Using a Pot to vary resistance to ground?? HELP!!

Status
Not open for further replies.
K

kenyan06

New Member
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
 
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?
 
**broken link removed**
**broken link removed**

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

thank's a lot for your help
 
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
 
>>> 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.
 
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)
 
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.
 
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. https://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.
 

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?
 
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.
**broken link removed**
 
Last edited:
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?
 
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
 
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.
 
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?
 
hi kenyan,
What format is the drawing and what size??

Sometimes it appears to take a long time.
 
hi,
Looked over the circuit I dont seem any major problems in the PT to AM88 ADC inputs.
Its a pity you didnt post this dwg earlier, I could have given a more precise answer.

As you now say the supply to the PT detector is only +5V and I expect the AM/ADC inputs are able to
accept 0 thru +5V, then you will not over drive the ADC inputs.

If you examine the datasheet for the A88 device it will give you a maximum acceptable value of the ADC inputs source impedance.
For example the PIC 16F877 has a max recommended input impedance of 10K0, so if you have a load resistor greater than about 10K it could give you acquistion time problems in the ADC part of the A88.

Ideally I would have chosen to have an opa between the PT detector load and the A88 input, this would have given a greater control over the detectors output range.

You may find that overall the output from the PT has a low voltage swing.

Does this help?

EDIT: downloaded the ATMEGA88 datasheet, its not very helpful at all, unless I have got a simplified version??
 
Last edited:
Yea ive gone through all this already and it also mentions an input impedance of 10K is what its designed for. But does what ambient is saying make sense because i observed the behavior he describes when i was testing the sensors in the lab without the atmega88. So does it mean that if i implement the circuit mentioned by ambient with resistance higher than 10k i will have problems with the ADC?
 
hi,
Ref to 'ambient' dwg, if you have a fixed resistor of 10K0 in series with say another 10K variable, then if you turn the pot to say max, total 20K0 then you are working outside the input impedance limit for the AT88.

This will not harm the AT88, but will effect the ADC conversion values to some extent. If you chose a higher value pot it makes it worse.

The voltage across the combined fixed/variable will of course increase.
If you can accept the effect of the acquisition time change ???
its no problem??

EDIT: Out of interest, why are you using the ADC to detect a 'pulse' from the PT detector when the satellite is detected???
 
Last edited:
Status
Not open for further replies.

Similar threads

J
  • Question
Where to add a potentiometer to vary input voltage?
Replies
15
Views
1K
Jim Kailey
J
A
  • Question
Using latches to capture 8086 bus
Replies
3
Views
846
rjenkinsgb
U
  • Question
Help to identify a component
Replies
4
Views
2K
Nigel Goodwin
C
  • Question
Is there a way to convert SGL8022W touch switch to momentary switch?
Replies
4
Views
767
call_sign_diesel
C
K
  • Question
Want to convert from 9v alkaline to solar rechargeable
2
Replies
20
Views
2K
ZipZapOuch
Menu
Log in
Register
Cookies are required to use this site. You must accept them to continue using the site. Learn more…