V+ and V- from 24v dc - Page 1
Forum section: electronics / beginners

author: loop123
author_rank: Frequent Contributor
timestamp: January 19, 2024, 11:00:09 pm
content: Hi,Is it easy to create +12v and -12v from 24v dc source to supply amplifiers with V+ and V- for the voltage swing? What methods are usually used?

author: Zero999
author_rank: Super Contributor
timestamp: January 19, 2024, 11:14:41 pm
content: Yes, the easiest way is to use a rail splitter/virtual ground circuit.The TLE2426 is an ICE, specifically designed for this.https://www.ti.com/lit/ds/symlink/tle2426.pdf]Another way is to use an op-amp.How much current does your circuit need?Is it just AC output/input such an an audio amplifier? If so, use a single supply topology and AC couple the signals.

author: MarkT
author_rank: Frequent Contributor
timestamp: January 19, 2024, 11:33:04 pm
content: A DC-DC converter typically....

author: brucehoult
author_rank: Super Contributor
timestamp: January 19, 2024, 11:54:50 pm
content: It's all relative. If nothing has to be connected to real ground then you can call your two 24V DC terminals 0 and +24, or -24 and 0, or -12 and +12, or -5 and +19 or whatever you want.If you need some intermediate voltage at significant current then you can use two resistors in the right ratio to make a reference voltage, and connect that to the input of a sufficiently beefy opamp in voltage-follower configuration (i.e. connect the output to the -ve input).

author: themadhippy
author_rank: Super Contributor
timestamp: January 19, 2024, 11:55:35 pm
content: if the current demand is low a just 2 resistors will do the job

author: TimFox
author_rank: Super Contributor
timestamp: January 20, 2024, 12:12:56 am
content: The circuits discussed by Zero999 above work well for many applications.Note that the op amp or similar device needs to supply the difference between the currents at +Vout and -Vout.

author: loop123
author_rank: Frequent Contributor
timestamp: January 20, 2024, 01:39:40 am
content: Quote from: brucehoult on January 19, 2024, 11:54:50 pmYou mentioned -5 and +19V. Here is the strange thing.In the Din output, there is a -7.2V and +7.2V DC. This occurs when 2 sets of 7.2V rechargeable battery packs are installed). But when the battery packed are removed. The terminal measures 3.96V and -18.20V with respect to terminal ground (the unit is not connected to the real ground in the AC outlet)! Sum of the 3.96V and -18.20 is 22.17V (closed to the 24v main DC source of the circuit) which is what I measured if you will directly measure the terminals. So where did 3.96V and -18.20 come from?

author: loop123
author_rank: Frequent Contributor
timestamp: January 20, 2024, 10:59:42 am
content: Instead of using circuits to create +Vs and -Vs to power op-amps, etc. and swing theem. Would it be easier to just use batteries. + and - for the +Vs and another - and + battery for the -Vs?

author: Andy Chee
author_rank: Super Contributor
timestamp: January 20, 2024, 11:05:26 am
content: Quote from: loop123 on January 20, 2024, 10:59:42 amYes, this is very easy solution.However, depending on the circuit, it may draw power unevenly from each rail. One battery may go flat first.

author: brucehoult
author_rank: Super Contributor
timestamp: January 20, 2024, 12:02:50 pm
content: Quote from: loop123 on January 20, 2024, 10:59:42 amSure. Two 12V batteries, or two 12V floating power supplies, or a 24V transformer with a center tap -- would all be easy solutions. But the OP says he has a 24V supply. There was no mention of actually needing a 0V connection, but it seems probable it is wanted. If not then all is very simple.

author: Zero999
author_rank: Super Contributor
timestamp: January 20, 2024, 03:57:43 pm
content: The 0V/earth/ground part of a circuit is somewhat arbitrary. It's just a reference point, from where all other voltages are measured from. Here's a circuit showing two 9V batteries connected in series, with the 0V reference connected to different points. Note that the voltage i.e. potential difference between all of the points remains the same.

author: loop123
author_rank: Frequent Contributor
timestamp: January 30, 2024, 08:27:04 am
content: Guys I encountered a circuit where the potential difference of the wires 6 inches away are totally different. How can this occur? If you measure top battery directly. It's 7.95V, but 6 inches away with plain wires (all continuous from battery) to the circuit terminal. The terminal measures 3.95V only. How can this be possible?And what kind of voltage converter can produce this effect where they use the same rails for Vs, -Vs and even ground yet the voltages 6 inches away vary without any significant resistance different from a typical ordinary wire.

author: Zero999
author_rank: Super Contributor
timestamp: January 30, 2024, 10:40:44 am
content: The sum of the output volages is near enough the input 8 + 8 = 12 + 4, which is indicative of a bad 0V connection.

author: loop123
author_rank: Frequent Contributor
timestamp: January 30, 2024, 12:03:21 pm
content: Quote from: Zero999 on January 30, 2024, 10:40:44 amWhat you mean bad 0V connection? How do I find it? The entire pcb measures 3.95V from V+ to ground, and -11.95V from -V to ground. Yet it still works. Couldn't it be designed like that? What component can add or subtract the voltage from the main battery of two 7.95V with common like in the illustration.

author: Zero999
author_rank: Super Contributor
timestamp: January 30, 2024, 12:07:40 pm
content: Quote from: loop123 on January 30, 2024, 12:03:21 pmI mean the resistance between the 0V of your device and power supply is much higher than it aught to be. It might work if it hardly uses the 0V connection and doesn't need a regulated power supply, so 4V and -12V is enough for it to be functional.

author: loop123
author_rank: Frequent Contributor
timestamp: January 30, 2024, 01:00:38 pm
content: Anyway, how do you fix the high resistance between the 0V of the device and power supply of the above unit? (if it's really the problem).Also something puzzles me. When there are no batteries and I used only 24V ac/dc adaptor. What voltage should appear between +Vs and ground and between -Vs and ground if there is no problem of high resistance? I was getting the same 3.95V between +Vs and ground and higher -18.20 between -Vs and ground when only ac/dc adaptor used.Why the same 3.95V between +Vs and ground irregardless what voltage is inputted? Can I test using multimeter the high resistance between ground and the positive and negative rail?

author: Zero999
author_rank: Super Contributor
timestamp: January 30, 2024, 01:42:39 pm
content: Your ground connection must be non-existant.The voltages add up.When you used the mains supply, there was no ground. 3.95 + 18.2 = near enough 22.15V, as was the case with the batteries.If you connected it to the batteries again, you'll find the voltage between the battery 0V and the 0V in your device is around 4V.I don't know why the positive is always 3.95V because you haven't posted a schematic of the device, so it's impossible to know.You fix this by finding out where the ground wire is broken and reconnecting it. You need to remove the battery and power supply, before checking the continuity of the 0V connection with a multimeter.

author: loop123
author_rank: Frequent Contributor
timestamp: January 30, 2024, 02:04:20 pm
content: Quote from: Zero999 on January 30, 2024, 01:42:39 pmI opened the case and measured the resistances (I have to set up ESD pad, wrist band and wire to ground to avoid static (that's why it is not easy to just open it up and measure anytime).With all batteries disconnected.The resistance between ground and +Vs is 12.69 kOhmsThe resistance between ground and -Vs is 21.44 kOhmsThe resistance between +Vs and -Vs is 8.75 kOhms.Can the above tell you anything? What is supposed to be the resistance between ground and +Vs for example in normal circuit?Btw.. In an opamp where there is only +Vs, -Vs, how is it connected to ground if there is no ground?I'm still trying to trace it. I'm just careful because there would be many multimeter probing to avoid ESD as it costs $1600 new.

author: loop123
author_rank: Frequent Contributor
timestamp: January 30, 2024, 02:41:30 pm
content: Quote from: Zero999 on January 30, 2024, 01:42:39 pmI'm puzzled by most of your sentences. Why do you say the ground is non-existent? What should you expect if there is a ground?QuoteFirst this is the circuit of the charger inside it.By removing the battery. I exposed the circuit to much higher voltage of 22V instead of just 15V. Therefore I was worried some components could have broken (those that accept 15V or below).Why do you say the the voltage between the battery 0V and the 0V in the device is around 4V? The battery 0V and the device 0V is supposed to be connected. You mean with all batteries terminal connected. I'd remove the ground between them and measure the potential differences?QuoteWhat has this got to do with the ground wire being broken? It's all PCB copper inside and the chassis is connected to ground. You mean the PCB trace is broken? Why did you say this? What must be the normal voltages if there is no broken ground? And what has this got to do with the +VS to ground appearing 3.95V instead of uniform 7.95V?I'm concerned of it because the +Vs and -Vs must be both 7.95V to power the opto-isolator unit plugged to it at front.

author: Zero999
author_rank: Super Contributor
timestamp: January 30, 2024, 03:07:29 pm
content: Quote from: loop123 on January 30, 2024, 02:04:20 pmThat will tell you nothing.I meant, measure the ground connection between the device ground and the ground on the battery holder. Obviously do this, without any batteries in it.

author: loop123
author_rank: Frequent Contributor
timestamp: January 30, 2024, 03:22:24 pm
content: Quote from: Zero999 on January 30, 2024, 03:07:29 pmOf course the ground are connected because I bought a new pair of slots and soldered the wires and tested the continuity using multimeter.I followed the convention in the device where the ground is between the red and black in the 3 terminal Moldex KK connector. So I'm sure the ground between battery and device is connected with brand new wire.With this. What is your other theory? How could ground and supply have high resistance? Where to find it and fix it?

author: loop123
author_rank: Frequent Contributor
timestamp: January 30, 2024, 03:50:25 pm
content: Quote from: Zero999 on January 30, 2024, 03:07:29 pmMy God. Your theory worked! I tried what you did and there was no ground connection. I used the old Moldex connector and the ground pin must have corroded. I connected chassis ground to battery ground using jumper and it measured 7.95V and -7.95 alreadySo I didn't use the ground the whole time. But still. Why did +Vs measures 3.95V (and -V measures -11.95V) to chassis ground without battery ground? Does this also occur in other circuits, maybe because of differences in resistance between +Vs,-Vs and chassis ground?Too much bad I didn't ask you this earlier. I already cut my isolation module power to feed a balance power to it. Could have saved all the purchase of 4 slots and 24 batteries.My BCI project would be more delayed but good to go. Many thanks.

author: Zero999
author_rank: Super Contributor
timestamp: January 30, 2024, 05:43:50 pm
content: Quote from: loop123 on January 30, 2024, 03:50:25 pmExactly. In fact the whole point of the 0V connection is because the currents taken from the +V and -V rails differs.Remember the 0V node is just a reference point, from where all voltages are measured from. Here are the voltages, when there's no ground connection between the batteries and load. The top drawing is with the 0V point, i.e. the meter's black probe between the batteries and the bottom, with the black probe on the 0V point of the load.Connecting the 0V solves the problem.

author: loop123
author_rank: Frequent Contributor
timestamp: January 30, 2024, 10:23:47 pm
content: Thanks a lot!While it was -11.95V the whole week. If there is a component that is only rated up to -8V. Then could the component be damaged from overvoltage?When using just the charger and without the battery and the whole circuit seeing 22.15V with the negative rail reaching -18.2V, could it have damaged component that is rated up to say + - 15V only? I'm afraid the Isolation unit I connected to the front with ISO122 rated up to -15V may be damaged because it's not working.

author: Zero999
author_rank: Super Contributor
timestamp: January 30, 2024, 11:27:42 pm
content: The data sheet says an absolute maximum of ±18V, so it should be fine.https://www.ti.com/lit/ds/symlink/iso122.pdf

author: loop123
author_rank: Frequent Contributor
timestamp: January 31, 2024, 01:35:35 am
content: Quote from: Zero999 on January 30, 2024, 11:27:42 pmBut remember a pcb contains all other components. Usually what components can't survive past 18V (+, -)? Transistor? diodes? capacitors? resistors? ordinary ic? etc?

author: Zero999
author_rank: Super Contributor
timestamp: January 31, 2024, 02:36:10 pm
content: Quote from: loop123 on January 31, 2024, 01:35:35 amIt's impossible to answer that question without reverse engineering it to some degree.There are parts on the board which might be damaged by 18V, but it depends on how they're connected. For example there are parts which are only rated to 5V, but there appears to be a voltage regulator: VRx, the TO-92 part, on the left hand side of the board, near the red, yellow and brown wires cabled tied together.

author: brucehoult
author_rank: Super Contributor
timestamp: January 31, 2024, 11:58:06 pm
content: It's a type of package. Many different parts come in this package style.https://en.wikipedia.org/wiki/TO-92Looks like LM336, a 2.5V "reference diode".

author: loop123
author_rank: Frequent Contributor
timestamp: February 01, 2024, 10:54:13 pm
content: Guys. While you still remember a bit the circuit and before you forget them. I want to know if any noises can couple to the main unit and is it better to have a separate power supply for an isolation head that connects beween main units and electrodes? I'm fixing the connectors of the isolation head and need to decide whether to create separate dc pack for this instead of getting it from the main circuit shown earlier. Thank you.