Calculating Current from a Bipolar Potentiometer - Page 1
Forum section: electronics / projects

author: Chris Willocks
author_rank: Newbie
timestamp: December 09, 2019, 06:38:23 pm
content: Hi everyone,I'm currently designing a CEM3340 (the datasheet and other info can be found here:https://electricdruid.net/cem3340-vco-voltage-controlled-oscillator-designs/) based VCO Eurorack synth module.Being a Eurorack module, the power supply is a split +12V/GND/-12V supply. I am using a 100K linear potentiometer to control the coarse frequency control voltage into pin 15 of the CEM3340, with -12V being fed into pin 1 and +12V into pin 3 of the potentiometer. Pin 2 goes through an 100K resistor to pin 15 of the CEM3340.I'm struggling to understand how to calculate the current flowing out of pin 2 of the potentiometer with it utilising both -12V and +12V? Is there a specific formula that can be used to calculate current for bipolar potentiometers or voltage dividers? The frequency decreases linearly to zero as expected when the potentiometer is turned from the 100% position (+12V) down to the 0% position (-12V). Would the current be 0A when the potentiometer is set to the 0% position i.e. -12V? The fact that the frequency decreases to zero when the potentiometer is set at the 0% position would indicate that the current is 0A or near to 0A.I understand that voltages are all relative to each other. Apologies if this sort of question has already been asked, but I've been searching for a day or so, trying to work this out and haven't had any luck.Many thanks,Chris Willocks

author: james_s
author_rank: Super Contributor
timestamp: December 09, 2019, 06:46:13 pm
content: Ohms law is all you need. The current flowing between the pot and the device is going to depend on the impedance of the input on the device it is feeding but I suspect it will be negligible. The fact that the supply is bipolar is irrelevant, you've got a 24V supply across a 100k resistor (the pot) so it's trivial to calculate the current flowing through the pot, assuming its driving a high impedance input.

author: Chris Willocks
author_rank: Newbie
timestamp: December 09, 2019, 08:58:05 pm
content: Thanks for the reply and explanation. I believe the input impedance of the CEM3340 is negligible.Just so I'm clear; for ease of calculating, would it be satisfactory to say that pins 1 and 3 on the pot can be treated as 0V and 24V respectively, instead of -12V and +12V?Therefore, assuming negligible impedance in the wire and into the IC, the total current across the pot i.e. pins 1 and 3 is:I = V/R = 24V/100K = 0.24mA.Therefore, when the tap is at the 100% position, as there is no resistance between pins 2 and 3, the current output through pin 2 would be 24A, however, this would have a mixing resistor of 100K in series with the output, so the current into the IC would also be 0.24mA, as per the above.When the tap is at the 0% position with the 100K mixing resistor again included in series with the pot, the current would be,I = 24V/(100K + 100K) = 24V/(200K) = 0.12mA.Does this sound correct or am I being stupid?Many thanks,Chris

author: james_s
author_rank: Super Contributor
timestamp: December 10, 2019, 02:00:54 am
content: Yes the current through the pot will be 240uA. For current to be drawn from the wiper it would have to be connected to a node sitting at some specific potential and then the current will depend on that potential relative to the pot setting and the source impedance of the node. In your case the 100k resistor in series is going to limit the max current, and I expect the impedance of the pin on the VCO is very high so any actual current flow will be negligible.

author: Chris Willocks
author_rank: Newbie
timestamp: December 12, 2019, 12:34:44 am
content: Okay, thanks. That sort of makes sense.Just to add a bit of context as to why I was trying to find out the current; the original schematic is powered by a +/-15V supply and pin 15 of the CEM3340 i.e. the frequency CV control, usually has a 100K pot (RV1) and 100K mixing resistor (R1), with a 360K pull-up resistor (R2) in parallel with these. R2 sets the base frequency of the VCO. The schematic for the +/-15V version is shown in the attachments below. I've omitted other components on the schematic for simplicity.I'm trying to convert this schematic to suit a +/-12V supply and in doing so, I want to maintain the same base frequency into the VCO i.e. by changing the value of the 360K pull-up resistor (R2) accordingly. I'm going to leave the values of the 100K pot (RV1) and 100K mixing resistor (R1) the same, as these set the overall frequency range of the VCO, which will be negligible between +/-15V and +/-12V. I'm more concerned about maintaining the same base frequency and mid-point frequency of the pot i.e. when the pot is halfway.Am I correct in saying that to find the equivalent value of R2 for the +/-12V supply, that all I have to do is reduce the overall resistance by x0.8 i.e. 12V/15V = 0.8?So in other words, find out the overall resistance of the +/-15V version using 1/Rt = 1/(RV1 + R1) + 1/R2 and find out Rt. Then reducing Rt by a factor of x0.8 and then subbing the new value of Rt into the equation and solving for R2.In doing this, I found the ideal value of R2 for a +/-12V supply to be about 210K or so. So I'd probably use the nearest value resistor available i.e. a 200K for the +/-12V version. I've attached the schematic for the +/-12V version as well below.Is this correct? Have I done my calculations correctly? I suppose current isn't really important and it's more about the ratio of resistances?Hopefully this makes sense?Many thanks,Chris

author: Chris Willocks
author_rank: Newbie
timestamp: December 12, 2019, 07:15:08 pm
content: Anyone?