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JohnTT Occasional Visitor
Joined: Nov 27, 2006 Posts: 55
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Posted: Fri Apr 20, 2007 10:37 am Post subject: Charging voltage? and crowbar protection |
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I've tried searching but can't find it. Does anyone know what the correct charging voltage for a TomTom ONE V2 should be?
My original TomTom car charger outputs 5.2V on the mini USB but another charger I bought outputs 5.0V and is it my imagination? - it seems to take ages to charge. I might be wrong. I'm not sure if the voltage difference is just tollerance variation though, as the spec might be 5V+/-5% which would allow 4.75V to 5.25V.
I bought the other charger because it is neater and allows dissassembly to incorporate a crowbar protector http://www.mitedu.freeserve.co.uk/Design/overvoltage.htm . There have been a few TomTom chargers go faulty and output 12V which blows the TomTom up . |
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Andy_P Pocket GPS Moderator
Joined: Jun 04, 2005 Posts: 19991 Location: West and Southwest London
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Posted: Fri Apr 20, 2007 11:00 am Post subject: |
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I wouldn't have thought that amount of variation in voltage would make a difference, but the maximum current capacity of the charger might.
As the TomTom One uses the USB socket to charge, it will conform to the USB power specifications.
Wikipaedia has a good layman's guide HERE
("power" is about half way down).
The main jist is:
Quote: | The USB specification provides a 5 V (volts) supply on a single wire from which connected USB devices may draw power. The specification provides for no more than 5.25 V and no less than 4.35 V between the +ve and -ve bus power lines. Initially, a device is only allowed to draw 100 mA. It may request more current from the upstream device in units of 100 mA up to a maximum of 500 mA. |
So maybe the TT is asking for current that the charger cannot provide?
Finally, I think the fault where TT supplies went overvoltage was restricted to a batch of very early ones. I haven't heard of any problems with them for a long time. |
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JohnTT Occasional Visitor
Joined: Nov 27, 2006 Posts: 55
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Posted: Fri Apr 20, 2007 11:20 am Post subject: |
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Thanks for that. Yes it should conform to the USB standard but sometimes manufacturers tweak things a bit. The lead for the PC is of course reliant on the PC for voltage.
Good point about the current. I'll measure the current draw with both chargers to compare.
It still might be worth fitting a crowbar in the unit as with all these things they are made cheaply so are as basic as possible. They use buck regulator switchers and the failure mode of these devices is often a shorted switch which will connect input to output. There is a fuse but that will only blow after the damage is done. |
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Andy_P Pocket GPS Moderator
Joined: Jun 04, 2005 Posts: 19991 Location: West and Southwest London
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Posted: Fri Apr 20, 2007 11:41 am Post subject: |
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JohnTT wrote: |
It still might be worth fitting a crowbar in the unit .... |
You're dead right there... |
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JohnTT Occasional Visitor
Joined: Nov 27, 2006 Posts: 55
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Posted: Tue May 01, 2007 10:57 pm Post subject: |
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I measured the following charger current draws from 12V:
TomTom turned OFF:
Standard TomTom charger = 250mA
Ebay TomTom charger...... = 250mA
TomTom turned ON:
Standard TomTom charger = 350mA
Ebay TomTom charger...... = 270mA
Clearly there is little spare power left to charge the battery with the TomTom turned ON when using the ebay charger as the current only rises by 20mA.
Presuming 85% efficiency that would make the battery charging current at 5.2V on the Standard TomTom charger ~490mA with the current rising to ~686mA to both charge and run the TomTom.
Further testing showed that the 5.0V output of the ebay charger drops to 4.55V under load with the TomTom on or off. Clearly it needs some re-design!
If anyone's interested here's the Mini USB pinout of th eebay charger:
Pin 1 = V BUS = 5V
Pin 2 = D- = joined to Pin 3
Pin 3 = D+ = joined to Pin 2
Pin 4 = ID = 20k resistor to GND
Pin 5 = GND = GND
Although the TomTom charger has the voltage output on pins 1 + 5, I could not find any evidence of the resistor and joined pins. |
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JohnTT Occasional Visitor
Joined: Nov 27, 2006 Posts: 55
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Posted: Thu May 10, 2007 8:54 pm Post subject: |
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Right, I changed the inductor in the switch mode for a slightly larger one, so it won't saturate, reduced a resistor in the current limit and changed the output voltage to match the Tom Tom charger. The switcher employed is a MC3406A.
It now draws 360mA @ 12V and maintains 5.2V out so the output current is around 750mA.
I also added a 5W 5.6V zener across the output, so there is no way the output can go over voltage and destroy the Tom Tom. You don't need to include the thyristor shown in the crowbar circuit elsewhere on this site as the zener is more than capable of blowing the fuse by itself.
I wish I could give the ebay seller some negative feedback after all that effort. I might as well have made one myself! |
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Oldboy Pocket GPS Moderator
Joined: Dec 08, 2004 Posts: 10642 Location: Suffolk, UK
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Posted: Thu May 10, 2007 9:04 pm Post subject: |
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JohnTT wrote: | It now draws 360mA @ 12V and maintains 5.2V out so the output current is around 750mA. |
As it's a DC to DC Voltage Regulator, I'm mystified as to how there can be more current out, than the current in _________________ Richard
TT 910 V7.903: Europe Map v1045
TT Via 135 App 12.075: Europe Map v1120 |
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JohnTT Occasional Visitor
Joined: Nov 27, 2006 Posts: 55
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Posted: Thu May 10, 2007 9:56 pm Post subject: |
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Oldboy wrote: | JohnTT wrote: | It now draws 360mA @ 12V and maintains 5.2V out so the output current is around 750mA. |
As it's a DC to DC Voltage Regulator, I'm mystified as to how there can be more current out, than the current in |
It's a switch mode regulator, not a linear regulator, so is much more efficient. It runs at about 90% efficiency so the calculation geoes like this:
Power input = 12V x 360mA = 4.32W
Efficiency (n = 90%)
Output power = 4.32W x 0.9 = 3.89W
Power lost = 4.32W - 3.89W = 0.43W
Output voltage set to 5.2V
Output current @ 5.2V = 3.89W / 5.2V = 747mA
This is the beauty of switch mode supplies. Little power is lost, so with a step down supply (12V to 5V), because nearly the same power is available for output, the output current is higher than the input current and it runs cool .
A linear regulator would require at least the same current input as output so in this example the power lost would be:
(12V - 5.2V) x 750mA = 5.1W
Power in = 12V x 750mA = 9.0W
So efficiency = 3.89W / 9.0W x 100 = 43%
It would require a heatsink to keep from overheating and would run very hot. |
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Oldboy Pocket GPS Moderator
Joined: Dec 08, 2004 Posts: 10642 Location: Suffolk, UK
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Posted: Fri May 11, 2007 7:31 am Post subject: |
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JohnTT wrote: | This is the beauty of switch mode supplies. Little power is lost, so with a step down supply (12V to 5V), because nearly the same power is available for output, the output current is higher than the input current and it runs cool . | My understanding of the workings was taken from this.
You are correct in saying that it runs cool, because the output is either on, or off. However I can't see how the Input and Output currents can be different, other than the efficiency factor. _________________ Richard
TT 910 V7.903: Europe Map v1045
TT Via 135 App 12.075: Europe Map v1120 |
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JohnTT Occasional Visitor
Joined: Nov 27, 2006 Posts: 55
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Posted: Fri May 11, 2007 8:39 am Post subject: |
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That document does not make it clear and indeed the fact that a step down or "Buck" regulator type switch mode can produce more current out than it takes in is often overlooked.
Here's a basic Buck regulator. Energy is only lost when current and voltage appear together through and across the same component. As you've pointed out, the switch is either conducting with no volts across it or is off with volts across but no current (so power loss is zero). The inductor and capacitor only store energy and no power is lost. The diode only looses a little and nearly all the power is delivered to the load:
You need to look at it in terms of energy and power. There is a basic principal that you can't get more energy out than you put in and you can only approach near to 100% efficiency without actualy quite getting there - there are always some losses - hense 80-95% is typical from a switch mode supply.
If a power supply draws 6 Watts from the 12V battery and is 100% efficient (let's be perfect for a moment) - then it must be both drawing 0.5 Amps (12V x 0.5A = 6W) and delivering 0.5 Amps at 12V (12V x 0.5A = 6W).
But let's say you halve the output voltage to 6V. If it's still drawing 0.5A (12V x 0.5A = 6W) and is still 100% efficient then it must still be outputting 6W (i.e. no energy losses), so the current must be greater (6V x 1A = 6W)
i.e. Power in = Power out
12V x 0.5A = 6V x 1A
That's how payphones you see on the street manage to run microprocessors and electric motors and modems from a telephone line current of 20mA. They step up the current by lowering the voltage.
See here for more information:
http://www.hills2.u-net.com/electron/smps.htm
http://www.onsemi.com/pub/Collateral/SMPSRM-D.PDF |
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Skippy Pocket GPS Verifier
Joined: 24/06/2003 00:22:12 Posts: 2946 Location: Escaped to the Antipodies! 36.83°S 174.75°E
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Posted: Fri May 11, 2007 8:54 am Post subject: |
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Oldboy wrote: | I can't see how the Input and Output currents can be different, other than the efficiency factor. |
Perhaps you can identify with this in mechanical terms:
You can get more torque by gearing a motor down but you get less speed so the total power output remains the same (minus transmission losses). _________________ Gone fishing! |
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Oldboy Pocket GPS Moderator
Joined: Dec 08, 2004 Posts: 10642 Location: Suffolk, UK
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Posted: Fri May 11, 2007 12:19 pm Post subject: |
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JohnTT wrote: | They step up the current by lowering the voltage. | I can see that that would apply if you were driving a transformer from the MC3406A, and then rectifying the final output.
In the switch mode, if the 12V in is switched at a 1:1 mark/space, and it was drawing an average current of 0.5A, the peak current would be 1A nominally.
At the output the average, and the peak, current would be the same as they are not being switched.
skippy wrote: | You can get more torque by gearing a motor down but you get less speed so the total power output remains the same (minus transmission losses). | That is the mechanical equivalent to a transformer.
I would liken it to a tank of water with an output of 1 gallon a minute.
If I have a 2 gallon bucket, I need only to top it up every 2 minutes. But it's still only 1 gallon a minute. _________________ Richard
TT 910 V7.903: Europe Map v1045
TT Via 135 App 12.075: Europe Map v1120 |
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JohnTT Occasional Visitor
Joined: Nov 27, 2006 Posts: 55
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Posted: Fri May 11, 2007 2:57 pm Post subject: |
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Oldboy wrote: | JohnTT wrote: | They step up the current by lowering the voltage. | I can see that that would apply if you were driving a transformer from the MC3406A, and then rectifying the final output.
In the switch mode, if the 12V in is switched at a 1:1 mark/space, and it was drawing an average current of 0.5A, the peak current would be 1A nominally.
At the output the average, and the peak, current would be the same as they are not being switched.
skippy wrote: | You can get more torque by gearing a motor down but you get less speed so the total power output remains the same (minus transmission losses). | That is the mechanical equivalent to a transformer.
I would liken it to a tank of water with an output of 1 gallon a minute.
If I have a 2 gallon bucket, I need only to top it up every 2 minutes. But it's still only 1 gallon a minute. |
Effectively it works like a transformer - it's the same principal of energy transformation with little loss. |
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Skippy Pocket GPS Verifier
Joined: 24/06/2003 00:22:12 Posts: 2946 Location: Escaped to the Antipodies! 36.83°S 174.75°E
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Posted: Sun May 13, 2007 8:25 pm Post subject: |
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Oldboy wrote: | skippy wrote: | You can get more torque by gearing a motor down but you get less speed so the total power output remains the same (minus transmission losses). | That is the mechanical equivalent to a transformer. |
Re-reading it, I see your point. Now I'm a bit puzzled too!
Ahh, perhaps it's because the load is powered using the energy stored in the inductor and capacitor not directly from the input source. _________________ Gone fishing! |
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JohnTT Occasional Visitor
Joined: Nov 27, 2006 Posts: 55
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Posted: Sun May 13, 2007 8:52 pm Post subject: |
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Skippy wrote: | Oldboy wrote: | skippy wrote: | You can get more torque by gearing a motor down but you get less speed so the total power output remains the same (minus transmission losses). | That is the mechanical equivalent to a transformer. |
Re-reading it, I see your point. Now I'm a bit puzzled too!
Ahh, perhaps it's because the load is powered using the energy stored in the inductor and capacitor not directly from the input source. |
That's correct - because the current and voltage are always 90 degrees out of phase with each other in an inductor and capacitor, no energy/power is lost in these components and energy can only be stored or transferred in and out.
Effecively, the inductor and capacitor form a low pass filter to smooth out the high frequency switching 12V square wave voltage at the input and the DC output is the filtered average voltage. So, a 50:50 mark space 12V swithed voltage becomes a 6V DC voltage. Regulation is achieved by altering the mark space ratio |
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