DC > DC converter problems please help!!

[Moha99]

Hey all!


I have a DC/DC converter that I bought a week ago.
What im doing is using it in some project, using a battery as the power supply and the converter as a step up transformer and sometimes a stop down.

Problem is when i use the converter to power a small DC motor, it works for a few seconds and just stops! and i try to connect it again dosen't work.
On the converters board there is a small red LED that indicates the power!

When i connect the battery it lights up, i connect the output with a higher voltage from 12volts/IN to 13 -16/DC volts/OUT to the motor the LED light fades away slowly and the motor stops.

I said to myself their could be a problem with the charge or something... So I used a capacitor at 16 volt 2200 MicroFarads and a 3 AMP diode following this diagram exactly!:


http://imageshack.us/photo/my-images/853/130320121349.jpg/



So what really is the problem here? Im trying to power the DC motor using this converter all details in the link:

http://www.ebay.com/itm/DC-DC-Boost...ps=63&clkid=6967086345806141991#ht_2942wt_986


And the DC/Motor type and specs:

http://www.ebay.com/itm/33060431467...NX:IT&_trksid=p3984.m1439.l2649#ht_651wt_1219

I say that there's a im not sure what they call it in the "electric world" im kinda new it I think there a BACKFIRE of current maybe it flowing back to the capacitor or something?

*Note: I connected everything correctly interms of the + AND THE - terminals, I need to use the converter in this project it a MUST.
+ I used more than one type of converter with similar spec's still has the same problem. + The power supply is perfect no problems with it at all.

 

[Harald Kapp]

Your links to Ebay are dysfunctional. The image on imageshack is rather small. Do you have working links or information and maybe a larger image?

Harald

 

[Moha99]

Your links to Ebay are dysfunctional. The image on imageshack is rather small. Do you have working links or information and maybe a larger image?

Harald

Yea sorry about that.

everything is functional now.

 

[Harald Kapp]

What I understand is this:
Your battery is 12 V. This is connected to the IN-Pins of the DC/DC converter.

Your Motor can be operated anywhere between 2V and 18V. This is connected to the OUT-pins of the DC/DC converter.

Right?

To which output voltage is the DC/DC converter set?
What type of battery are you using?
Is the motor loaded or free running?
Have you measured the current drawn by the motor during the time it is running? Have you monitored (measured) the battery voltage? What happens to the battery voltage when the motor starts spinning down/slowing?


Harald

 

[Moha99]

*Your Motor can be operated anywhere between 2V and 18V. This is connected to the OUT-pins of the DC/DC converter.

Right?: Yes that it correct.

*To which output voltage is the DC/DC converter set? between 13 to 16 volts/DC I adjust depending on the use.

*What type of battery are you using? Im made a battery bank combining a 2 Duracell DCELLS at 1.5 Volts and a 9Volt Duracell to make a 12 volt battery set.

*Is the motor loaded or free running? Im not really sure what you mean by that... But when I connect the converters OUT. to the motor it works for a few seconds then stop when I reconnect it it does not work at all... When I leave it for a while then re-connect it does the same process all over again.

*Have you measured the current drawn by the motor during the time it is running? Well I tried to measure the current my volt meter has some issues... Im not sure what it is. I try to test the amperage of the project but somethings wrong with my voltmeter...

*Have you monitored (measured) the battery voltage? What happens to the battery voltage when the motor starts spinning down/slowing? Yes I have when the motor is connected the OUTPUT of the converter drops from 14Volts DC to 2 or 3 volts DC even when i disconnect the motor stays that way and the RED LED indicator goes very low. Even the battery INPUT drops down.

 

[(*steve*)]

Im made a battery bank combining a 2 Duracell DCELLS at 1.5 Volts and a 9Volt Duracell to make a 12 volt battery set.

There's your problem right there.

Use 8 D cells in series, or even 8 AA cells.

I suspect that if you measure the voltage across this battery while you try to operate the circuit it rapidly drops, or worse, even reverses.

 

[Moha99]

There's your problem right there.

Use 8 D cells in series, or even 8 AA cells.

I suspect that if you measure the voltage across this battery while you try to operate the circuit it rapidly drops, or worse, even reverses.

You and I always differ about the batteries What's the problem with them?

 

[(*steve*)]

I suspect that if you measure the voltage across this [9V] battery while you try to operate the circuit [you'll find] it rapidly drops, or worse, even reverses.

It is simply very poor practice to place batteries in series that have vastly different capacities.

edit:

Most battery chemistries allow serial and parallel configuration. It is important to use the same battery type with equal capacity throughout and never mix different makes and sizes. A weaker cell causes an imbalance. This is especially critical in a serial configuration and a battery is only as strong as the weakest link.

edit2:

You and I always differ about the batteries

I suspect one of us is going to have to change

 

[davenn]

You and I always differ about the batteries What's the problem with them?

As Steve is trying to tell you..... You CANNOT mix different types of batteries
THAT is the problem

Dave

 

[Moha99]

As Steve is trying to tell you..... You CANNOT mix different types of batteries
THAT is the problem

Dave

It is simply very poor practice to place batteries in series that have vastly different capacities.

edit:

edit2:

I suspect one of us is going to have to change

Ok I'm going to listen to what you all say. And use 4 DCELL rechargeable at rated at 1.2 volts and about 2600mAH and boost the voltage up from about 4.8 to 12 volts using the converter how about that?

 

[(*steve*)]

2600mAH is pretty small for a D cell (I'd be expecting 9000+). I suggest that it's just an AA cell in a bigger cylinder.

But ignoring that, yes, a boost SMPS should work with this. The 4 cells will last less than half the time that 8 would though -- just letting you know. The advantage (with the boost SMPS) is that the voltage will not droop as much as the cells discharge.

 

[Moha99]

2600mAH is pretty small for a D cell (I'd be expecting 9000+). I suggest that it's just an AA cell in a bigger cylinder.

But ignoring that, yes, a boost SMPS should work with this. The 4 cells will last less than half the time that 8 would though -- just letting you know. The advantage (with the boost SMPS) is that the voltage will not droop as much as the cells discharge.

Thanks Steve for you're help so far!

Im just using 4 for now... But later I'll be using more. The D Cell that i bought said it has about 2600mAH x by 4 thatsabout 10K mAH.

Thats a lot of current for that converter. it has a Min of 1 Amp and a Max of 3 Amps.
I was thinking if I would use 2x 9Volt batteries instead.


but i found most of them between 150mAH to 650 mAH max.... You think there are more types?

 

[Moha99]

I also heard that the dc converter would decrease the current as a consequence of increasing the voltage.

So I guess having a hight current power supply is good right..?

 

[jackorocko]

I also heard that the dc converter would decrease the current as a consequence of increasing the voltage.

So I guess having a hight current power supply is good right..?

You can't create power from thin air. You can only convert one quantity of power to another quantity of power, but the overall power won't change from in to out (except for the losses in the circuit.)

Power = voltage x current.

The answer to your question is yes if you want to make a boost dc - dc converter.

 

[Moha99]

You can't create power from thin air. You can only convert one quantity of power to another quantity of power, but the overall power won't change from in to out (except for the losses in the circuit.)

Power = voltage x current.

The answer to your question is yes if you want to make a boost dc - dc converter.

Exactly. Since I'll be using a battery to power the converter I wanted to calculate the current loss, do you have an equation for it or so...?

 

[BobK]

There is not a current loss. The current used on input side times the input voltage must equal the current used on the output times the output voltage, except that is is not 100% efficient. So the equation is:

Vin * In * efficiency = Vout * Iout

Bob

 

[jackorocko]

Exactly. Since I'll be using a battery to power the converter I wanted to calculate the current loss, do you have an equation for it or so...?

What is the efficiency of the boost converter? It is probably in the datasheet for the IC. This will give you a very close approximation to what is lost in the circuit.

You put in 100W of power and get out 95W of power, then the efficiency of the circuit is 95%.

 

[Moha99]

What is the efficiency of the boost converter? It is probably in the datasheet for the IC. This will give you a very close approximation to what is lost in the circuit.

You put in 100W of power and get out 95W of power, then the efficiency of the circuit is 95%.

There is not a current loss. The current used on input side times the input voltage must equal the current used on the output times the output voltage, except that is is not 100% efficient. So the equation is:

Vin * In * efficiency = Vout * Iout

Bob

Max Efficiency: 86%

Im putting about 4.8Volts x 10Amps = 48W x 0.86 = 41.28W

Problem is there is a of 1 A Min current and a 3 A Max on the converter I fear to supply higher current...

So I thought with a lil bit of current lost that would give me a chance.

The batteries im using each have about 2500mAh which is about 2.5 Amps and about 1.2 Volts so I have to supply the converter with about 3.5Volts in so 1.2 x 4 = 4.8V/DC so now we have 2.5x4 = 10Amps
which is 7 amps above the MAX. range...

Also In what way can a DC TO DC converter increase charge? Now I know charge is being lost how can we increase it then...?

lets say I have about 4 volts with 160 mAh if I step it up to about 20 Volt wont the current also increase?

 

[Moha99]

I was thinking since the current is high in the input And I will increase the voltage from 4.8 to 13V/DC the current will drop significantly how about that?

 

[jackorocko]

What is your output voltage?
What is your input voltage?
How much current does the converter need to supply on the output?

And I will increase the voltage from 4.8 to 13V/DC

~42W/13V = 3.23A available on the output. Should probably figure 3A max.