistance wont it?
i thought that if you increases the value of the resistor behind the point where it goes to the atmega88 you would get a higher voltage according to V=IR please let me know if this makes any sense, because i need to get a PCB manufactured asap

ericgibbs · 13th May 2007, 01:37 PM

>>> but i want to use a POT to vary the amount of voltage getting to the atmega88 (changing the sensitivity).

If you adjust the pot in my drawing it will reduce the voltage going to the am88 from the IR sensor.... will it not??

If you not convinced, just put together a quick test circuit and try it.

kenyan06 · 13th May 2007, 01:45 PM

Yea thats fair enough i know that what you are saying is true but its not i don't think thats the effect i'm looking for. When I was testing the light sensors in the lab it we got less voltage for a given distance with say a 10k resistor than a 51k resistor. Does that make sense? does it have something to with the amount of current getting to ground or why did i observe this? Its too late to test it now because i have to submit the pcb design tomorrow morning (8hrs)

ericgibbs · 13th May 2007, 01:58 PM

hi,
Can you post any info on the IR device, drawing, specification or web link,
so that we can suggest an idea?

Without knowing what the IR source is, its impossible to give a good answer.

kenyan06 · 13th May 2007, 02:10 PM

Ok a little background info on the project. It is simulation of a satellite defence system. There is an infrared target that has to be detected with these light sensors. http://www.farnell.com/datasheets/71965.pdf

The voltage from the light sensor is then read by the ACDC converter on the Atmega88 to determine when it is looking directly at it and then fire a laser. The satellite is rotated using a motor and a flywheel. Here is a circuit schematic soon.

ericgibbs · 13th May 2007, 03:27 PM

Quote:

Originally Posted by kenyan06

Yea thats fair enough i know that what you are saying is true but its not i don't think thats the effect i'm looking for. When I was testing the light sensors in the lab it we got less voltage for a given distance with say a 10k resistor than a 51k resistor. Does that make sense? does it have something to with the amount of current getting to ground or why did i observe this? Its too late to test it now because i have to submit the pcb design tomorrow morning (8hrs)

I have looked at the PD drawing.

The amount of light falling on the PD will effect the resistance of the PD.
The stronger the light the greater the current [within the limits of the PD] and reduced light will give a lower current..

If you have a 10K say with 1mA flowing thru the PD you will get a voltage drop of about 10Vdp across 10K0.
If you use a 50K0 resistor you get a drop of 50Vdp across the 50K

If your Vsupply is say 24V, then the PD will voltage limit.

The choice of resistor is a balance between the range of light intensity you are expecting and your supply voltage.

Remember the ADC has a input impedance which may be lower than your PD's
load resistance, so it will have an effect on the output voltage.

Is this clear?

Ambient · 14th May 2007, 02:50 AM

To vary the resistance using a pot, just use the wiper and one of the other pins. As you turn the dial the resistance will change. But be VERY careful doing that if it is the only resistor to ground. If you turn all the way to 0 ohms you can damage the sensor, so put another resistor in series and use the pot to fine tune the resistance. the other resistor should be able to protect the device from too much current by itself if the pot is at 0 ohms.

If it is an IR "photo transistor" then increasing the resistance will decrease the base (IR) current that is needed to saturate the transistor. Therefore more of the voltage drop will be across the resistor. So increasing the resistor on the emitter side (if an npn detector) will make the sensor more sensitive.

If it is an IR "photo resister" then it will simply act like a voltage diode as ericgibbs described.

The transistor models the sensor if it is a photo transistor type.

Ambient · 14th May 2007, 03:18 AM

Actually I just looked at the spec sheet. It is a photo-transistor. The spec sheet does not list which lead is the emitter and which is the collector. Oddly, no IR photo-transistor spec sheets I have seen tell you either.

I do not know if it is true for this detector, but the ones I have use the long lead for the emitter and the short lead for the collector, the opposite polarity of an LED. It was very frustrating when I found out that was why I couldn't get a signal. Try them in both directions.

Is this going to be for a DC IR signal or will it be oscillating?

kenyan06 · 14th May 2007, 04:52 AM

damn if only you had posted before i had to submit the PCB design

i implemented it as suggested above but now that i think about it i think you are correct, i was planning to do what u mentioned above but then redesigned it. Oh well i guess i could still change it around somehow with a few cut tracks and a bit of wire....(not pretty but oh well) Thanks for all your help

ericgibbs · 14th May 2007, 06:56 AM

hi keyan,
The information I posted to you still stands.

Look at the datasheet and you will see that the current flowing thru the PD is
dependent upon the light falling on it.

It is not a linear function for a PD or PT, but as stated, the load resistor must be chosen to suit the range of light intensity expected and the voltage applied across the PD/PT and load resistor.

If you chose a high value of load resistor with a supply voltage that is too low to allow the required voltage drop across the load resistor, then the output voltage will limit. Also if for example, you have a 50K0 load and the input impedance to your Am88 is say also 50K0, then the effective load resistance is 25K0.
You should always consider the effect of the 'sinks' impedance, if necessary use a buffer amp.

Post the circuit you have submitted for the project, add any voltage values
and we can see if its screwed up.

kenyan06 · 14th May 2007, 07:07 AM

Hey i am trying to post a schematic but im having a lot of trouble trying to upload the image to image shack or photobucket and it just says uploading forever. Is there any other way i could post it?

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13th May 2007, 09:37 AM	(permalink)
kenyan06	Using a Pot to vary resistance to ground?? HELP!! Hi guys. I know this might sound like a stupid question but how do i use a Pot to vary the resistance to ground. The reason i ask is because i am doing a project at university where the output from an infrared transistor goes first to an atmega88 's ADC an then to ground through a resistor (normally 10 k ) but i want to use a POT to vary the amount of voltage getting to the atmega88 (changing the sensitivity). Could someone please explain how i am meant to hook the pots up to achieve this. Cheers

13th May 2007, 10:11 AM	(permalink)
ericgibbs	hi, The 10K0 pot has three terminals, one of the outer terminals goes to the voltage output of your IR device, the other outer terminal of the pot goes to 0Volt. The centre terminal of the pot goes to the atmega88. If you find that when you turn the pot 'clockwise' to increase the output it goes down, just swop over the two outer terminals of the pot. Is that clear? __________________ Eric "Good enough is Perfect" PIC tutorials: Gramo's: www.digital-diy.net/ Bill's: www.blueroomelectronics.com/

13th May 2007, 10:35 AM	(permalink)
kenyan06	Just to make sure i understood correctly. Is this what you ment? thank's a lot for your help

13th May 2007, 11:44 AM	(permalink)
ericgibbs	hi, This way. __________________ Eric "Good enough is Perfect" PIC tutorials: Gramo's: www.digital-diy.net/ Bill's: www.blueroomelectronics.com/ Last edited by ericgibbs; 7th July 2008 at 11:19 AM.

13th May 2007, 01:29 PM	(permalink)
kenyan06	im still not 100% convinced because if i do that its just going to divide the voltage i get from the sensors and reduce it with increasing resistance wont it? i thought that if you increases the value of the resistor behind the point where it goes to the atmega88 you would get a higher voltage according to V=IR please let me know if this makes any sense, because i need to get a PCB manufactured asap

13th May 2007, 01:37 PM	(permalink)
ericgibbs	>>> but i want to use a POT to vary the amount of voltage getting to the atmega88 (changing the sensitivity). If you adjust the pot in my drawing it will reduce the voltage going to the am88 from the IR sensor.... will it not?? If you not convinced, just put together a quick test circuit and try it. __________________ Eric "Good enough is Perfect" PIC tutorials: Gramo's: www.digital-diy.net/ Bill's: www.blueroomelectronics.com/

13th May 2007, 01:45 PM	(permalink)
kenyan06	Yea thats fair enough i know that what you are saying is true but its not i don't think thats the effect i'm looking for. When I was testing the light sensors in the lab it we got less voltage for a given distance with say a 10k resistor than a 51k resistor. Does that make sense? does it have something to with the amount of current getting to ground or why did i observe this? Its too late to test it now because i have to submit the pcb design tomorrow morning (8hrs)

13th May 2007, 01:58 PM	(permalink)
ericgibbs	hi, Can you post any info on the IR device, drawing, specification or web link, so that we can suggest an idea? Without knowing what the IR source is, its impossible to give a good answer. __________________ Eric "Good enough is Perfect" PIC tutorials: Gramo's: www.digital-diy.net/ Bill's: www.blueroomelectronics.com/

13th May 2007, 02:10 PM	(permalink)
kenyan06	Ok a little background info on the project. It is simulation of a satellite defence system. There is an infrared target that has to be detected with these light sensors. http://www.farnell.com/datasheets/71965.pdf The voltage from the light sensor is then read by the ACDC converter on the Atmega88 to determine when it is looking directly at it and then fire a laser. The satellite is rotated using a motor and a flywheel. Here is a circuit schematic soon.

14th May 2007, 02:50 AM	(permalink)
Ambient	To vary the resistance using a pot, just use the wiper and one of the other pins. As you turn the dial the resistance will change. But be VERY careful doing that if it is the only resistor to ground. If you turn all the way to 0 ohms you can damage the sensor, so put another resistor in series and use the pot to fine tune the resistance. the other resistor should be able to protect the device from too much current by itself if the pot is at 0 ohms. If it is an IR "photo transistor" then increasing the resistance will decrease the base (IR) current that is needed to saturate the transistor. Therefore more of the voltage drop will be across the resistor. So increasing the resistor on the emitter side (if an npn detector) will make the sensor more sensitive. If it is an IR "photo resister" then it will simply act like a voltage diode as ericgibbs described. The transistor models the sensor if it is a photo transistor type. Last edited by Ambient; 14th May 2007 at 03:06 AM.

14th May 2007, 03:18 AM	(permalink)
Ambient	Actually I just looked at the spec sheet. It is a photo-transistor. The spec sheet does not list which lead is the emitter and which is the collector. Oddly, no IR photo-transistor spec sheets I have seen tell you either. I do not know if it is true for this detector, but the ones I have use the long lead for the emitter and the short lead for the collector, the opposite polarity of an LED. It was very frustrating when I found out that was why I couldn't get a signal. Try them in both directions. Is this going to be for a DC IR signal or will it be oscillating?

14th May 2007, 04:52 AM	(permalink)
kenyan06	damn if only you had posted before i had to submit the PCB design i implemented it as suggested above but now that i think about it i think you are correct, i was planning to do what u mentioned above but then redesigned it. Oh well i guess i could still change it around somehow with a few cut tracks and a bit of wire....(not pretty but oh well) Thanks for all your help

14th May 2007, 06:56 AM	(permalink)
ericgibbs	hi keyan, The information I posted to you still stands. Look at the datasheet and you will see that the current flowing thru the PD is dependent upon the light falling on it. It is not a linear function for a PD or PT, but as stated, the load resistor must be chosen to suit the range of light intensity expected and the voltage applied across the PD/PT and load resistor. If you chose a high value of load resistor with a supply voltage that is too low to allow the required voltage drop across the load resistor, then the output voltage will limit. Also if for example, you have a 50K0 load and the input impedance to your Am88 is say also 50K0, then the effective load resistance is 25K0. You should always consider the effect of the 'sinks' impedance, if necessary use a buffer amp. Post the circuit you have submitted for the project, add any voltage values and we can see if its screwed up. __________________ Eric "Good enough is Perfect" PIC tutorials: Gramo's: www.digital-diy.net/ Bill's: www.blueroomelectronics.com/

14th May 2007, 07:07 AM	(permalink)
kenyan06	Hey i am trying to post a schematic but im having a lot of trouble trying to upload the image to image shack or photobucket and it just says uploading forever. Is there any other way i could post it?