# Easiest way to get 5v and ~1V

1. ## Easiest way to get 5v and ~1V

Hello. I am attempting to follow this circuit. I want to keep things small and cheap. How should I go about getting 5V DC for Vsupply and -1V DC (or less) for Vref? I was hoping to do this from a 9V DC source, but I am open to other ideas. I looked at voltage regulators but am not sure if I'd need a heat sink. If I went the battery route, I could use a 4/5AA for 1.2V, but that's a bit higher than I'd like, and also what would I do for the 5V? Please help a noob out!

2. ## Re: Easiest way to get 5v and ~1V

Hi,

A battery voltage varies....and the same magnitude in % your sensor signal will vary, too.
--> high error.

You want it simple...
Especially: where does the Opamp output signal go to?

Then I'd modify the circuit a little bit.
I'd use a voltage divider from ADC_VRef to get +1V. Add a capacitor to stabilize this voltage.
Then feed it to the Opamp+In (instead of GND).
Connect the sensor-VRef to GND (instead of -1V).

The benefit:
* most simple circuit
* no negative supply needed
* very precise

The drawback:
* You loose 1V of your ADC_input range. This sounds bad, but the gain in precision will compensate this. You will gain in overall performance.

Klaus

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3. ## Re: Easiest way to get 5v and ~1V

+1 What Klaus just mentioned.

Before the advent of rail to rail opamps, one had to procure a negative supply, a big hassle for battery operated products.

Nowadays, with the proliferation of R2R opamps, both input and output, that is no longer necessary.

As Klaus mentions, generate a "virtual ground" 1 volt above the battery's negative voltage.

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4. ## Re: Easiest way to get 5v and ~1V

Originally Posted by KlausST
You want it simple...
Especially: where does the Opamp output signal go to?
My apologies, I should have mentioned that. I want Vout of the opamp to vary between 0v and +5v which will connect to a Control Voltage input of a synthesizer audio effect, which will then control a specific parameter depending on the effect (e.g. volume, speed of an LFO, cutoff frequency of a low-pass filter). Therefore, the amount of pressure you apply to the force sensor will control the output voltage from 0v to +5v.

I thought I could get away with not having to use negative voltage, but in the notes of that circuit it says that Vref should be opposite polarity of Vsupply. There is another recommended circuit (here) which is single source and supposedly has an equally linear response. This looks like it more resembles the modified circuit you described. Is that so? For this one I would just need e.g. +1V for Vref and Vsupply I suppose should be 5v and used with a rail-to-rail output op amp such as MCP601?

5. ## Re: Easiest way to get 5v and ~1V

Hi,

The circuit is exactly what I suggested. Did you recognize it?

Klaus

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6. ## Re: Easiest way to get 5v and ~1V

Originally Posted by KlausST
Hi,

The circuit is exactly what I suggested. Did you recognize it?

Klaus
Originally Posted by KlausST
Hi,

The circuit is exactly what I suggested. Did you recognize it?

Klaus
I partly recognized it. I know you are right, but to my newb eyes I can't reconcile some things.

1) The capacitor C1 is in the negative feedback loop connected to the sensor and Vout. You suggested I add a capacitor to the voltage divider at the non-inverting input. So the circuit would have two capacitors, correct?
2) I can use the voltage divider to get +1V at Opamp+In from a +9V wall wart, but I still don't know the best way to get the +5v for Vsupply from the +9V. It seems like my options are to use a linear voltage regulator or a buck converter, but I don't know if I would need a heat sink for this (considering it's a low current circuit).
3) I don't understand why I'd lose 1V of Vout range (I believe you are referring to Vout when you say ADC_input). I should still be able to get 0V to +5V at Vout as long as Vsupply = +5V and the op amp is R2R output, correct?

7. ## Re: Easiest way to get 5v and ~1V

Hi,

1) you are right ... and not.
* Your second circuit shows one capacitor. Just for the function and as low pass filter.
* My circuit shows a second capacitor, because all noise at +IN node is amplified with "1 - pressure_signal_gain".
Thus one should keep +IN node quiet. You need to know that supply voltages are noisy, this additionally makes the capacitor necessary.
* But indeed you need additional capacitors for the power supply. These often are not shown in the circuits, (they just focus on the described function).
But mind this general rule: put a ceramics capacitor on every supply pin of every IC (often used 100nF, X7R), plus a big bulk capacitor.

2) many wall wart supplies are very noisy. They are cheap and dirty....
I don't know your overall power requirement, but yes, for low currents use a linear regulator.
Heatsink or not depends on circuit current, input voltage but also on regulator package, mounting and air flow.
Linear regulators usually ate less noisy than switching regulators. Especially at low load current, when they enter hickup mode.

3) the output signal at zero input signal will be +IN voltage = 1V (output offset voltage). Thus your output signal range will be 1V...5V.
When you look for a "difference amplifier circuit" you see they use an additional resistor.
Make it adjustable ... with it you are able to compensate the output offset voltage.

Klaus

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8. ## Re: Easiest way to get 5v and ~1V

One diode pointing downward provides 0.6V regulated voltage. Two diodes yield 1.2V. (Similar principle to a zener diode.) A safety resistor is needed.

If you want a specific reference voltage, then attach a resistor divider or potentiometer across the diodes.

9. ## Re: Easiest way to get 5v and ~1V

Hi,

True, the diode solution will be an improvement for noisy supply voltages.
The drawback is the known temperature drift of about -4.2mV/°C.

If you are interested in low temperature drift, then you may use a shunt reference, like ZXRE125.

But when you use the "difference amplifier circuit", with compensated offset, then a diode or shunt reference solution may be counter productive.

In detail it depends on the whole circuit, including power supplies and expected temperature range...

Klaus

10. ## Re: Easiest way to get 5v and ~1V

Originally Posted by KlausST
Hi,
Klaus
Ok! Here it is. My first schematic ever. How did I do? Did I include all necessary capacitors, and in the right places? I wasn't sure where to place the "big bulk capacitor." Also I'm not sure I placed the adjustable offset resistor in the right place.

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11. ## Re: Easiest way to get 5v and ~1V

Originally Posted by shivasage
Ok! Here it is. My first schematic ever. How did I do? Did I include all necessary capacitors, and in the right places? I wasn't sure where to place the "big bulk capacitor." Also I'm not sure I placed the adjustable offset resistor in the right place.

no capacitors block dc
put it like this

as close as possible to supply pin

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12. ## Re: Easiest way to get 5v and ~1V

Originally Posted by wondrous

no capacitors block dc
put it like this

as close as possible to supply pin
Ok.

1) Do I also need that configuration at the non-inverting input, like this (C1 and C5)?

2)I didn't add a 100nF cap to the non-inverting side. Do I need one there as well?

3) Where should I put the a bulk capacitor and what would be its value?

Thanks!!

13. ## Re: Easiest way to get 5v and ~1V

brother that capacitors add to pass noise
so in general everywhere you need a clean
dc use capacitor
to have clean rail

14. ## Re: Easiest way to get 5v and ~1V

Originally Posted by wondrous
brother that capacitors add to pass noise
so in general everywhere you need a clean
dc use capacitor
to have clean rail
Ok! How's it look?

[At ibb . co / NFybFfV]?

15. ## Re: Easiest way to get 5v and ~1V

Hi,

No need for making it complicated.
Try this.

Klaus

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16. ## Re: Easiest way to get 5v and ~1V

Originally Posted by KlausST
Hi,

No need for making it complicated.
Try this.

Klaus
Thanks for that, Klaus! Just wondering if the capacitor on the non-inverting input is X7R or bulk?

I'm also a bit confused by your offset variable resistor. You connected it to the inverting input instead of non-inverting input... Okay, that makes sense, I think. But it's also connected to the +9V power source? Well, I guess that makes sense too. I think it's time I start testing these things.

17. ## Re: Easiest way to get 5v and ~1V

Hi,

Capacitor for -IN:
* X7R are used for high current peaks and/or high frequency..because they have low ESR (at high frequencies)
But at this nod there neither us high current, nor high frequency. Thus the capacitor type is not critical.

"Bulk" just means "high capacitance", it's no capacitor type. If you have installed all the X7Rs, then the bulk type is not critical, too.

*********
Offset correction:
The OPAMP output voltage is positive.....and you want it to be more negative (in direction to zero).

Generally if you want the OPAMP output to be more negative you have two options:
* (try to) make the non inverting more negative
(Non inverting means "+"; more negative means "-" --> "+" times "-" gives "-")
But this reduces sensor voltage, which you don't want.

* or (try to) make the inverting input more positive.
(inverting means "-"; more positive means "+" --> "-" times "+" gives "-")
This is what the variable resistor does.

Klaus

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18. ## Re: Easiest way to get 5v and ~1V

Originally Posted by KlausST
Hi,
Klaus
Alright. I'm going to follow this schematic. It's identical to yours, I simply added LM7805 (hopefully I wired the ground connection correctly). I chose 100uF for the bulk capacitor at the supply pin and 10uF for the stabilizing capacitor at the non-inverting input.

19. ## Re: Easiest way to get 5v and ~1V

Forgot to mention that I'll be using MCP601 for this circuit.

20. ## Re: Easiest way to get 5v and ~1V

Hi,

hopefully I wired the ground connection correctly
Like every electronics designer you need to read datasheets.

Did you read a bout the currents?
* An 7805 is able to supply more than 3000 of MCP601 Opamps.
* An 7805 draws maybe more than 20 times the current of an MCP601.

...but it should work.

Klaus

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