Rheostat act as a potentiometer? - Page 1
Forum section: electronics / beginners

author: brownt
author_rank: Regular Contributor
timestamp: June 14, 2021, 01:06:26 am
content: Is it possible to make a rheostat act as a potentiometer with the addition of some components?A circuit provides 5 volts to a rheostat and the current is fed back to control something, but I want the same rheostat, with the addition of whatever is required, to provide 0 - 5 volts. The rheostat cannot be physically altered. The current available is in the microamp range.I suppose it is a current to voltage circuit that is needed. Is there some arrangement of resistors, or something that will make this possible, even to an approximate degree of a solution? Some voltage loss is ok.

author: ledtester
author_rank: Super Contributor
timestamp: June 14, 2021, 02:13:11 am
content: What's the resistance of your rheostat?Note that if you only have, say, 10 microamps, you'll need a resistance of 500K ohms to produce a voltage drop of 5V.

author: Brumby
author_rank: Unknown
timestamp: June 14, 2021, 03:04:28 am
content: This is the first question:Quote from: ledtester on June 14, 2021, 02:13:11 ambut there is a second question:What is the resistance of the load that will be utilising the 0-5V?.... and for completeness, a third question:What is the open circuit voltage and ESR of the "5V" source?Dave's latest video will take you directly into the things you will need to consider

author: brownt
author_rank: Regular Contributor
timestamp: June 14, 2021, 03:36:13 am
content: I see. I really don't know much about these things.The rheostat is 10kohm.The resistance of the load I am not sure but it might be about 10kohm.You can see the schematic here. It's a guitar expression pedal circuit.I would like to use a rheostat instead of a potentiometer because that's what is in the expression pedal that I have. I would prefer not to tamper with the pedal directly and instead wanted to make a conversion circuit. thanks

author: james_s
author_rank: Super Contributor
timestamp: June 14, 2021, 05:01:17 am
content: Rheostat and potentiometer are words that are typically interchangeable. Technically a potentiometer has all three terminals used while a rheostat is a simple variable resistor with only two terminals however I don't know that I've ever seen one with only two terminals, if you don't need the third you just leave it hanging or tie it to the wiper pin. Maybe you can elaborate on what you mean when you refer to a rheostat and a potentiometer?

author: WattsThat
author_rank: Frequent Contributor
timestamp: June 14, 2021, 05:16:50 am
content: A potentiometer has three terminals and is an adjustable, ratiometric voltage divider. A rheostat is a two terminal device that is adjustable from zero (end resistance) to the rated element resistance.You can use a potentiometer as a rheostat by using the wiper and one element connection. You cannot use a rheostat as a potentiometer as-is.The OP’s schematic shows a potentiometer. A rheostat would only work with current to voltage conversion or voltage level shifting.

author: Ian.M
author_rank: Super Contributor
timestamp: June 14, 2021, 06:07:10 am
content: It could be done in this specific application with a dual RRIO OPAMP, one half configured as a Howland current source feeding the rheostat to get a voltage proportional to its position and the other amplifying the voltage from the rheostat to get close to a full 0-5V output swing. However the OPAMP needs a regulated supply voltage (for the current source), and the 1K resistor in the effects unit's positive feed to the pedal (CTRL2) from AVDD is likely to be problematic.

author: brownt
author_rank: Regular Contributor
timestamp: June 14, 2021, 06:27:13 am
content: Hi thanks,Yes I mean a rheostat, one terminal tied to the wiper. Opamp yes I guessed so. However, there is negligible current available to power anything. So, if at all possible I guess it would be some type of parallel resistor arrangement, or diodes or ..... Hence the question because I do not know that the answer if there is one.But to repeat the question I would like to use a two-terminal rheostat rather than a potentiometer. There is virtually no current available to run any active circuitry and there is no access to the rheostat other than through its two terminals

author: Ian.M
author_rank: Super Contributor
timestamp: June 14, 2021, 06:50:02 am
content: In that case, there's little hope. Putting the rheostat in series with a fixed resistor to make a potential divider will either give an extremely non-linear response if you try for a large output swing, and draw significant current when near the min. resistance position, or if you make the fixed resistor a high enough value to get a reasonably linear response, you'll get a very small output swing, and a high output impedance.

author: Gyro
author_rank: Super Contributor
timestamp: June 14, 2021, 09:47:14 am
content: A constant current source could be used do the job. An LM334 has an adjustment range of 1uA to 10mA and a dropout voltage of 0.9V. You could maybe place this between AVDD and the top of the rheostat (in place of R2 in the schematic?).The LM334 just needs one current setting resistor. It would at least keep the output linear, even if you couldn't quite achieve full voltage swing.

author: brownt
author_rank: Regular Contributor
timestamp: June 14, 2021, 09:55:52 am
content: hmmmm that might do. Linearity is not a huge problem, and neither is full swing. Many expressionable effects don't sound good at full swing. I mean tremolo, phasing and others don't sound so good when they are wavering so fast. And also the multi-type effects often have a calibrate function so swing and linearity don't matter anyway as the calibration will spread it evenly and maximise the expression based on whatever swing is available.Good one Ian. Simple as. thanks.Now that you know how unexacting the circuit can be, are there any other simple modifications possible that might increase the swing or improve the linearity.

author: brownt
author_rank: Regular Contributor
timestamp: June 14, 2021, 09:57:21 am
content: Thanks. Are you saying that the few microamps that are available will be enough to power the opamp?I don't want to modify the effect unit, rather I am looking at an inline solution. Does that fit with what you are describing?

author: Gyro
author_rank: Super Contributor
timestamp: June 14, 2021, 10:31:15 am
content: Hmm, If you don't want to modify the pedalandyou don't want to modify the effects unit, then it does limit things. Looking at the schematic you just drew, putting the LM334 between tip and ring (in place of the 10k fixed resistor) ought to work. It would hopefully give you 0-4.1V (0.9V LM334 dropout voltage), if set to source around 500uA.

author: brownt
author_rank: Regular Contributor
timestamp: June 14, 2021, 10:46:34 am
content: I am not sure about that. From the data sheet I guess the diagram is the one I have attached, and using the formula I get about 470 ohms RSET. Is that correct?

author: Gyro
author_rank: Super Contributor
timestamp: June 14, 2021, 10:55:56 am
content: Yes, that looks right. Remember that there will be a tolerance on the LM334 and on the rheostat resistance so you might end up with a fine tweak for maximum output swing but the result should be better than a simple resistor.

author: brownt
author_rank: Regular Contributor
timestamp: June 15, 2021, 12:02:04 am
content: I see. Why 500uA?If the rehostat was a different value, would a different current be required?

author: Gyro
author_rank: Super Contributor
timestamp: June 15, 2021, 08:23:52 am
content: The 500uA to give approximately 5V across the 10k rheostat at its maximum resistance. I suppose more accurately it would be 410uA to account for dropout voltage of the LM334 (which will prevent the rheostat from reaching the maximum 5V) but the current setting only needs to be approximate anyway due to loading of the Effects unit - for instance there is an internal 470k resistor from the wiper connection to AVDD. The aim is just to get the biggest and most linear voltage swing.Yes for a different value rheostat, you would chose the current setting to achieve the desired voltage at its maximum resistance.

author: tooki
author_rank: Super Contributor
timestamp: June 15, 2021, 04:19:35 pm
content: Quote from: james_s on June 14, 2021, 05:01:17 amPlease don’t suggest the terms are interchangeable. They two words exist precisely to keep them apart.What’s true is that a) every potentiometer can be used as a rheostat, but not every potentiometer can be substituted with a rheostat, and because of this b) true rheostats are very rare.

author: brownt
author_rank: Regular Contributor
timestamp: June 20, 2021, 07:49:57 am
content: What is it about the attached circuit that prevents a 10kohm potentiometer from fully modulating an effect.The manufacturer recommends a 100Kohm potentiometer. My experiments show at least 20Kohm is required.Note that the change is not linear. For example, 10kohm does not quite fully modulate, but 15kohm is no better. 20kohm and above provide full modulation depth.

author: mikerj
author_rank: Super Contributor
timestamp: June 20, 2021, 10:09:54 am
content: Resistor R1 introduces a voltage drop that will be determined by the total resistance of the potentiometer. A lower value potentiometer will draw more current through R1, causing a larger voltage drop and limiting the maximum voltage attainable on the wiper.

author: brownt
author_rank: Regular Contributor
timestamp: July 05, 2021, 09:46:42 am
content: LM334's have turned up, so I am ready to try the circuit, but the calculation to determine Rset, eludes me now. I cannot understand how I calculated 470 ohms.Ignoring the voltage drop for now and so assuming 5 volts, how did I arrive at 470 ohms for Rset, using a 10kohm potentiometer?

author: Gyro
author_rank: Super Contributor
timestamp: July 05, 2021, 06:17:47 pm
content: Oops, yes. That value is wrong isn't it.Rsetis approximately VR/ Iset(by rearranging the simplified equation on page 5 of the datasheet). At a temperature of ~20'C, VRis about 63mV (graph on page 4 of the datasheet),The dropout voltage of the LM334 is about 0.8V, so with a 5V supply, you have a maximum possible of 4.2V across the rheostat. So you want the current to be 4.2V/10k = 420uA.This gives an Rsetof 63mV/420uA = 150 OhmThere are tolerances involved, both in the LM334, the rheostat track resistance, the supply voltage and the current setting resistor, but 150R is conveniently an E12 standard value, and will be good enough for practical use.

author: brownt
author_rank: Regular Contributor
timestamp: July 06, 2021, 02:09:32 am
content: I do not know what I am doing wrong here. I cannot get it to work at all.Even in its most basic configuration, I cannot measure any current.It must be something obvious.Here is a quick video demonstrating that it doesn't workhttps://youtu.be/orm8cXLy18II must be having a moment of the proverbial. Any help would be appreciated.thanks

author: Gyro
author_rank: Super Contributor
timestamp: July 06, 2021, 10:12:05 am
content: Oh dear, this seems to be one of those situations, it clearly should be working but it just isn't.Unless my eyesight has gone too far then it looks as if you have the pinout correct according to the datasheet. I don't know where you sourced the LM334s from, but if it is from a Chinese ebay seller then it is quite possible that all 10 could all be faulty - or rather re-marked something else.The other thing that comes to mind is that you are using a breadboard with various thickness wires. I never rule out contact faults. I noticed when you tested the setup with the 10k resistor, you touched the resistor to the red crock clip rather than the V+ device pin, so you haven't eliminated a disconnect there. Also, if either end of the resistor is disconnected, then the LM334 will drop to sub uA quiescent leakage current which wouldn't register on that meter range. For such a simple test I would normally just spread the legs of the device slightly and tack solder the resistor between the R and V- pins, everything is clearly visibly and solidly connected then.I was momentarily confused by the schematic as you have put -Ve at the top, but it is logically correct.Sorry, I can't give you a definitive answer, but itlooksright to me, so it must either be a device issue or dodgy connection.

author: brownt
author_rank: Regular Contributor
timestamp: July 06, 2021, 12:09:12 pm
content: Thanks, yes I did check continuity between all pins. Chinese yes. I was thinking that the pinout might have been non-standard, but no luck there.Maybe source some other ones.

author: Gyro
author_rank: Super Contributor
timestamp: July 06, 2021, 01:20:35 pm
content: Yes, get a refund and go to a local reputable distributor. This sort of thing has got to epidemic (pandemic) levels!

author: brownt
author_rank: Regular Contributor
timestamp: July 13, 2021, 10:44:56 am
content: Ordered in LM334 from RS, and they work.So, the 10 from aliexpress are all faulty or are not actually LM334Z.I opened them up, but nothing except sand inside. I guess they were not quite cooked.Ok, so now I have some devices I can get on with testing.

author: Gyro
author_rank: Super Contributor
timestamp: July 13, 2021, 01:09:45 pm
content: Quote from: brownt on July 13, 2021, 10:44:56 amQuoteOoh, photo please!I hope you got a refund - I realise it wouldn't be much but punishment of such behaviour is important if thing are ever going to change.