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'[EE] Low voltage variable power supply'
2006\07\28@181012 by Mauricio Jancic

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face
Hi,
       I'm in the need to make a very reliable power supply. It has to
supply two different voltages 200mV and 900mV. Those voltages are to be
applied to a sensor and can be RMS or DC.

       In case that wasn't clear enough, I mean to say that I can supply
the sensor either with 200/900 mV DC or a square wave of 200/900mV rms.

       I have to control the voltages within 10~20 mV.

       The input of the system will be obtained from a transformer and
passed trough a rectier.

       I currently have the system working with a PWM signal of about 8ms
and a peak value of 4.3V fed to the sensor trough a bipolar transistor.

       The problem I have is that the VRMS is a square root function, ence,
it has big steps per each PWM on the lower values. For example:

Ton(units of PWM)     Vpeak      VRMS
1                     4.3V       134 mV
2                     4.3V       190 mV
3                     4.3V       232mV
4                     4.3V       268 mV
.
.
.
43                    4.3V       881mV
44                    4.3V       891mV
45                    4.3V       901mV
46                    4.3V       911mV
47                    4.3V       921mV


As  you can see, the difference between values decreaces with the PWM. This
is pretty obvius if you graph the vrms function:

Vrms = Vhigh * sqrt( Ton / Period )


Well, then, I would like to have a way to control the RMS value more
precisely on both ends. Any ideas on how to do it?

Mauricio

2006\07\28@192132 by Bob Axtell

face picon face
Mauricio Jancic wrote:
{Quote hidden}

I've had to do this several times while building test equipment. There
is ONE reliable way:

1. Build an utlra-reliable regulated source with a low impedance. Say a
1A 0-5V supply
with 10mV resolution.

2. Drive a precision resistor voltage divider. Divide it  8:1 or even
10:1. Use low value
resistors, and provide a GOOD heat sink. Possibly a fan is needed.

3. Adjust the MAIN supply, NOT the divided one.

You can obtain 1mV resolution if the wiring is heavy and finely sized.

NOTE: DC is better to work with than RMS. It is difficult to measure RMS
precisely at
very low values.

--Bob

2006\07\28@192250 by Russell McMahon

face
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> I'm in the need to make a very reliable power supply. It has to
> supply two different voltages 200mV and 900mV. Those voltages are to
> be
> applied to a sensor and can be RMS or DC.

You call this a "power supply".
How much power do you need from it and how efficient does it need to
be?

If the allowed efficiency is low you can divide the PWMs level down so
full scale is 900 mV. Follow the divider with a unity gain power
buffer. The RMS function will still make the bottom end less
desirable, but less so.

1    eg if you have 4.3V peak PWM and want 0.9V max then dividing the
PWM signal (or the resultant DC after smoothing) by 4.3/.9 = about 4.7
= say 4 gives you 4 times the dynamic range. If you have 8 bit PWM
then your minimum voltage step now becomes 900 mV/256 =~ 3.5 mV. This
is the equivalent of going from 8 bit to 10 bit PWM.

2    Also, you could have a coarse divider (a 1 bit output) which gave
either "about" 200 mV when low and "about" 900 mV when high and then
fine tune that with N bit PWM. As long as you don't need anything in
between then the result could be extremely fine around the two
required voltages. In fact as fine as you like - the smaller the steps
the less the available range.

3    If allowed response time is low you can have output vary slowly
from eg an integrator and use A2D feedback with a 'many' bits
converter using eg Sigma Delta. You can get as many bits of accuracy
as anyone would sensibly want with enough cost and effort, or 12 to 16
bits without too much of either.

4.    If you have A2D feedback you could use a non-linear response
curve (diode shapers or whatever) shaped to complement the
characteristic sof the RMS response.

5    Using PWM to control a current source rather than a voltage
source may improve response shape.


       Russell McMahon


2006\07\28@194506 by Gerhard Fiedler

picon face
Mauricio Jancic wrote:

>        I'm in the need to make a very reliable power supply. It has to
> supply two different voltages 200mV and 900mV. Those voltages are to be
> applied to a sensor and can be RMS or DC.

How much current do you need? What are the output impedance requirements?

Gerhard

2006\07\28@200241 by Mike Singer

picon face
Mauricio Jancic wrote:
...
> Well, then, I would like to have a way to control the RMS value more
> precisely on both ends. Any ideas on how to do it?

Use PIC's A/D input to measure the output signal and so adjust PWM
values dynamically.

MS

2006\07\28@202710 by Mauricio Jancic

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part 1 2834 bytes content-type:text/plain; (decoded 7bit)

Sorry, I said "Power supply" but is really not that "Power". It has to
supply 120 mA max.

The whole system is working with batteries and has to be as cheaper as
possible, but reliability will be payed.

I already have an easy system like the one in the picture. It works fine,
but we would like to improve the accuracy, so, I'm looking for different
aproaches.

Mauricio

> {Original Message removed}

2006\07\28@202757 by Mauricio Jancic

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> How much current do you need? What are the output impedance
> requirements?

~120mA and don't know how much impedance.

Mauricio

2006\07\28@203000 by Mauricio Jancic

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> Use PIC's A/D input to measure the output signal and so
> adjust PWM values dynamically.

Yes, that's what I'm doing now. The problem is that I need to work on the
lower part of the range and there, the accuracy, is not so good.

Mauricio

2006\07\28@203237 by Mauricio Jancic

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> You call this a "power supply".
> How much power do you need from it and how efficient does it
> need to be?

120 mA. It will be supplied from batteries or from a trasnformer, so as much
as possible.

> If the allowed efficiency is low you can divide the PWMs
> level down so full scale is 900 mV. Follow the divider with a
> unity gain power buffer. The RMS function will still make the
> bottom end less desirable, but less so.

You suggest to make a voltage divider and then buffer it to the load?


Mauricio Jancic
Janso Desarrollos
http://www.janso.com.ar
spam_OUTinfoTakeThisOuTspamjanso.com.ar
(54) 11-4542-3519
 

2006\07\28@205930 by Mark Rages

face picon face
On 7/28/06, Mauricio Jancic <.....infoKILLspamspam@spam@janso.com.ar> wrote:
> Sorry, I said "Power supply" but is really not that "Power". It has to
> supply 120 mA max.
>
> The whole system is working with batteries and has to be as cheaper as
> possible, but reliability will be payed.
>
> I already have an easy system like the one in the picture. It works fine,
> but we would like to improve the accuracy, so, I'm looking for different
> aproaches.
>
> Mauricio
>

At 120ma, a 0.1 ohm resistance in the connection to the load will
result in an 12 mV voltage error.

You didn't describe your connection to the load, but if it goes
through a connector or long wires, you might consider using a Kelvin
(4-wire) connection to the load.

Regards,
Mark
markrages@gmail
--
You think that it is a secret, but it never has been one.
 - fortune cookie

2006\07\29@002027 by Mauricio Jancic

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face
The load is sometimes connected directly to the board and, in the worst
case, 150cm away in 0.75 mm2 wire.

Mauricio Jancic
Janso Desarrollos
http://www.janso.com.ar
infospamKILLspamjanso.com.ar
(54) 11-4542-3519


> {Original Message removed}

2006\07\29@021458 by Russell McMahon

face
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A battery based device with a 120 mA current drain at 900 mV (or less)
could reduce the effective load of the sensor on the battery by a
factor of 3 or more by using a switching regulator to provide a
preregulated sensor supply of say 1.1 Volt that you then derive your
sensor voltages from. Whether this is useful depends on how much of
the device supply current the 120 mA represents.

       Russell

2006\07\29@021909 by Vasile Surducan

face picon face
One of the most common "mistakes" I heard from the piclist peoples, is
that they need to do everything with a microcontroller. (sometimes I'm
wander if they even make love using a PIC 16 or a PIC 18 ?) :)

The problem you have it's a very simple one in the analogic world. You
need a calibrator not a power supply, with 200-900mV DC in step of
less 10mV and
200-900mV AC square (with unknown frequency based on your request) in
step of less 10mV, with max outpuit current of 120mA right ? That
means it could be a continuous reglage as well.

For the DC you need a stable reference (I did something similar using
a real chemical reference of 1.01417V) a common operational amplifier
and a serial output transistor. The standard negative feedback
connection is necessary like on any linear supply.
For the AC use instead of the unregulated input DC voltage, the square
wave generated by your PIC (with a transistor as a current booster) or
by a 555 (which is able to deliver 100-120mA) with the same regulator
scheme as used in DC mode. Take care a bout the frequency bandwith of
the regulator (assuming you need KHz as well not only 50Hz or 60Hz)

900-200 = 700mV/256 = 2.7mV or
700mV/1024 = 0.68mV
so a 10 bit PWM should be enough

About the problem you have is curious because you don't need to
compute any RMS for a square wave 50% dutycycle  voltage. So what
square root are you talking about?

greetings,
Vasile


On 7/29/06, Mauricio Jancic <.....infoKILLspamspam.....janso.com.ar> wrote:
{Quote hidden}

> -

2006\07\29@022205 by Vasile Surducan

face picon face
On 7/29/06, Russell McMahon <EraseMEapptechspam_OUTspamTakeThisOuTparadise.net.nz> wrote:
> A battery based device with a 120 mA current drain at 900 mV (or less)
> could reduce the effective load of the sensor on the battery by a
> factor of 3 or more by using a switching regulator


Russel,
if he is designing what I'm thinking about, any switching must be avoided.
(including the PWM ripple).

greetings,
Vasile

2006\07\29@085254 by Mauricio Jancic

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I'm using the PIC because it is already there :)

The RMS value is calculater because since the peak voltage of the signal is
around 5 volts, you usualy need a (aprox) 5% square wave to generate the 0.9
vrms.

What do you think I'm designing?

:>

Regards,

Mauricio

2006\07\29@092917 by Vasile Surducan

face picon face
On 7/29/06, Mauricio Jancic <infospamspam_OUTjanso.com.ar> wrote:
> I'm using the PIC because it is already there :)

Ok I've seen now your schematic.
>
> The RMS value is calculater because since the peak voltage of the signal is
> around 5 volts, you usualy need a (aprox) 5% square wave to generate the 0.9
> vrms.

I still didn't understood. Your signal on the load is
pseudo-sinusoidal made from PWM ? You said it's a square wave.
You're not using the whole PWM resolution for the 200-900mV range with
that schematic. Neither DC on the load.

>
> What do you think I'm designing?

As like it looks on the picture, I have no ideea. Could be a part of a
sensor for gas analyses.

Vasile
>
> :>
>
> Regards,
>
> Mauricio
>
> -

2006\07\29@124125 by Mauricio Jancic

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> As like it looks on the picture, I have no ideea. Could be a
> part of a sensor for gas analyses.

You are right. Its exactly that.

> I still didn't understood. Your signal on the load is
> pseudo-sinusoidal made from PWM ? You said it's a square wave.
> You're not using the whole PWM resolution for the 200-900mV
> range with that schematic. Neither DC on the load.

A square wave also has a RMS value. You might be refering to the average
value, which is another thing. Is that so? The load is purely resistive, so,
the signal at the load is a square wave.

A 5% duty cicle signal that has a peak of 5V, has an RMS value of 1.12V and
an average value of 0.25V. I need to have 0.9 Vrms because that's the value
that will define the power that's applied to the load.

Regards,

Mauricio

2006\07\29@132013 by Vasile Surducan

face picon face
On 7/29/06, Mauricio Jancic <@spam@infoKILLspamspamjanso.com.ar> wrote:
> > As like it looks on the picture, I have no ideea. Could be a
> > part of a sensor for gas analyses.
>
> You are right. Its exactly that.
>
> > I still didn't understood. Your signal on the load is
> > pseudo-sinusoidal made from PWM ? You said it's a square wave.
> > You're not using the whole PWM resolution for the 200-900mV
> > range with that schematic. Neither DC on the load.
>
> A square wave also has a RMS value.

Sure! But the RMS of a square it's a constant multiplied with the pulse level.
So you don't need any complexe computations just measure the pulse level.

You might be refering to the average
> value, which is another thing. Is that so? The load is purely resistive, so,
> the signal at the load is a square wave.
>
> A 5% duty cicle signal that has a peak of 5V, has an RMS value of 1.12V and
> an average value of 0.25V. I need to have 0.9 Vrms because that's the value
> that will define the power that's applied to the load.

That's because you're trying to do it without any auxiliary hardware.
You'll never get the required resolution in that way.

greetings,
Vasile

2006\07\29@154419 by Mauricio Jancic

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face
> That's because you're trying to do it without any auxiliary hardware.
> You'll never get the required resolution in that way.

OK, I see that. I'm trying to change an old design and improve something,
like resolution and various features of the equipment.

> Sure! But the RMS of a square it's a constant multiplied with
> the pulse level.
> So you don't need any complexe computations just measure the
> pulse level.

What do you mean? I use this formula:

VRMS = Vh * SQRT( Ton * F )

The magnitude that I measure is Vh (peak voltage). F and VRMS are fixed. Ton
is given by:

Ton = ( (VRMS/Vh)^2 )/f

Then

Ton = (VRMS * VRMS) / ( Vh * Vh * f )

I define  constant:

K = (VRMS * VRMS) / f

So this leads to:

Ton = k * Vh^2

This is the formula I acually use, it's a parabola (or should I spell
"parabolae"?).

If I'm wrong, can you clarify your point?

Thanks!

Mauricio

2006\07\29@182402 by Mike Singer

picon face
Mauricio Jancic  wrote:
> Yes, that's what I'm doing now. The problem is that I need to work on the
> lower part of the range and there, the accuracy, is not so good.

Change not only duty cycle, but also period of PWM to get the needed resolution.

MS

2006\07\30@025448 by Vasile Surducan

face picon face
Would you be upset if we'll start again from the begining with the
measuring principle ?

The gas sensor is a two-port (four pins) device. I presume yours too.
On one port you're injecting a current inside (or apply a constant
voltage), on the other  port you're reading a voltage (or measuring a
current). It may be variations from this rule depending by the sensor
type.

There are two major problems (as you pointed):
1. the voltage applied to the sensor is not equal with the voltage you
may generate from the PIC PWM with full resolution

You have to add hardware (OP AMP) to generate 200-900mV on the sensor
with a PWM output of 0-5V (in 1024 or 256 step depending the requested
resolution, assuming you still want doing that using the  PWM)
Don't expect this may be done with one transistor (you've tested already).

2. the voltage generated by the sensor is compatible with the full
range of the PIC ADC (full ADC resolution)

Again you need maybe a differential amplifier and a reference
(assuming the sensor  has a DC offset) to change the sensor's signal
into full ADC range signal.


If the problem above are solved, I think you may forget your parabolic
equation (which I still don't understand it, what is the "ton" ?).

BTW, why the sensor will react different if polarised with VRMS than
polarised with Vpp (peak to peak voltage)?


greetings,
Vasile


On 7/29/06, Mauricio Jancic <KILLspaminfoKILLspamspamjanso.com.ar> wrote:
{Quote hidden}

> -

2006\07\30@025553 by Vasile Surducan

face picon face
On 7/30/06, Vasile Surducan <RemoveMEpiclist9TakeThisOuTspamgmail.com> wrote:
{Quote hidden}

2006\07\31@080851 by Mauricio Jancic

flavicon
face
> > Would you be upset if we'll start again from the begining with the
> > measuring principle ?

Of course not!

> > The gas sensor is a two-port (four pins) device. I presume
> yours too.

Actually, mine has 3 pins heater, common, and output. It is like a
potentiometer, and it actually varies its resistance with gas concentration.

{Quote hidden}

I wanted to keep the circuit as simple as it is now, but that's not a
mandate. What I certanly don't want to do is to make a circuit that has very
low efficiency, since it will have to run on batteries

> > 2. the voltage generated by the sensor is *not* compatible with the
> > full range of the PIC ADC (full ADC resolution)
> >
> > Again you need maybe a differential amplifier and a reference
> > (assuming the sensor  has a DC offset) to change the
> sensor's signal
> > into full ADC range signal.

That's not a problem actually. I'm using a 2.5V reference and the accuracy
of the measurement is good enough.

> > If the problem above are solved, I think you may forget
> your parabolic
> > equation (which I still don't understand it, what is the "ton" ?).

The vrms value of a square wave is given by the square root of the division
between the active pulse and the period, all that multiplied by the peak
value of the wave. Ton is the "on time" of the signal, the high time, or
whatever you'd like to call it.

In my system, the PIC measures the peak voltage at the sensor (given by the
battery level at the colector of the transistor) and then adjust Ton to give
a 0.9 Vrms square wave. That's way I calculate Ton and that's what Ton is.
Its actually the duty cicle.


> > BTW, why the sensor will react different if polarised with
> VRMS than
> > polarised with Vpp (peak to peak voltage)?

No, it won't. polarizing it with VRMS is a way to keep the efficency higher.
I asume that by vpp you mean a DC signal of 0.9 V, right?

Regards,

Mauricio


'[EE] Low voltage variable power supply'
2006\08\01@002219 by Vasile Surducan
face picon face
On 7/31/06, Mauricio Jancic <TakeThisOuTinfoEraseMEspamspam_OUTjanso.com.ar> wrote:
> I wanted to keep the circuit as simple as it is now, but that's not a
> mandate. What I certanly don't want to do is to make a circuit that has very
> low efficiency, since it will have to run on batteries

I've understand now. Maybe you could try the following trick:
-use an LC filter between the transistor and the heater, and a high
frequency range
on the PWM
-instead of PWM regulated voltage supply change the config for current supply
(you need a pnp and a few resistors), then you'll not be forced to use
a small range of the PWM resolution and the variation of the supply
(which I understand it's a battery) in quite large limits will not
influence the measurement accuracy as it probably does now.

You probably know there are OP AMP with very small sourcing current,
single power supply and rail to rail output, made just for battery
applications...

greetings,
Vasile

2006\08\01@081804 by Gerhard Fiedler

picon face
Vasile Surducan wrote:

> On 7/31/06, Mauricio Jancic <RemoveMEinfospamTakeThisOuTjanso.com.ar> wrote:
>> I wanted to keep the circuit as simple as it is now, but that's not a
>> mandate. What I certanly don't want to do is to make a circuit that has
>> very low efficiency, since it will have to run on batteries

> -instead of PWM regulated voltage supply change the config for current
> supply (you need a pnp and a few resistors), then you'll not be forced
> to use a small range of the PWM resolution and the variation of the
> supply (which I understand it's a battery) in quite large limits will
> not influence the measurement accuracy as it probably does now.
>
> You probably know there are OP AMP with very small sourcing current,
> single power supply and rail to rail output, made just for battery
> applications...

Isn't both of this (a linear regulator providing a lower voltage for the
PWM, op amp driving the output) low on the efficiency scale? I still think
that a switcher that provides a low voltage plus maybe PWM sounds like the
better deal. What's the problem with this?

Gerhard

2006\08\01@083246 by Mauricio Jancic

flavicon
face
Hi,
       If you replied with that answer I'm sorry but I missed it. The
onlyone I have from you is the one asking power requirements.

       I think it's a good idea, I just don't want to kill the other
options and hear everyone that has an idea, then decide the best one.

       I think the efficiency of the system that you are proposing can be
very good, and also the accuracy.

       Thanks

Mauricio Jancic
Janso Desarrollos
http://www.janso.com.ar
infoEraseMEspam.....janso.com.ar
(54) 11-4542-3519


> {Original Message removed}

2006\08\01@101154 by Vasile Surducan

face picon face
On 8/1/06, Gerhard Fiedler <EraseMElistsspamconnectionbrazil.com> wrote:
{Quote hidden}

Noise and price I think. There are only a few small noise switchers on
the market, most of them made by Linear Technology. But probably there
are others.
My opinion is that you want an accurate sensor, then avoid noise.

greetings,
Vasile
>
> Gerhard
>
> -

2006\08\01@103139 by Gerhard Fiedler

picon face
Mauricio Jancic wrote:

>> I still think that a switcher that provides a low voltage plus maybe PWM
>> sounds like the better deal. What's the problem with this?

> If you replied with that answer I'm sorry but I missed it. The onlyone I
> have from you is the one asking power requirements.

I think it was Russell who beat me to it :)  Anyway, it's where I'd start
working on a solution.

Gerhard

2006\08\01@132046 by Gerhard Fiedler

picon face
Vasile Surducan wrote:

>> Isn't both of this (a linear regulator providing a lower voltage for the
>> PWM, op amp driving the output) low on the efficiency scale? I still
>> think that a switcher that provides a low voltage plus maybe PWM sounds
>> like the better deal. What's the problem with this?
>
> Noise and price I think. There are only a few small noise switchers on
> the market, most of them made by Linear Technology. But probably there
> are others. My opinion is that you want an accurate sensor, then avoid
> noise.

I understand that. But I don't know the particular sensor, and Mauricio
says that he wants to drive the sensor with unfiltered PWM... I thought if
that is compatible with the measurement principle and required precision, a
bit of ripple or noise can't hurt either.

Gerhard

2006\08\01@142442 by Vasile Surducan

face picon face
On 8/1/06, Gerhard Fiedler <RemoveMElistsspam_OUTspamKILLspamconnectionbrazil.com> wrote:
> Vasile Surducan wrote:
>
> >> Isn't both of this (a linear regulator providing a lower voltage for the
> >> PWM, op amp driving the output) low on the efficiency scale? I still
> >> think that a switcher that provides a low voltage plus maybe PWM sounds
> >> like the better deal. What's the problem with this?
> >
> > Noise and price I think. There are only a few small noise switchers on
> > the market, most of them made by Linear Technology. But probably there
> > are others. My opinion is that you want an accurate sensor, then avoid
> > noise.
>
> I understand that. But I don't know the particular sensor, and Mauricio
> says that he wants to drive the sensor with unfiltered PWM... I thought if
> that is compatible with the measurement principle and required precision, a
> bit of ripple or noise can't hurt either.

You have right. But I feel that Mauricio is far away from the ppm range.
I'm wrong ?

Vasile

2006\08\01@152404 by Mauricio Jancic

flavicon
face
>  You have right. But I feel that Mauricio is far away from
> the ppm range.
> I'm wrong ?

Yes, by choice. We have 2 different models. One has 3 leds that indicate
low, medium and high concentration, so accuracy is not an issue.
The other model has an LCD and shows the concentration, currently in 50ppm
steps.

I don't really know how are we going to do this at the moment. I have 2 week
to carefouly think about it and decide.

Thanks everyone for your inputs.

Mauricio Jancic

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