3
\$\begingroup\$

I am attempting to lower a 3.3V source to 3V for use with a sensor (which has a rating of 3V +/- 0.1V). Based on the other answers in here, I believe that placing two resistors in series would achieve this goal.

The 3.3v Vin is coming from an iPhone and the 3v Vout is going to an MR513 gas sensor.

The sensor requirements are:

Working Voltage: 3.0V +/- .1V
Working Current: 100 mA +/- 10mA

Using the math from this accepted answer:

Vin = 3.3V
Vout = 3V

I get that two possible values for R1 and R2:

R1 = 200 Ohms
R2 = 2k Ohms

Can someone confirm that this is correct? Also, will the Vout be affected if I connect the ground from the load at Vout to the same ground as this circuit?

UPDATE:

The accepted answer from stevenh recommends use of a LDO regulator manufactured by Torex (XC6210) which supplies the required 3.0 voltage and 100mA current. I ordered the exact part from DigiKey and to my surprise it showed up and was about the size of a small pill (had never ordered a surface mount before). I have been searching for a viable alternative that is "through hole" mounting type, but for some reason there does not seem to be a solution. Am I missing something or do I need to have the circuit printed professionally to test?

Really appreciate any assistance.

\$\endgroup\$
  • 1
    \$\begingroup\$ Can you link to some specs of the sensor? How much current it draws is important. \$\endgroup\$ – Cybergibbons May 11 '12 at 20:32
  • \$\begingroup\$ Sorry, totally forgot to add that in. It draws 100 mA of current. The sensor is a MR513 gas sensor and the full data sheet is here:futurlec.com/Datasheet/Sensor/MR513.pdf \$\endgroup\$ – Richard May 11 '12 at 20:43
  • \$\begingroup\$ Also, the Vin is coming from an iPhone which is 3.3v (not sure what the current is). \$\endgroup\$ – Richard May 11 '12 at 20:44
  • \$\begingroup\$ @Richard Add those details into your question. Also bring any relevant details out of the datasheet and into the question. \$\endgroup\$ – Kellenjb May 11 '12 at 20:53
  • 1
    \$\begingroup\$ @Richard - sorry you feel disappointed about the SMD. Maybe this is a good start to get used to it. You'll find that more and more parts only exist in SMT. Try to solder thin wires to it with a fine soldering tip. image, image \$\endgroup\$ – stevenvh May 17 '12 at 7:56
14
\$\begingroup\$

Your math for the resistor divider is correct, but the output voltage will change it you connect anything to it. You can't draw 100mA from it. The output will sag to zero at a 16mA load.

Working Voltage: 3.0V +/- .1V

That's 3%, that requires already precision components. Standard components are often 5%. Also keep in mind that the 3.3V may have a 5% tolerance as well!

A series resistor was suggested. That's the cheapest solution. And the worst. It's a Bad Habit™ you never should adopt. The voltage drop across the resistor changes with the load, so that you don't get a properly regulated 3V.

edit
I'm getting some critique on this. The reason I'm against a series resistor as a voltage regulator is that basic specs for a regulator, like line regulation and load regulation are very bad. But! This seems to be a special case. The load is a Wheatstone bridge which has a rather constant resistance, probably far more constant than the 10% the spec mentions (which is probably process tolerance, not measurement variation). A constant load means a constant voltage drop across the resistor, so that fixes the load regulation question. Then line regulation. Input is 3.3V from an iPhone. The voltage suggests this is regulated too, but we don't know for sure. If it's regulated then there's no line regulation issue either, and the series resistor can be seen as half of a resistor voltage divider, dividing a fixed voltage. In that case a series resistor will work. I have no problem admitting that. I still have to see the 3.3V confirmed, though. If it can vary the resistor is still bad, regardless of the constant load! In most cases either the input voltage or the load (or both) is variable, and then a series resistor won't give you a properly regulated, fixed output voltage.
Figures: the series resistor will give you a line regulation of 910mV/V. Compare with an LM7805, which has a typical line regulation of 0.2mV/V.

The series Schottky diode's voltage also varies a bit with current, but there's also process variation. Tony's graphs show typical and maximum voltage drops, but if you want to use one it's the minimum voltage drop which is important as well. Check it before use. Especially if the 3.3V would have a +5% tolerance (=3.47V). Tony's diode only drops 250mV typical, at this input voltage that would be too little. This diode only drops 150mV at 100mA, so that wouldn't even do at the nominal input voltage.

The zener diode that Kaz proposed also suffers from process variations. At 5mA this 3V zener can have a reverse voltage between 2.85V and 3.15V, which is outside the limits you specify. Most zener diodes will have this kind of variation (5%). Precision zeners (2%, 1%) exist, though. Your zener will have to draw at least 10mA, preferably 15mA (5 to keep it stabilizing, 10 to cover the 10mA tolerance on the load), which may or not may be within your budget.
A zener with a differential resistance of 30\$\Omega\$ drawing 15mA has a line regulation of 860mV/V, so that's only as good (or bad) as the resistor.

The neatest solution is an LDO regulator. It's a bit more expensive but the specs are worth it. My favorite for low power applications is the Seiko S-812C (only 1\$\mu\$A ground current!), but that can't supply the required 100mA. The Torex XC6210 however can, even up to 700mA. 2% accurate, only 50mV dropout voltage and only 35\$\mu\$A ground current.

edit
Typical application for the XC6210:

enter image description here

Don't forget the capacitors, especially the output cap. And don't forget to connect the Chip Enable (CE) to \$V_{IN}\$.

\$\endgroup\$
  • \$\begingroup\$ This seems to make the most sense so far (and the price difference is negligible considering it is a 100% solution). To confirm, the LDO will need to go in series before the sensor (load) and it will simply regulate the voltage at 3v (within a 2% accuracy this leaving 1% of wiggle room for the unforseeable. Using this component, how do I control the current and ensure a 100mA ± 10mA flow to te sensor? I understand how this component regulates the voltage but not entirely sure how it controls the amperage \$\endgroup\$ – Richard May 12 '12 at 15:12
  • \$\begingroup\$ @Richard - All you have to do is supply the 3V, the current will then flow automatically. Just like a resistor: if you apply a voltage to it there will be a current I=V/R. There's no need to push the current through it. Don't forget the 1uF capacitor on the output of the regulator. It needs it to avoid oscillating. \$\endgroup\$ – stevenvh May 12 '12 at 15:17
  • \$\begingroup\$ Thank you for that explanation, got a bit confused but totally understand now. Also, thank you for the capacitor reminder, totally missed that initially. Im ordering all the parts from digikey right now, going to try to throw it all on a breadboard this week. Needless to say, I am extremely excited to try it out. Thank you again for your help \$\endgroup\$ – Richard May 12 '12 at 15:27
  • \$\begingroup\$ @Richard - I added a typical application schematic from the datasheet to my answer. Also note the remark on the CE pin! Success! \$\endgroup\$ – stevenvh May 12 '12 at 15:37
  • \$\begingroup\$ Great schematic, thanks! So from what I can tell, I have the Vcc going in parallel to a capacitor (1uF) and to the regulator (both the Vin and CE pins). On the output side, I have Vss going to ground and the Vout pin going in parallel to the Sensor and another 1uF capacitor. Neither NC pin is used. If that sounds about correct I think i am ready to try it out! Just ordered all the components! \$\endgroup\$ – Richard May 12 '12 at 16:35
3
\$\begingroup\$

You need only one resistor in series with the device. That resistor has to drop 0.3 volts, so we know V. You need to know how much current the device draws when on a 3V supply. That gives you I. Then you can solve V = IR for R.

If the device draws variable current, then these resistor based methods will not provide a "stiff" voltage of 3V.

In that case the thing to try would be a 3V Zener diode in series with a resistor (similar to your two resistor circuit diagram, but the bottom resistor is replaced with the Zener). You take the voltage across the Zener to supply the sensor.

The resistance has to be such that the voltage will not drop below 3 under the maximum possible current draw.

Example: suppose the device draws up to 0.1 mA. Assume that when the device is drawing this maximum current, the voltage is at 3V, just on the threshold of closing off the Zener diode. I.e. the device is getting the entire 0.1 mA and nothing is going through the diode. The resistor has to drop 0.3V, which at that current calls for 0.3V/0.0001A = 3000. You need a 3K resistor on top of the 3V Zener.

Suppose the device draws less than 0.1 mA at times. In fact, suppose the worst case: the device shuts off, and behaves like an open circuit. In that case, 0.1 mA will still flow through the resistor because the Zener clamps the voltage at 3V. But now the current is going through the Zener instead of the device.

That's the idea: the Zener keeps the voltage at 3, and picks up the excess current.

What if your device draws more current than the 0.1 mA design assumption? This means that the voltage across the resistor will drop more than 0.3V and so the voltage starts to fall below 3V. The Zener diode is then not conducting and is out of the picture. Further demands on current just drop the voltage. The circuit no longer provides regulation.

\$\endgroup\$
  • \$\begingroup\$ This answer was amazing, thank you so much for describing it in extreme detail. I am not sure about the variable draw of the sensor, so I think what i may do is set up the first circuit on my breadboard and measure the draw to see how much it changes. Either way, it sounds like the second method would be much safer in that it can handle even the unexpected. I had to google Zener Diode (yes, total newb), but I understand how it works and am definitely going to buy a few to test with. Thank you again for your help. \$\endgroup\$ – Richard May 11 '12 at 21:03
  • 1
    \$\begingroup\$ Okay great. So the temp rise is negligible (in theory) and the only worry is that the voltage range is exactly the same as the tolerance on the sensor. Seems like i have some testing to do. Thank you again for your help. P.S. You mention to rate my acceptance with points (think that was directed towards me). I clicked to accept your answer but not sure how to assign it points. \$\endgroup\$ – Richard May 11 '12 at 22:27
  • 3
    \$\begingroup\$ NEVER use a series resistor to regulate a voltage! \$\endgroup\$ – stevenvh May 12 '12 at 5:14
  • 1
    \$\begingroup\$ @stevenvh, I have seen it done with success, it just does not really regulate very well over varying loads. I use them to regulate the output for my LEDs all the time. Luckily they are relatively immune to the poor regulation a resistor provides. \$\endgroup\$ – Kortuk May 12 '12 at 13:45
  • 2
    \$\begingroup\$ This sensor may very well have a constant current draw in which case a series resistor is perfectly appropriate since it will drop a predictable voltage. Since the current draw of the sensor is fairly high, the conditions for making a stiff voltage divider (say, at least 10:1 ratio of divider current to device current) call for resistor values that would be ridiculously wasteful or power, dissipating about 3.3W. So the only reasonable purely resistor-based solution is a series 3 ohm. Never say NEVER; evaluate the actual conditions. \$\endgroup\$ – Kaz May 12 '12 at 17:11
1
\$\begingroup\$

The simple answer is use a series Schottky Diode rated for 1 A continuous on the 100mA load to drop the 3.3 to 3.0 +/-0.1V

This hot wire sensor needs an accurate voltage to function. You could use a diode drop of 0.25 V from a 1 Amp Schottky diode and this would work over a limited temperature range as diodes are also great thermometers with a negative voltage vs temp. characteristic so V drop can range from 0.15 to 0.35V over the military temperature range.

If you define your operating temperature range of the ambient and temp rise inside the case to where this diode could be mounted, then you can do a worst case tolerance analysis to ensure the supply will meet the sensor specs or conversely test it and have calibration error curve to correct the results on the iPhone software, taking into consideration the iPhone battery voltage.

The MBR120VLSFT1 costs $0.11 on tape & Reel.
Many alternate parts both leaded and Surface mount depending on total requirements.

See the variation on each graph for TYP to MAX variation at the bottom of the axis (0.1A) with temperature and consider what is your operating range. MBR120VLS

\$\endgroup\$
  • \$\begingroup\$ Okay, so i would still use the the 3ohm resistor in series) to drop the Vin initially by .3V to 3V. Then in series after the resistor I would need a Shottky Diode which woudl regulate the voltage and maintain the 3v required. So that I am entirely understanding this (please forgive my ignorance), the Vdrop across the Shottky will vary on 1)the draw of teh sensor and 2) on the temperature of the diode (which you recommend i test within the casing). \$\endgroup\$ – Richard May 11 '12 at 21:45
  • \$\begingroup\$ A more complex solution can be made that is 3.0V +/- 50mV using a 2.5V Band gap reference, a comparator and transistor to regulate the total load so that the voltage drop on the diode is exactly 3.0V to the load. The diode I chose has an ESR of 1Ω so you dont really need the 3 Ω unless your worst case calculations improve the result without resorting to shunt regulator below digikey.ca/product-detail/en/MAX6102EUR%2BT/MAX6102EUR%2BTCT-ND/… SMT $0.50 on tape \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 May 11 '12 at 21:48
  • \$\begingroup\$ I think the idea here is that all you have is the diode in series with your sensor. It drops 0.25-0.35 volts for you and that is that. \$\endgroup\$ – Kaz May 11 '12 at 21:50
  • \$\begingroup\$ Yes exactly.. low ESR of 1Ω helps regulate but temp factor is large, so shunt regulator with Band Gap and current sink to regulate the series drop to exactly 3.0V at load. is best design. but this may work at room temp OK with maybe another 2Ω in series or so. Do the math. then verify. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 May 11 '12 at 21:55
  • \$\begingroup\$ To clarify, there are two possibilities here: 1) place a Schottky Diode in series with the sensor to lower the voltage .25-3.5v (with no other resistor), but there is a risk of overheating. 2) Use the same diode but also add a Band Gap and current sink to regulate the voltage at 3.0v (with less worry of overheating). Does that sound about right? \$\endgroup\$ – Richard May 11 '12 at 22:09
0
\$\begingroup\$

The accuracy of the system is totally dependant on the calibration of the unit and the error in assumptions. A thorough design review of any potential solution would include;

None of the solutions offered did the above, so if I were conducting a Design Review, all presentations (including my own ;) would have been rejected, and reconvened when completed. As this forum is not a formal process, many valuable suggestions have been made but many valid and invalid criticisms have been made. The worst ones are the ones which are only half-baked, meaning that they drew a conclusion and presented an un-finished argument without full analysis which might (IMHO) contradict the (possibly false) assumption due to lack of full Sensitivity Analysis.

( I offer this Answer to reflect my real-world design review experience having done this for large and small commercial engineering/manufacturing corporations in Aerospace, Telecom and AMR industries.)

A reverse engineering analysis of the sensor is required and its specifications.

  • The MR513 Hot-wire type gas sensor spec. MR513.pdf has no date nor revision, so it is untraceable and the supplier must be called upon to supply a dated document with a revision code, so we treat this as "Preliminary".
  • The above spec. illustrates a "wheatstone" bridge configuration with the hotwire detector,D, and temperature compensation, C, across the 3.0Vdc power supply which has NO POLARITY markings!! Can we assume that it is a bipolar semiconductor device? Maybe, but, we should verify that.
  • D & C is the path which draws 99% of the 100mA +/10mA (10%) rated current using a 3.0Vdc +/-0.1V (3.3%). If we wanted to design an instrument more accurate or as much as this component suggests, a calibration test to null offset and gain is needed before each use. Otherwise an annual drift is given as +/-10mV @100mV test condition with an example given with a nominal value from 96~112 mV when 100mV was expected for 100ppm ethanol and +/-3mv for air where the cyclical variation was not correlated with the cyclical error in Gain of about 1~2 cycles per year.

    • One cannot assume the test conditions were nominal or ideal, since they are not fully stated such as temperature, input voltage drift, humidity. A full report should be requested from vendor including MTBF data and Root Cause Failure Analysis reports with suitable NDA.

      • A bridge sensitivity to Voltage error only affects gain and not offset since it is a ratio-metric bridge (a.k.a. a Wheatstone bridge) with internal null adjustment. But one may still choose to provide an auto null adjustment for air to perform rapid air vs volatile gas measurements. The sensor gain sensitivity is 50% which is due to the 50% R ratio in the bridge.
      • The sensor has a 500Ω null adj. pot. presumably with a laser tuned 500Ω adjustment range at the factory. This does not affect gain.

      - Since there is only a minimum gain spec of >100mV in 100ppm of ethanol, you must add an external gain adjustment if you want precise gain control of overall design.

      • This is only the minimum sensitivity and you may or may not appreciate that more is better but there is no maximum specified nor any precautions about aging or contamination or cleaning or solder heat precautions.

Now let us examine the solutions offered to meet the 3.0Vdc supply

  • The sensor has an effective series resistance of V/I = 3V/0.1A= 30 Ω**
  • The load current is constant with the exception of minor temperature compensation for detector sensitivity adjustments and may be considered a current sink passive device.
  • Shunt Zener = rejected because the high ESR (20Ω) of low voltage Zeners and excessive bias current to meet regulation requirement.
  • Series Zener = rejected {by stevevh} for 0.25V drop in a high 3.3V state.
  • Active shunt regulator with low cost, low current precision reference and differential current sink to regulate the .3V drop.
  • The MAX610x series precision bandgap chip is the best solution for build a precision regulator. It can source 5mA or sink 2mA to guarantee 3.00V out, regardless of 3.3V tolerance. It is spec'd @ 0.4% with 125µA max current and a low temperature coefficient (tempco.)of 75ppm/deg'C for a low cost ~$1.
  • Actually the series drop part is not important , as it is clear a regulator is needed to maintain say 1% accuracy to provide a 0.15V to 0.45V drop to 3V, assuming ~5% of 3.3V input.
  • Series Resistor of 3Ω with constant resistance load of 30Ω will have a sensitivity to gain wrt. supply of +/-10mA *3Ω= +/-30mV or 1% in addition to 3.3V tolerance, so this fails. But if the 3.3V supply was 1% then this solution is as acceptable as @Stevenvh's Torex XC6210 which is 2%, not as reliable in case of failure of 3.3V regulation.

Conclusion:

  • Passive solutions do not offer the regulation needed to drive this sensor.
  • System level calibration must be done to correct gain errors inherent to power supply voltages and sensor combined.
  • Stability of the system is dependant on many things but in 1 year a 10mv variance per 100mV reading is achievable if the supply is stable.
  • The use of a band gap reference is inherently the most reliable means of regulating a supply with only 4µA/V variation.
  • The MAX610x series.and can achieve 0.2% at room temp. is considered by the industry as "best in class".

    • Alternative solutions with 2% are viable but sacrifice gain error by 1% unless compensated by calibration. Most important is to determine what the design goal is for overall use, accuracy and calibration.
    • In order or an instrument brand like HP to gain notoriety, they do all these things and more in order to gain your trust and offset the need to worry about its accuracy & precision for ease of mind and dependability of use. Lack of this diligence is what creates [failures] [Also consider, redundancy]17 by design, is what makes fault tolerant systems more reliable.18.

Last minute additional solution... While looking for a 100mA buffer for BG ref, I came across an integrated solution (a.k.a. another LDO) Microchip TC2015 0.4% $0.31 on reel ww1.microchip.com/downloads/en/DeviceDoc/21662e.pdf Spec. ww1.microchip.com/downloads/en/DeviceDoc/21060z.pdf Power Mgmt family

Now go do your SA and DVT plan before you finish your Designs.

- I am not doing all your homework for you!
- Did anyone wonder, how hot a Hotwire Sensor actually gets with 300mW?
- Did anyone wonder, how gain changes with ambient temperature?
- Did anyone know hot wire sensors are used in cars to detect flow rate?
- How does turbulence affect gain for this gas sensor?

**Fun Fact: Did you know Charles Wheatstone invented an Encryption method in 1854? What was it called? Now be marvelous and "play fair" with the hard working Engineers who contribute their spare time & opinions, based on experience to this pro bono site!!

\$\endgroup\$
  • 1
    \$\begingroup\$ The MAX6103 can source 5mA, as you quoted yourself. We need 110mA. \$\endgroup\$ – Federico Russo May 14 '12 at 13:59
  • \$\begingroup\$ I did not spoon feed the entire solution. A current amplifier is needed. eg open drain comparator to regulate current to desired voltage. Sorry I was expecting too much... while looking up a p/n I came across a device with both integrated to an LDO solution. TC2015 0.4% $0.31 on reel ww1.microchip.com/downloads/en/DeviceDoc/21662e.pdf from ww1.microchip.com/downloads/en/DeviceDoc/21060z.pdf Power Mgmt family \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 May 14 '12 at 16:34
  • \$\begingroup\$ @Tony, thanks for the great response above. It took me a while to understand it all, but i read through he datasheets (as well as the fun facts!) and what it seems like is that while the MAX component is higher accuracy, its not ideal for this application because of the discrepency in max current and required current. From what i gather, it can be built to work, but its not a simple "plug-n-play" component like the Torex. Please feel free to tear that apart, im just trying to decide the best (simple yet effective) design for this circuit. Thank you also for SA & DVT info, def going to do that \$\endgroup\$ – Richard May 14 '12 at 18:07
  • \$\begingroup\$ You are most Welcome Richard, I was focusing more on procedure than design. 1st Spec, 2nd design, 3rd Analyze (SA, Sim etc.) 4th Test plan & DVT (redo until meet spec or budget or time limit, whichever comes 1st and ensure Program Mgr knows the risks and ask for Priorities to decide tradeoffs) Last minute update to my original thread near end ... Microchip TC2015 0.4% $0.31 on reel 100mA. 3V from 3.2V \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 May 14 '12 at 22:24
  • \$\begingroup\$ @Tony, going to purchase a few TC2015 to test with as well. Trying to consider all options before i make a final decision (and needless to say, testing the circuit and alcohol sensor should be fun). Also, i just replied to stevenh above about the possibility of replacing the power source with a AAA battery. Found another LDO regulator here that i think may work farnell.com/datasheets/89922.pdf. Would love your thoughts. Anyhow, parts are on their way and getting excited! \$\endgroup\$ – Richard May 15 '12 at 19:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.