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DC-DC converter

Buck Converter

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Speedy Builder529/03/2020 15:54:04
1985 forum posts
139 photos

I am using a Buck converter to step down 12v DC to 1.5 volts to drive some digital callipers (Modified for the Z axis of the mill). The display on the calliper wanders up and down +- 0.003" even with the quill locked. The battery is removed from the calliper and the 1.5 volts is fed into the PCB by soldered connection. Any ideas please ?

Amazon part Ref

https://www.amazon.co.uk/LM2596-Converter-3-0-40V-1-5-35V-Supply/dp/B01GJ0SC2C/ref=sr_1_8?crid=2HX53WG8RS38I&dchild=1&keywords=buck+converter+12v+to+5v&qid=1585493171&sprefix=buck+converter%2Caps%2C171&sr=8-8

Steviegtr29/03/2020 16:19:52
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1163 forum posts
99 photos

Do we assume it works fine with the battery in place. So just under converter power it fluctuates.

Steve.

Howard Lewis29/03/2020 16:22:32
3146 forum posts
2 photos

Unregulated DC or with a ripple?

Try putting 1.5 V regulator across the output and see what happens.

Howard

Neil Wyatt29/03/2020 16:28:04
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17722 forum posts
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Fit a small 10 or 33uF capacitor in the battery compartment.

Worked when I had similar issues!

Neil

SillyOldDuffer29/03/2020 16:42:54
5631 forum posts
1157 photos

Or it might need a resistive load. A Digital Caliper only draws microamps while the buck converter is designed to output up to 3A. Could be the buck converter can't regulate fully into a tiny load. Try fitting a 220 ohm resistor (not critical, up to 2.2k might do) across the output with, or instead of, Neil's capacitor.

Dave

Speedy Builder529/03/2020 16:59:53
1985 forum posts
139 photos

Yes, worked fine with a battery, but the battery contacts are a bit corroded as the battery leaked.

Thanks for the ideas of resistors and capacitors

Martin Kyte29/03/2020 17:47:24
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1804 forum posts
33 photos

Hi Speedy.

data sheet here.

**LINK**

Only has a quick look but the device you have will exhibit 2 switching modes continuous and discontinuous. In the discontinuous mode the output voltage will typically exhibit high ripple as the controller turns off for significant periods and then on again. For best results you should operate in continuous mode which will entail quite a high load current, enough to maintain the core saturation in the inductor. Fig 30 indicates around 1 Amp with a supply voltage of 12-15Volts. So you would need a shunt resistor of 1.5 ohms at 2 Watts.

You could as has been said run the thing in discontinuous mode with a simple R-C filter to smooth the ripples. But the unit is really not suited to the job.

I would suggest you would be better to consider a simple Low Dropout Regulator as here

**LINK**

best wishes Martin

Les Jones 129/03/2020 17:59:52
2121 forum posts
146 photos

They take almost no current so there is no need to use a switching regulator to save power. There is probably noise on the output a the switching frequency. I suggest using using an LM317L linear adjustable regulator and also fitting a few uf tantalum capacitor mounted in a dummy battery in the battery compartment of the calliper.

Les.

Martin Kyte29/03/2020 18:10:03
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1804 forum posts
33 photos

Yes. That would do it fine. LM317LZ is more than apt.

Martin

Neil Wyatt29/03/2020 18:24:56
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17722 forum posts
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I've done it before now by taking the 1.22V across an LED, the current is in microamps so regulation isn't an issue. the capacitor is needed because the dro is such high impedance the supply wires are prone to picking up noise.

Neil

Clive Steer29/03/2020 18:43:14
23 forum posts

As has already been stated the callipers only use microamps of current so even an LM317 may be overkill. All that is required is a 10K ohm series resistor and 2 forward biased silicon diodes in series plus as Neil has recommended a capacitor across the diodes to suppress noise.

Clive

Speedy Builder529/03/2020 18:58:26
1985 forum posts
139 photos

I started off with a circuit diag for an LM317, but no 317s in the electronics box, but found a couple of L200s. Then I found a diag for the L200 and away we went, only to find that I couldn't regulate it below 3.9volts.

Then on Amazon I see that the Buck converters were as cheap as chips, so that is where I am. Tomorrow is another day and will see what capacitors and resistors will do.

Andrew Johnston29/03/2020 20:42:41
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5410 forum posts
627 photos
Posted by Les Jones 1 on 29/03/2020 17:59:52:

They take almost no current so there is no need to use a switching regulator to save power. There is probably noise on the output a the switching frequency. I suggest using using an LM317L.....

I agree entirely, although the LM317 is a bit old hat. There will be better, and smaller linear regulators available.

The selected DC-DC converter is the wrong device. The LM2596 is rather out-dated, although it was very popular at it's introduction by National Semiconductor. One of the first completely integrated and easy to use buck converters. No need to understand Bode plots and feedback loops!

I'm not going to bore the pants off people with buck converter theory. It basically outputs a PWM waveform which is then filtered to give a DC voltage lower than the input voltage. It is normal to operate buck converters in continuous mode as the design is simpler and output ripple is less. What that means is that the current in the inductor does not fall to zero during the off period of the switch. In this way the current ripple in the inductor, and hence output voltage ripple, can be controlled as part of the design. For low output currents and/or high input voltages, which implies low PWM duty cycles, the converter operates in discontunuous mode, ie, the current in the inductor falls to zero on each cycle. The converter in the link will be operating in continuous mode. I suspect that what is happening is that the converter does a few cycles at a low on time then whoopsie the output voltage is too high. So the converter turns off for a period until the output voltage falls as current is drawn off. Then the converter fires for a few cycles and whoopsie again the output voltage is too high. The result is relatively large output voltage ripple. This is not a mode the LM2596 is really intended for, a lot of the graphs in the datasheet for output current versus something start at several hundred milliamps.

In newer more complex ICs operation at very low currents is catered for and is often known as hiccup or burst mode.

I'm going to get a glass of wine; you may need to do the same after that. smile

Andrew

Michael Gilligan29/03/2020 20:59:46
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15478 forum posts
668 photos

Out of mere curiosity, Andrew, may I ask:

Although extravagant, would it be practical [in the pursuit of stability] to run the existing buck converter module at a higher voltage and current, and follow it with a simple potential divider ?

MichaelG.

Maurice Taylor29/03/2020 21:06:46
76 forum posts
9 photos

Could you try an AA battery with switch ,see if it works if it don't ,use regulator etc. This seems easiest and cheapest.

 

Edited By Maurice Taylor on 29/03/2020 21:08:54

not done it yet29/03/2020 21:32:22
4503 forum posts
16 photos

Agree with you, Maurice. Then perhaps recharge the battery cell on E7🙂🙂?

An AAA cell might even cope with the load🙂🙂. Save funds and omit the switch, perhaps?

Martin Kyte29/03/2020 22:33:51
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1804 forum posts
33 photos
Posted by Michael Gilligan on 29/03/2020 20:59:46:

Out of mere curiosity, Andrew, may I ask:

Although extravagant, would it be practical [in the pursuit of stability] to run the existing buck converter module at a higher voltage and current, and follow it with a simple potential divider ?

MichaelG.

It's the output current you require in order to put the regulator into it's designed mode. If you are happy dumping an an Amp into a load resistor then fine, but the module is not designed to generate fixed voltage ouput at low current. The whole basis of this regulator is using the magnetic field in the inductor to supply energy to the load when the output driver is off and replenished when the sense voltage falls below threshold. They work best when the inductor is near or in saturation and that implies high load currents. As Andrew said when the inductor current falls to zero there is no stored energy in the field and thats why you get the high ripple.

regards Martin

Michael Gilligan29/03/2020 23:27:13
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15478 forum posts
668 photos
Posted by Martin Kyte on 29/03/2020 22:33:51:
Posted by Michael Gilligan on 29/03/2020 20:59:46:

Out of mere curiosity, Andrew, may I ask:

Although extravagant, would it be practical [in the pursuit of stability] to run the existing buck converter module at a higher voltage and current, and follow it with a simple potential divider ?

MichaelG.

It's the output current you require in order to put the regulator into it's designed mode. If you are happy dumping an an Amp into a load resistor then fine, […]

.

Thanks Martin,

I’m happy with the basic principle, but didn’t know it would need an Amp for stability.

I was thinking in the region of 10-100mA

MichaelG.

Edited By Michael Gilligan on 29/03/2020 23:28:00

not done it yet30/03/2020 08:05:13
4503 forum posts
16 photos

In addition to the KISS principle (use a 1.5V rechargeable cell), why would anyone even consider using goodness how many times the electrical power, for no good reason, a sensible option

A rechargeable AAA cell should last several(?) weeks and actually only need a couple hours charging at 100 mA after that time, even if left switched on all the time. Presumably draining 100mA to waste (along with the other power losses involved in the converter) is an acceptable waste of high grade energy for some people? It may not be much for each individual but repeated millions of times, by all and sundry, becomes a million times bigger!

If run off the mains, some form of data retention may be needed when switched off or if powered by a 12V battery, that generally requires considerable extra cost over a simple AAA cell?

Seems to me like buying and using a large bus to take one child to school just half a mile down the road.🙂

On a separate note (as a diversion) how many calipers are required for one mill? Like, is it one caliper or a pair of same - only only one jaw is movable (not quite like scissors), after all.🙂

Martin Kyte30/03/2020 09:16:49
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1804 forum posts
33 photos
Posted by Michael Gilligan on 29/03/2020 23:27:13:
Posted by Martin Kyte on 29/03/2020 22:33:51:
Posted by Michael Gilligan on 29/03/2020 20:59:46:

Out of mere curiosity, Andrew, may I ask:

Although extravagant, would it be practical [in the pursuit of stability] to run the existing buck converter module at a higher voltage and current, and follow it with a simple potential divider ?

MichaelG.

It's the output current you require in order to put the regulator into it's designed mode. If you are happy dumping an an Amp into a load resistor then fine, […]

.

Thanks Martin,

I’m happy with the basic principle, but didn’t know it would need an Amp for stability.

I was thinking in the region of 10-100mA

MichaelG.

Edited By Michael Gilligan on 29/03/2020 23:28:00

That was taken from the oerating region diagram Fg 30 in the data sheet for the inductor fitted.

regards Martin

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