Positive/Negative power supply circuits

[NHSA]

Hi,

I need Positive/Negative power and have been looking at circuits in Internet and I am a bit confused about the value of the two used resistors for the voltage splitter.

The values go from 4K7 up to 220K and so my question is, how do I decide which resistor I use ?

Do this values depend on the current consumption of my attached circuit ?

There is a big difference between using 2x 4K7 and 2x 220K in regard to the current flowing when in rest but what about if my current is 100mA and the resistor to the common ground is 220K. How do I get the 100mA when a 220K resistor lets flow for example in a 12V divider which is 6V / 220K = 0.027 µA, not even milliamp.

Also the used capacitor values do vary a lot from let me say 10 µF to 220 µF. I guess here this is due to eventual spikes in current consumption and the capacitor do deliver the necessary spike current, correct ? The bigger the better the spike will be covered, correct ?

Thanks in advance
Rainer

[rstofer]

42 views and no replies.

Maybe a link or a schematic would get more feedback.

[TimFox]

No offense intended, but this must be the most ill-posed question here in years.
No mention of voltage or current requirements.
No mention of what the resistors in question do in the circuit.
No mention of the circuit.  Hint that one was attached, but no link or attachment.
Etc.

[Ian.M]

Potential dividers make *LOUSY* rail splitters.  To be even marginally useful you need an order of magnitude more current through the divider resistors than you can draw from the tap (the node between them).  For any pretense at precision you need two or more orders of magnitude more current through them.    This limits them to small fractions of a milliamp load current, preferably only a few microamps.

If you need more than that you need an active rail splitter.  200mA load would be well into active rail splitter territory, and depending on the average total power may even be more than is reasonable from a linear rail splitter, demanding a switched mode solution.

[NHSA]

Hi,

I apologize and yes Tim Fox is right. This was the most Ill-posted question here and no excuses.

On the other side reading TimFox answer ,scares away any beginner from asking anything, Thank You.

In order to answer TimFox question.

Power supply is 12 V and I want +6V and -6V as well as a common ground ...

For the next question I wrote in my first post " ... two used resistors for the voltage splitter. " OK, it was just a hint for the circuit.

Then I also wrote about the capacitors ... another hint but OK ... it was ill posted --- I admit it

OK. let us forget it because I got a satisfactory result after 42 views and 3 answers     and a lot of experimenting (exchanging trim pots with different values and capacitors) and measuring the output, I know it is not highly professional nor the voltage is stable like a three legged chair, but it serves my purposes. I am not measuring the cardiac frequency of a human 

I took a 1K trim pot and two 220µF capacitors. I have voltage variations but it works and the total power consumption of the attached circuit is 20mA. While looking at the attached circuit working I adjusted the trim pot in order to get nearly equal voltages for the op amps I am using. The biggest consumer is the 10 LED bar for the output.

The circuit I was talking about is the one depicted below (not with the resistor and capacitor values depicted in the image)

I used a 1K trimpot adjusted in order to get nearly similar voltage output which swings ... but acceptable for me and the biggest capacitor I have which proved to stabilize the voltage swing better then the 10µF capacitor I used before ...

Thanks and regards Rainer

PD, @TimFox in future, if there is a future for me in electronics     , I will try to give more info from the start on ... 

and thanks to rstofer and Ian.M for the teaching answers.

[TimFox]

I meant no offense, but it is important to pose a question with enough detail to get a meaningful answer.
Your circuit obviously works to develop equal voltages with respect to common or ground, with no load current or equal load currents on both outputs.
What output current is required for your circuit, and how much difference is there in the two currents?
If you are powering an op-amp that drives a high-resistance load, the simple resistor divider works well.
However, any output current from the op-amp to ground will unbalance the currents from the positive and negative terminals, which will unbalance the voltages.
Whether this is important or not is a quantitative question for the design.  The capacitor values are not important if the input voltage is "clean" (for example, from a battery), but the capacitors may be useful if there is a strong AC current at the op-amp output.  These are more details that are required to give a complete answer to your question.

[NHSA]

I meant no offense, but it is important to pose a question with enough detail to get a meaningful answer.

Hi,

It is OK, I understand, we are all humans and different and not all answer questions in the same way ...

Attached the circuit in question which poses the difficulty that the second stage e.g. a LM3914 and a 10 LED bar unbalances the whole enchilada and is the one using the most current for driving the LED bar.

Here my first posts in regard to the developmnet of this circuit which is a vibration sensor.

https://www.eevblog.com/forum/beginners/problem-with-full-wave-rectifier-using-tlp082/msg3598826/#msg3598826

https://www.eevblog.com/forum/beginners/lm3914-led-bar-driver/msg3605845/#msg3605845

OK, the first test was using two 9V batteries and here one battery drains faster then the other one. OK, a solution is to just exchange positions once in a while. This device will not work 24/7 but is just a test device for astronomical mounts in order to see how much the tripod vibrates and so show the telescope user that the vibrating telescope is not a good thing.

On the other side two 9V batteries make the whole thing much bigger. I still have the power supply woth two batteries incorporated and will most certainly leave it there so the user can decide what power supply he uses. On the other hand as you cans ee there is also the possibility to piggy back it on an arduino in order to graph, if the user wnats to do it. In this case it will be better to use one power supply, a 12V power brick with a Y splitter in order to have the same ground and avoid eventual ground loops.

Then I came up with the idea to use a 12V power brick as nearly 100% of the telescope mounts are powered with 12V and so I decided to incorporate the 12V supply and there then I need to have a voltage splitter to get positive / negative voltage.

OK, I will keep experimenting.

Thanks everybody
Rainer

[TimFox]

If you can tolerate the wasted power in the pot, then you can adjust the pot to get balanced voltages under the actual load, so long as the unbalanced load currents are reasonably stable.
As a human, I can only answer a question when the questioner gives some information about what he needs.  If you were referencing an earlier post where you gave those details, you should cite that post.
You seem to not understand why I could not answer your question initially.  Re-read your original post by itself and ask yourself how you would start to answer it.

[NHSA]

If you can tolerate the wasted power in the pot, then you can adjust the pot to get balanced voltages under the actual load, so long as the unbalanced load currents are reasonably stable.
As a human, I can only answer a question when the questioner gives some information about what he needs.  If you were referencing an earlier post where you gave those details, you should cite that post.
You seem to not understand why I could not answer your question initially.  Re-read your original post by itself and ask yourself how you would start to answer it.

Hi,

... then you can adjust the pot to get balanced voltages under the actual load, so long as the unbalanced load currents are reasonably stable.

That is what I did and it is acceptable for my output result.

My question really was why so many same circuits with so many different values for the resistors as well as the capacitors but playing around with different values answered my question.

OK, ok, you are right. I learned something about how to better express myself and next time I will try to give more information.

Have a nice weekend.

Rainer

[TimFox]

I would guess that the different examples you found varied from each other by the current levels and imbalances.  The difference in currents with a resistive derived ground will shift the voltages according to the resistors.  Again, it's all quantitative.

[Rick Law]

NHSA, I think you are missing one or two pieces of the puzzle causing it to be difficult to understand.

On both the top and bottom, you have 10K so the voltage is evenly divided.  When you use a DMM to measure, you will get approx 6V for both bottom and top side, that is all fine and good.  But once a load is connected, the load changes the combined resistance.  So, your top side resistance may not equal bottom side resistance.  As a result, the divider may no longer be dividing it to half each.

Because of such changes, the easiest way probably is to use a potentiometer and adjust it by actually measuring the divided voltage after the loads are attached.

But, problem is not totally solved yet.  The load may draw different current during run time, so the division may go out of wack.  So, it could be a moving target.  In a nut shell, the bigger the resistance in the potentiometer, the less current will be going through the potentiometer (dividing resisters).  You would be wasting less, but it also means less effective the potentiometer controls the division as the loads' current drawn changes.

If you like to understand that in more detail, you have to look at the math:

Focusing on the top side (your +6V side) to explain the issue:

Say you connect a load to to the top side, and let say this load always takes exactly 500mA (0.5A) and you are running it at 6V.
V=IR, so R=V/I.  Your load has an effective resistance of R=6V/0.5A=12ohms.

Once you connect this load to the top side, you can see that this 12ohms resister (your load) is in parallel to the the 10K resister and the capacitor.  The capacitor also in parallel has impedance/resistance, but leave that out for the time being.  So you are left with 12ohm paralleling your 10K resister in the divider.  It is no longer 10K anymore.

To calculate the effective resistance of 10K and 12 ohm in parellel:
R = 1 / (1/10K + 1/12) = 1 / (0.0001 + 0.0833) = 1/0.0834 = 11.9904 ohms

Your voltage divider on top went from 10,000 ohms to 11.9904 ohms by connecting this load.  You are no longer evenly dividing the voltage.

Now, if both your top and bottom load draws exactly 0.5A, both top and bottom are 10K parallel with 12ohm, they are both 11.9904ohms and you are evenly divided.

But -- Let say the bottom load draws only 100mA and the top load is the same, drawing 500mA as example above.

The bottom load with 100mA (0.1A) draw at 6V is R=V/I = 6/0.1 = 60 ohms
You can see that R=1/(1/10000 + 1/60) = 1/(0.0001+0.0167) = 1/(0.0168) = 59.5238 ohms

So, you can picture your divider (assuming those loads always draw exactly the same current, 0.5A and 0.1A), the top side resistance is 11.9904 and the bottom is 59.5238.

Top side voltage would be:
V = 12*11.9904/(11.9904+59.5238) = 12*11.9904/71.5142 = 2.0120V
Bottom side voltage would be:
V = 12*59.5238/(11.9904+59.5238) = 12*59.5238/71.5142 = 9.9880V

Totally out of wack with each have 1/2 of 12.  You can change that by altering the top and bottom side resistance and re-balance it to 6V each.  But once the current drawn by either the top/bottom side changes, the voltage will again change.  The more current goes by way of the dividing resisters (instead of the loads), the lesser the loads would alter the effective resistance of the divider hence less change to end voltages of the division.  But of course, the more current going through the resisters in the divider, the more power you are wasting.

I hope this is helpful in your understanding how this works...

[NHSA]

NHSA, I think you are missing one or two pieces of the puzzle causing it to be difficult to understand.

On both the top and bottom, you have 10K so the voltage is evenly divided.  When you use a DMM to measure, you will get approx 6V for both bottom and top side, that is all fine and good.  But once a load is connected, the load changes the combined resistance.  So, your top side resistance may not equal bottom side resistance.  As a result, the divider may no longer be dividing it to half each.

Because of such changes, the easiest way probably is to use a potentiometer and adjust it by actually measuring the divided voltage after the loads are attached.

But, problem is not totally solved yet.  The load may draw different current during run time, so the division may go out of wack.  So, it could be a moving target.  In a nut shell, the bigger the resistance in the potentiometer, the less current will be going through the potentiometer (dividing resisters).  You would be wasting less, but it also means less effective the potentiometer controls the division as the loads' current drawn changes.

If you like to understand that in more detail, you have to look at the math:

Focusing on the top side (your +6V side) to explain the issue:

Say you connect a load to to the top side, and let say this load always takes exactly 500mA (0.5A) and you are running it at 6V.
V=IR, so R=V/I.  Your load has an effective resistance of R=6V/0.5A=12ohms.

Once you connect this load to the top side, you can see that this 12ohms resister (your load) is in parallel to the the 10K resister and the capacitor.  The capacitor also in parallel has impedance/resistance, but leave that out for the time being.  So you are left with 12ohm paralleling your 10K resister in the divider.  It is no longer 10K anymore.

To calculate the effective resistance of 10K and 12 ohm in parellel:
R = 1 / (1/10K + 1/12) = 1 / (0.0001 + 0.0833) = 1/0.0834 = 11.9904 ohms

Your voltage divider on top went from 10,000 ohms to 11.9904 ohms by connecting this load.  You are no longer evenly dividing the voltage.

Now, if both your top and bottom load draws exactly 0.5A, both top and bottom are 10K parallel with 12ohm, they are both 11.9904ohms and you are evenly divided.

But -- Let say the bottom load draws only 100mA and the top load is the same, drawing 500mA as example above.

The bottom load with 100mA (0.1A) draw at 6V is R=V/I = 6/0.1 = 60 ohms
You can see that R=1/(1/10000 + 1/60) = 1/(0.0001+0.0167) = 1/(0.0168) = 59.5238 ohms

So, you can picture your divider (assuming those loads always draw exactly the same current, 0.5A and 0.1A), the top side resistance is 11.9904 and the bottom is 59.5238.

Top side voltage would be:
V = 12*11.9904/(11.9904+59.5238) = 12*11.9904/71.5142 = 2.0120V
Bottom side voltage would be:
V = 12*59.5238/(11.9904+59.5238) = 12*59.5238/71.5142 = 9.9880V

Totally out of wack with each have 1/2 of 12.  You can change that by altering the top and bottom side resistance and re-balance it to 6V each.  But once the current drawn by either the top/bottom side changes, the voltage will again change.  The more current goes by way of the dividing resisters (instead of the loads), the lesser the loads would alter the effective resistance of the divider hence less change to end voltages of the division.  But of course, the more current going through the resisters in the divider, the more power you are wasting.

I hope this is helpful in your understanding how this works...

Hi Rick,

Thanks. Very helpful.

So the problem in my circuit is the LM3914 with the attached 10 LED bar. Using it in DOT mode keeps the current draw more or less constant. If I now would switch to Bar mode then I would get into a bigger problem because every time an additional LED is switched on it would draw additional current e.g. when one LED is on it draws 10mA and every LED switched on in Bar mode would add 10mA and so when all 10 LED are lit then I would be drawing on that side 100mA.

So at the moment I am using a 500 Ohm trim pot 1/2 Watt and my PSU is telling me it draws a total of  ~30mA when it works signaling my vibration. I measured the total current draw just through the 500 Ohm, well my PSU does that, and I get 20mA and when the whole circuit is running it has 30mA and so it looks like everything is OK. 20mA waste plus 10mA LED lit gives 30mA.

After going down to the 500 Ohm trimpot my voltage also swings less and the circuit does what I expected. I know perhaps the op amp is working a bit weird as it gets variable positive and negative voltage but looking on the result of the amplification and the full wave rectification the result, I repeat myself, is acceptable.

Today I will measure how much does the op amp half draw in current and how much does the LED circuit draw to get a better understanding. Will also use my Frequency generator with a fixed frequency and so see better what is going on.

Will see what it does if I feed the LED circuit directly from the input with 12V and not from the positive half. I guess that could be a solution for avoiding the voltage swinging for the op amp part. Did not think about that. It just came to my mind.

Rainer

[tkamiya]

Hi....

As already mentioned, use of registive divider as such doesn't work very well in practice.  Unless the load on PLUS side and NEGATIVE side is exactly identical, output voltage won't be equal half.  Also, it won't hurt anything but capacitors are not necessary.

If I were doing this, I will just use two 6V batteries, or create a proper + side and - side regulated power source. 

[Rick Law]

A better way to get two 6V out of your 12V  power supply would be to use a pair of adjustable buck converter connected in parallel to the 12V.  Alternatively, you can use a pair of voltage regulators in place of the buck converters.  In both cases, you have your 6V output accurate to the accuracy of your selected buck or selected regulator.  5% can be easily accomplished.

The buck converters step the voltage down from 12V to whatever voltage you set (below 12V), in your case, 6V out.  Adding noise, but rather efficient.  You can find cheap buck or buck+boost on aliexpress for under $2USD (not including shipping and tax and stuff like that).   Buck+boost can actually boost below 6V input to 6V output.  Buck/boost is a power game.  You input enough power (power=current*voltage), it can output the voltage at that power (minus efficiency factor).  Boost would be much lower in efficiency.  Buck typically achieve better than 90% but boost would be far less efficient at 80% 70%-ish depending on how much you are boosting.  If your input voltage is very low (like around but barely above minimum required, you may see below 50% efficiency.

If in place of buck, you use linear voltage regulator, you have less noise in the output but far less efficient.  The voltage regulator in a manner act like a self-adjusting resister leaving the output to always 6V (or whatever the regulator output is designed to be, and in your case, you would make it 6V).

So while buck converts your 12V to 6V typically in the 90% plus efficiency.  Regulators on the other hand just burn away the power leaving you with 6v from the 12v input, that means half your power goes into warming your house with the regulator being the mini-space-heater.  So, make sure you have adequate heat-sink or you may end up giving the regulator a heat-stroke and kill it.

I would suggest using just buck if you don't plan to play around with it too much.

If you are playing with it (switching loads) much, use buck+boost with voltage adjustment and current-limit adjustment.  They are often marketed as CC/CV, with CC standing for Constant Current.  CC/CV would likely more than double the price, but provide a level of safety.   CC stands for Constant Current but it really isn't constant current - it would only be constant when your current draw is exceeding the setting, then the CC part held the current at the "constant" limiting level so it really is an upper current limit.  The CC part saved me from burning things out too often.  I set the current limit to be just a little above what I should be needing.  When the "CC" led on the boost/buck board lights up, it is telling me I am at current limit (which is above what I should need).  It is kind of like a warning led - I did something wrong so I am drawing more current than I should be needing...  So I can power off and go hunt for the error before something worst happens.

EDIT: typo correction, it should be "above" instead of "about".  Corrected "about what I should need" to "above what I should need"

[NHSA]

Hi tkamiya and Rick,

Thanks for the ideas. I already used two 9V batteries but it is too bulky. Works fine, one battery drains faster then the other one but here just exchanging them from time to time would help.

I like to learn more and therefore I came up with the idea of using  a 12V power brick and split it. Works OK now with slight voltage jumps but acceptable.

Will see what it does if I feed the LED circuit directly from the input with 12V and not from the positive half. I guess that could be a solution for avoiding the voltage swinging for the op amp part. Did not think about that. It just came to my mind.

The above idea did not work out. Looks like I get a negative voltage earth even being the power take off before the voltage splitter     Well, does not matter.

Will keep playing arond with this.

Rainer

[Rick Law]

Playing with it and learn is a great way to approach...  Besides playing around with voltage regulators, allow me to give you a couple more ideas to play and learn...

You could use a zener diodes.  6V zener diode is available.  LED's also have a "constant voltage" property up to a point.

If you use two 6V zener diodes instead of two 10K ohms...  Each zener will try to hold the voltage to 6V giving you a much nicer divider.  Once when a zener exceeds it designed voltage, it will "break down" and let current just pass on.  Look at this link to see how zener diodes can be used to regulate voltage.
https://www.electronics-tutorials.ws/diode/diode_7.html

Now for something more fun...

I said LEDs has a "constant voltage" property up to a point as well.  When the current is high enough for an LED to light, increasing the current doesn't increase the voltage linearly like a resister would.  It would increase voltage by some but not as much - till current is so high that it burns out.  So, a regular LED can be use as an improvise voltage regulator - not well, but to some degree.

Say you find yourself four 3V LEDs.  Connect the 4 in a series, top 2 adds up to 6V and bottom two add up to be 6v.  So you would have about 6V on the top two LEDs and about 6V on the bottom two LEDs. If your loads are not drawing a lot of current and they are rather close in current needs, that could work a lot better than dividing resistors.   Typical blue and white LEDs are in the 3.2-3.7 volt range, they will serve well in this experiment.

Now the fun part in making it work...

If you get some of those 700mA+ LEDs (over a dollar each instead of 5 cents), you wont need by-pass as all, but for the cheap 5-cents LEDs, you will need the by-pass.  Read on, you will understand.

To reuse my earlier example of 0.5A top load and 0.1A for bottom load, you have a delta of 0.4A.  So your two LEDs at the bottom must be able to carry that 0.4A that the 0.1A load isn't using, or your bottom LEDs will burn-out and probably cause other issues at well.  But, we need certain minimum going via the LEDs on the top load.  Figure typical 5-cents LEDs are about 1/4W to 1/8W, at 1/8W, and takes about 20mA to light up (bring it up to voltage), so the top load is about 0.52A.   Each of the two bottom LED in series are 3V, and it needs to carry that 0.42A excess current that the bottom load doesn't use.  Typical 5 cents LEDs are 1/4 or 1/8 watt, about 40mA max. 420mA excess is way too much for them to carry.  So one would need a by-pass resister so the excess current is not all for the LEDs to carry.  Bottom two LEDs adds to 6V, bypass current is 0.42A,  a 14 or 15ohms resister should do it nicely.

So, your top 10K ohm running the 0.5A load (the larger) is replaced by 2 typical blue (or white) LEDs (3.2 to 3.4 volt typically) in series.  3.4V+3.4V putting it to about 6.8Volt max.  Your bottom 10K with the smaller load is replaced by 2 blue LEDs in series then parallel by a 15ohm for by-pass...  This would work a lot better than a 10K at top and a 10K at bottom voltage divider for 0.5A top load and 0.1A bottom load.  Your top and bottom are both regulated to around 6.8V max, and bottom one use 0.4A less than top.  So unlike the bare 10K resister which changes the bottom to > 9Volt (as seen in my example in the earlier reply), this will fix it to around 6.8V max at the bottom.

Of course 0.5A and 0.1A are my examples and not your actual loads.  You will have to redo the math and do some experimentation with your actual load's current range -- that would be the fun part.  Depending on your load's current requirement delta and variation, this may not work, but trying and experimenting is the fun part.

Lots of ways to play with that to have fun and learn...

EDIT: I better add this:
If you do try this, make sure that your loads can handle some excessive voltage.  During the experimentation, you would also be burning out some of the LED's you use for voltage regulation.  So be prepare, you will kill some LEDs and you may damage your load.

[NHSA]

Hi Rick,

WOW and thanks   

You could use a zener diodes.  6V zener diode is available.

I was also thinking in that after I read your comment using voltage reagulators, and my zener diode kit just arrived by the friendly smiling company.

I will do some tests with 5.6V zener diodes. Not sure if the 6.1V zener would work as I do not know what power bricks my friends have but a test with with my PSU will shed more light into this.

The LED idea I will need to read it a few more times and experiment with that. BTW, we amateur astronomers like to live in the dark, eye sight adaptation, and some White or Blue shining LEDs would destroy our dark ambient adaptation. Need to check what voltage are the Red LEDs. Red light does not destroy our dark eye sight adaptation.

As this vibration detector detects really very fine vibrations, I am testing it with my rolling char on a hard flor and yes it detects and at that levels of produced voltage and amplification of the piezo sensors I see no voltage jumping.

Rainer

[NHSA]

Hi Rick,

If you use two 6V zener diodes instead of two 10K ohms...

Tested it with different Zener diodes starting with 1N4734 (5.6V) and then ended with 1N4732 (4.7V) and that one works fine.

With the 1N4734 = 5.6V I had on the positive feed where the LM3914 and the 10LED bar is connected a voltage drop down to 4.6V.

Then I tested the 1N4733 and still had a drip, less but significant.

The one which works OK is the 1N4732 but still ahve a Voltage drop of about 50mV.

The circuit is a 100 Ohm resistor in series with the two zener diodes. My PSU reports a current draw of 16mA which is more or less the total draw of the LM3914 plus the attached 10 LED bar.

Maybe I am missing something     . Adding two 220µF capacitors reduces the voltage drop to about 40mV.

So far I like it because it means less bulky parts   

Rainer

[Rick Law]

First, the LED as regulator is improvisation - more for fun and learning but less practical.  Being an improvisation, it offers more opportunity to learn and play.  The zener or typical voltage regulator IC would be far more practical and easier to implement.

Now to the main point.

The simplest solution really is just to use a pair of regulators (zener, or a regulator IC like the 78xx, ams1117, something like that but for 6V) and run them in parallel.  Use something like 9V power brick, two regulators in parallel and each regulate it down to 6V for its load.  Don't try to stack them to "split" the voltage.  Stacking them would create inter-dependency that complicate things.

If you are stacking two regulators like they are voltage dividers to split the voltage, bare in mind the same amount of current will go through the top 6V regulator+load combo and the bottom 6V regulator+load combo.  The one with the lower current load will be shunning more current via the zener and that needs to be considered in the design.

Stacking regulator IC is a different story.  The top part will be draining into the V+ of the bottom part.  So the bottom part has to be able to sink that current to ground -- via a voltage regulator.  Typically, for an amateur like me and perhap for pros as well, asking a regulator to sink current doesn't end well.

Either way, stacking creates interdependence that in-turn creates complications you don't need.

Now if you have them in parallel, the interdependence is minimal.  One circuit drawing some power may cause a slight voltage drop from the power brick and noise from one side may affect the other; but that is about it.

Voltage regulators always have a drop - that is kind of the ticket price you pay.  I found the regulator IC to be easier and more stable particularly with high variation in load current.  Since your 5V seem good enough for your load, try a 7805 or ams1117, they can hold 5V and needs just a few parts.

By the way, what size telescope are you using?

[NHSA]

Hi Rick,

Thanks. Will study this.

Below an image of my observatory.

regards Rainer

[Rick Law]

Good size equipment - don't screw around too much with your power for the sensors.  You don't want to have false alarms because your voltage divider (or stacked zener divider) is out of wack due to changing current drawn.

I am not very experience with EE or amateur astronomy, but I did enough to understand your battle.  I have a Meade 90 refractor (1250mm focal) and even with that I had vibration issues being >50 feet from nearest residential road.

If I were you, I would use a good power-brick, use an IC regulator like the 7805, ams1117, or something like that.  Step down to 5V from 8V-9V if you can avoid 12V to reduce heat...  If you can wait, just get a buck-board (only about 2$ USD purchasing from the USA before shipping and handling).  Bucking at less than 1Amp would have very little thermo where as a linear regulator is a bit hotter.  Last thing you need is to add another rising stream of warm air near the equipment, or worst yet, another heat sink fan creating vibration and more thermo instability.

You should also consider those cell-phone chargers that plugs directly into the wall with micro-usb output.  Just cut the USB plug and you have 5V there at hand.

Hey, good luck with it.  When you get that working and if you do a lot of longer exposures, you should probably consider using an ardunio or something like that so vibration and other detectable problems could be logged.

[NHSA]

Good size equipment - don't screw around too much with your power for the sensors.  You don't want to have false alarms because your voltage divider (or stacked zener divider) is out of wack due to changing current drawn.

I am not very experience with EE or amateur astronomy, but I did enough to understand your battle.  I have a Meade 90 refractor (1250mm focal) and even with that I had vibration issues being >50 feet from nearest residential road.

If I were you, I would use a good power-brick, use an IC regulator like the 7805, ams1117, or something like that.  Step down to 5V from 8V-9V if you can avoid 12V to reduce heat...  If you can wait, just get a buck-board (only about 2$ USD purchasing from the USA before shipping and handling).  Bucking at less than 1Amp would have very little thermo where as a linear regulator is a bit hotter.  Last thing you need is to add another rising stream of warm air near the equipment, or worst yet, another heat sink fan creating vibration and more thermo instability.

You should also consider those cell-phone chargers that plugs directly into the wall with micro-usb output.  Just cut the USB plug and you have 5V there at hand.

Hey, good luck with it.  When you get that working and if you do a lot of longer exposures, you should probably consider using an ardunio or something like that so vibration and other detectable problems could be logged.

Hi Rick,

Thanks. BTW, this vibration sensor in reality is not meant for me. My two mounts reside on concrete piers and each pier has a weight of about 7 tons ~ 15400 lbs and I ahve no vibration, it is more ment for friends who have their portable mounts and I want to show them how everything vibrates when it is not heavy enough.

The heat development of this circuit we can forget it.

Now I am using this pos/neg voltage supply using two zener 1N4732A and 1 resistor with 100Ohm.



Making a circuit like two voltage regulators as in the image below and joining together the center plus-minus poles does not work. Maybe I am too dumb in thinking that that would work but now I have to ask why does it not work ?   

[Rick Law]

...
Making a circuit like two voltage regulators as in the image below and joining together the center plus-minus poles does not work. Maybe I am too dumb in thinking that that would work but now I have to ask why does it not work ?   
...

Nah...  it is not because you are too dumb, no dumb person can operate the telescope you got.  I tried that too as experiment, but as I said in earlier post, voltage regulators just isn't designed to be good current sink.

"...each pier has a weight of about 7 tons..." wow, you are a serious astronomer!

Enjoy your nights.  The stars are waiting...

[Zero999]

The simplist solution is to use a rail splitter IC, such as the TLE2426. It can output more than enough current for your application and is much easier, than messing around with a switching regulator. It's also possible to do a similar thing with an op-amp, but it's much easier to use a dedicated IC.


https://www.ti.com/product/TLE2426

[Terry Bites]

Exactly!