The title serves to illustrate a problem frequently encountered when trying to power a device that operates with a high peak current: Your energy storage - or your power source - may have plenty of energy capacity, but not enough current capability!
One such example of a power source that has plenty of capacity - but rather limited power capability - is that of the "cigarette lighter" outlet in a typical vehicle: As long as the engine is running, you can pull power for as long as you want - but not too much at any instant. More than 10-15 amps is likely to blow a fuse and even if you were to replace the original fuse with, say, a 20-30 amp fuse (not smart!) the rather light gauge wiring would likely result in a voltage drop that would render a typical 100 watt HF rig unusable on voice peaks.
For another, more extreme example let us consider a set of alkaline "D" cells, referring first to some online data:
- The "Energizer" D-cell data sheet - link.
- "Duracell" D-cell data sheet - link. Does not include direct amp-hour rating, but such may be inferred from the graphs presented.
- "'D' Battery" Wikipedia page - link.
One thing that jumps out is that a "D" cell has somewhere between 5 and 30 amp-hours per cell, but if you study the graphs, you'll also note that this apparent capacity drops like a rock with increasing current. Why is this?
At least part of this is due to internal resistance. If we examine the data for a typical alkaline "D" cell we see that on a per-cell basis that the internal resistance is 0.15-0.3 ohms per cell when it is "fresh", but this increases by 2 or 3-fold near the end of life of the cell (e.g. <=0.9 volts/cell) and continues to do so dramatically - and very quickly - at still-lower voltages. Interestingly, the manufacturers' data used to include graphs of internal cell resistance, but these seem to have disappeared in recent years.
If we take a general number of 0.2 ohms/cell and expand that to a 10-cell series battery pack we get a resistance of 2 ohms which means that if we attempt to pull even one amp we will lose 2 volts - and this doesn't take into account the contact and wiring losses related to these batteries!
If you look at the graphs that relate battery capacity to discharge current you will notice something else: If you draw twice the current, your apparent capacity - in "run time" decreases by more than half and if you convert this to amp-hours, the more current drawn, the fewer available amp-hours.
These two facts together tell us two things:
- We cannot draw too much current or else resistive losses will cause excess voltage drop.
- Higher current consumption will cause a marked drop in available amp-hour capacity.
As you may have inferred from the title of the article, our particular application implies a usage where the typical, resting (or even average) current consumption is quite low but the peak current consumption could be very high. Clearly, with a string of "D" cells, alone, while we may have enough theoretical capacity to provide power for operation, we cannot tolerate the peak currents!
A power source with good longevity:
What we need is a low-impedance power source and as it turns out almost any type of rechargeable cell - whether it is lead-acid, NiCd, NiMH or lithium - has a lower impedance than alkaline cells so the question that one might ask is why not use one of those other types?
The obvious problem with lead-acid is that of longevity: If you own a lead-acid battery that is more than three years old (and certainly more than five!) it has likely lost a significant percentage (at least 30%) of its rated capacity - probably more unless it has been treated really well (e.g. controlled temperature at or below 70F, 21C) and always kept at a proper floating voltage when not in use (e.g. 13.5-13.7 volts for a "12 volt" battery at nominal room temperature.) Similarly, NiCd and NiMH cells need to be treated properly to have good longevity - and even the "low self-discharge" NiMH types will run themselves down if kept in storage for quite a while.
On the other hand, modern alkaline cells will retain the vast majority of the capacity for at least 5 years, just sitting on a shelf - a period of time that is beyond the likely useful lifetime of either lead-acid or most rechargeable lithium-ion cells. What's more is that alkaline cells are readily available practically anywhere in the world and they come "fully charged". To be sure, there are other types of "primary" (non-rechargeable) cells that have excellent shelf life such as certain types of lithium, but these would likely "break the bank" if you were to buy a set with comparable capacity and they are certainly less widely-available than standard "D" cells.
A low-impedance (voltage) source:
An advantage of Lead-Acid, NiCd, Lithium Ion and some NiMH cell types is that they have quite low internal resistance compared Alkaline cells: Even an aging lead acid battery that is near the end of its useful life may seem to be "OK" based on a load test as its internal resistance can remain comparatively low even though it may have lost most of its storage capacity.
One could ask, then, why not simply parallel Alkaline cells, with their ready availability, long shelf life and high storage capacity with one of these other cell types and get the best of both worlds? In theory you could - if you had some sort of charge control circuitry that was capable of efficiently meting out the energy from the alkaline pack and using it to supplement the "other" storage medium (e.g. lead-acid, lithium-ion, etc.) but you cannot simply connect the two types in parallel and expect to efficiently utilize the available power capacity of both types of storage cell - this, due to the wildly different voltage and charge requirements.
Even if you do use a fairly small-ish (e.g. 7-10 amp-hour) lead-acid or lithium-ion battery pack, even though its internal resistance may be low compared to that of alkaline packs, it may not be able to source the 15-20 amp current peaks of, say, a 100 watt SSB transceiver without excess voltage drop, particularly if it isn't brand new.
This is where the use of "Ultracapacitors" come in.
In recent years these devices have become available on the market and for reasonable prices. These capacitors, typically with maximum voltages in the range of 2.5-2.7 volts per unit, may have capacitance values as high as several thousand Farads in a reasonably small package while at the same time offering very low internal resistance - often in the units of milliohms. What this means is that from this capacitor one may pull many 10's of amps of current while losing only a small percentage of the energy in heat: Indeed, many of these capacitors have current ratings in the hundreds of amps!
What this means is that we can use these capacitors to deliver high, peak currents while our power source delivers a much lower average current.
The difference between peak and average current - and power:
With a simple "thought experiment" we can easily realize that with our transmitter, one second at 100 watts is (about the) same total power as ten seconds at 10 watts. If, in each case, we averaged the power over ten seconds, we would realize something else: In both cases, the average power is 10 watts per second.
Clearly the instantaneous power requirements for a radio operating at 10 watts output are different than at 100 watts: For the former you'll likely need 2-5 amps of current but for the latter you'll need 18-25 amps (at 12 volts, in each case.)
Here, we have a problem: If we have a given resistance somewhere in our DC supply, we lose much more power at, say, 18 amps than at 5 amps according to the equation:
P = I2R
Where:
I = Current consumption
R = Resistance of the power source including the battery, wiring, connections, etc.
P = The amount of power lost
(Notice that voltage is irrelevant in this case!)
In other words, the ratio of the power loss is equal to the square of the ratio of the current differences - in other words:
182 / 52 = 12.96 (or 13)
That means that with a given amount of power supply resistance, power losses at 18 amps are 13-fold worse than those at 5 amps.
Clearly the best way to mitigate this is with heavy cables to minimize resistance, but what if your power source is, by its very nature, fairly high in resistance in its own right?
This is where the "ultracapacitors" can be useful. By acting as a reservoir to handle peak currents, but rely on the battery - whatever form it may take - to make up the average.
With SSB (Single SideBand) transmission we have an (almost) ideal situation:
- There is no RF power during transmit when we aren't saying anything
- The amount of RF power is proportional with voice peaks, and
- Speech has comparatively rare peaks with a lot of empty spaces and lower-energy voice components interspersed.
- SSB is 6-12 times more power efficient in conveying voice than FM and occupying 1/3-1/4th of the space (bandwidth) as one FM signal.
Compare this with 100 watt FM transmitter:
- With an FM transmitter, when you are "keyed down" it is always putting out 100 watts, no matter what your voice is doing.
- 100 watts of FM is less effective in conveying voice than 10 watts of SSB and it takes at least 3 to 4 times as much bandwidth as an SSB signal.
For the purposes of this discussion the point that I was really trying to make was the fact that the use of these "ballast" capacitors is appropriate only for relatively low duty-cycle modes such as SSB or, possibly CW: If you tried this with FM or digital modes such as PSK31 the long duty cycle (e.g. key-down) would quickly drain the energy stored in the capacitors and they would "disappear", putting the load back on the battery bank.
This technique is not new: For many years now one has been able to buy banks of capacitors intended for high-power car audio amplifier installations that provide that instantaneous burst of current required for the "thud" of bass without causing a similar, instantaneous voltage drop on the power supply. Originally using physically large banks of "computer grade" electrolytic capacitors, these systems are now much smaller and lighter, using the aforementioned "Ultracapacitors".
There are also devices on the amateur radio market that do this: Take as an example the MFJ-4403 power conditioner (link). This device uses several ultracapacitors in series to achieve approximately 4 Farads of capacitance across the output leads of the device, allowing high peak currents to be pulled and "smoothing" out the average the instantaneous overall voltage drop that can cause a fuse to blow and/or the radio to malfunction due to too-low voltage.
Now, a few weasel words:
- The device(s) described on this page can/do involve high currents and voltages that could cause burns, injury, fire and even death if improperly handled if reasonable/proper safety precautions are not taken and good building techniques are not followed.
- The device(s) described are prototypes. While they do work, they may (likely!) have some design peculiarities (bugs!) that are unknown, in addition to those documented.
- There are no warranties expressed or implied and the author cannot be held responsible for any injury/damage that might result. Your mileage may vary, and do not taunt happy fun ball.
- You have been warned!
How this may be done:
For a variety of reasons (physics, practicality) cannot buy a "16 volt" ultracapacitor: Any device that you will find that has a voltage rating above 2.7 (or, in some cases 3.something) volts is really a module consisting of several lower-voltage capacitors in series. What is more, you cannot simply throw capacitors in series as there is no guarantee that the voltage will always divide equally amongst them unless there circuitry is included to make this so.
Another consideration is that if you have such a device - a large bank of capacitance - that is discharged, you cannot simply connect it across an existing power source because a (theoretically) infinite amount of current will flow from the power source into the capacitors, if their voltage is lower, to force equilibrium. Practically speaking, if you were connect the capacitor bank "suddenly" to the power source, if the resistance of the wires themselves didn't serve to limit the current, you'd likely blow the fuse, trip the breaker and/or cause the power supply to go into some sort of overcurrent (or shut down) mode - none of which are at all helpful.
Finally, this capacitor bank will be (theoretically) capable of sinking or sourcing hundred of amps if applied to a high current source/shorted out - perfectly capable of burning even heavy gauge wire, so some sort of protection is obviously needed.
The diagram in Figure 2, below, provides these functions.
Circuit description:
Capacitors C1-C6 are "Ultracapacitors": Their exact values are not important, but they should all be identical values and models/part numbers. In the example of the MFJ-4403, six 25 Farad capacitors are used yielding a total of 41/6 Farads while the drawing in Figure 2 depicts six 350 Farad capacitor being used to yield a total of 581/3 Farads. The greater the capacitance, the more energy storage, but also the longer the "charge" time for a given amount of current and, of course, the larger the size and the higher the initial cost of the capacitors themselves.
Zener diodes D1-D6, each being 2.7 volt, 1.3 watt units, are placed across each capacitor to help equalize the voltage. As is the nature of Zener diodes they will be at least partially conducting at lower voltage than nominal, the current increasing dramatically as their rated voltage is approached - which, itself, can vary significantly.
Originally, I experimented with the use of a series-connected resistor, diode and green LED across each capacitor to equalize the voltage as depicted by components Ra, Da and LEDa. In this circuit the LED, a normal, old-fashioned indicator-type "non-high brightness" green LED was used, taking advantage of its 2.1 volt threshold voltage along with the 0.6 volt drop of an ordinary diode with a 5.1 ohm resistor in series to provide a "knee". While this circuit did work, providing a handy, visual indication of a "full charge" state of each of the six capacitors, it did not/could not safely conduct enough current to strongly force equalization of the capacitors' voltages under typical charge conditions.
The Zener diodes, with their maximum current of more than 400mA as compared to 15-25mA for the LED-based circuit, seemed to be more effective. I left the LED-based circuit in place after constructing the prototype since there was no reason to remove them and the illumination of the LEDs served to indicate during testing that the capacitors are charging up with the equal illumination being generally indicative of equal charge distribution.
It will be noted that the "top" (positive side) of the capacitor bank is connected to the positive side of the power source, "straight through", at all times via TH1 and TH2, current-inrush limiters. Because of this, when this unit is "off" it is, for all practical purposes, transparent, consuming no current. These devices (TH1, TH2) act as self-resetting fuses, limiting the current to a sane amount - somewhere in the 30-50 amp region - if the output (or input!) is shorted: Ordinary "slow blow" fuses could be used here, but if so, the advice is to keep spares on hand!
It is only the "bottom" (negative side) of the capacitor string that is connected/disconnected to enable or disable the "ballast" (and filtering) capability of this circuit: When "off" the bottom of the capacitor bank is allow to float upwards.
If the capacitors are discharged when switch SW1 is turned ON while power is applied, the first thing that happens is that the gate of Q1 is turned on via resistor R3. When this happens, current flows through R1, but when its drop exceeds approximately 0.6 volts - a voltage commensurate with approximately 2.5 amps of "charge" current - transistor Q3 begins to be turned on, drawing down Q1's gate voltage. When this circuit reaches equilibrium only 2-3 amps will flow through Q1, R1 and TH3 and thus charging the capacitor bank comparatively slowly and preventing a "dead short" that would likely blow fuses and/or cause a power supply to "fold back".
While in this equilibrium state the gate voltage on Q1 is necessarily reduced to keep it partially "off" to maintain the current at approximately 2.5 amps and there will be a voltage drop across R3: This voltage drop is detected by Q5, a PNP transistor via R4 which, if Q2 is turned on, will also be turned on which, in turn, turns on LED2, the "Charging" LED which also turns on Q4 which, in turn pinches off the drive to the main capacitor bank switch, Q2, forcing it off.
(In the event of a circuit malfunction, self-resetting fuse TH3 limits the maximum current through Q1 to 5 amps before "blowing", at which point the current will be reduced to a few 10's or 100's of milliamps.)
Once the capacitor bank has become (mostly) charged and the voltage across it is nearly the same as the applied voltage, the charging current will begin to drop and as it does, transistor Q3 will start to turn off, causing less voltage drop across R3 in an attempt to make Q1 conduct more. At some point, when the capacitors have reached (more or less) full charge and current flow starts to decrease, Q1 will be "fully" on (e.g. its gate voltage approaching "V+_SW") and as this happens the voltage drop across R1 will have decreased to practically nothing. When this drop is less than approximately 0.6 volts, Q1 will turn off completely and the voltage at the "bottom" of R1 (the side connected to the gate of Q1) will be equal to that of "V+_SW" and this will cause transistor Q5 to turn off.
Once Q5 turns off the "Charging" LED it will also turn off as will Q4 and the voltage on the gate of Q2, being pulled up by R5 and "slowed" by capacitor C8, will start to rise. As the gate voltage on Q2 crosses the "on" threshold it will conduct strongly, connecting the bottom of the capacitor bank to ground with only a few milliohms of resistance.
If switch SW1 is turned off the voltage at "V+_SW" drops and via R5, C8 and the voltage at the gate of Q2 drops, turning it off and disconnecting the capacitor bank.
Construction details:
The ultracapacitors were wired together outside the enclosure (a Hammond 1590D die-cast box) using multiple, folded pieces of #12 AWG bare wire, both for low ohmic resistance and mechanical support. Additional pieces of wire were used on the capacitors' electrically-isolated mounting pins for spacing and support when the two banks of three were arranged to be parallel to each other, terminals facing - but separated by a safe distance as seen in the pictures. The balancing circuits were also installed across the capacitors at this time.
Prior to mounting the capacitors in the bottom of the box the holes for the LEDs and switches were drilled/filed - this to eliminate the possibility of the tools damaging them during work. The two parallel banks of three series capacitors were prepared and then placed in the bottom of the box, held in place with RTV (Silicone (tm) seal).
There is no etched circuit board. The actual control circuitry is mounted on the lid, with Q1 and Q2 being bolted to the lid itself for heat-sinking using electrical insulating hardware (washer, grey insulating pads) that were scavanged from a dead PC power supply. The circuit itself was constructed on a piece of glass-epoxy circuit board material.
Without making a circuit board, there are several ways that the circuit could have been constructed, but I chose to use a variation of the "Manhattan" style which involved islands of copper. In some instances - such as for the large resistor(s) comprising the 0.25 ohm unit on Q1 and for the connections of the power in/out leads and TH1 and TH2, islands of copper were isolated on the board by first drawing them out with a pen and then slitting both sides of a narrow (1/16th-1/8th of an inch) trace with a sharp utility knife and straight edge and then using the heat from a soldering iron to aid in the lifting of the narrow strip to isolate the area of board.
For other portions of the circuit I used "Me Squares" available from QRPMe.com (link): These are small pieces of glass-epoxy circuit board material with nice squares etched on them that one glues down using cyanoacrylate (a.k.a. "super") glue and then used as points for soldering and connection.
The nice thing about these "Me Squares" is that they are very thin, look nice and are very flat - which makes them easy to solder and glue down, but one could also cut out squares of ordinary circuit board material and solder those down, instead, provided that they were also made flat on the back side and de-burred for maximal surface contact and adhesion. Finally, one could use a utility knife and isolate islands - or even use an "island cutter" tool - to produce isolated lands on the piece of circuit board material itself.
The main reasons for using this technique were that it was quick, and also that it was surface-mountable: Had I wired it on perforated board or even made a conventional circuit board with through-hole parts I would have had to stand it off from the lid to insulate the circuit from it: Using this technique the board itself was simply bolted flat to the lid so that it would fit in the limited space between it and the capacitors - and was quick to wire up.
For interconnects between the circuitry on the lid short lengths of #12 AWG stranded wire were used for the high-current leads connecting the capacitors and input/output leads and much smaller (#24-#28 AWG or so) for the wires that connect to the switch and LEDs. For the "outside world" connections 30 amp Anderson Power Pole (tm) connectors were used on 12 AWG cables in the standard manner.
A few caveats with this circuit:
- This circuit consumes some current (at least a few 10's of milliamps - maybe more) whenever it is set to "on", even after the capacitors have equalized. What this means is that it will slowly drain your battery if the switch is left in the "on" position and because of this it is recommended that one switch it to the "on" position ONLY when intermittent, high current is going to be needed. In other words, if you are going to receive (only), leave it "off", turning it "on" only if you plan to transmit, knowing that it may take several 10's of seconds for the capacitors to charge.
- If the power source cannot deliver the amount of current required during the "charge" cycle - or if it briefly "blinks" - the source voltage will sag and it will likely switch from "charge" mode to "operate" mode and connect the ultracapacitors directly across the power source. If the power source has limited current in the first place, this will simply mean a "dip" in voltage while the capacitor bank charges, but if this occurs with a high-current source that has had a momentary glitch, a premature switch to "operate" mode could, in theory, blow a fuse. This premature switching would likely happen if you had this connected via a cigarette lighter and started a vehicle while it was in a charge phase.
- This capacitor bank should be treated like a car battery: If it is shorted out you will get lots of current - more than enough to burn open small wires and blow fuses - or even burn open large wires or small tools if you do an "oops" on the unfused/unprotected side of the circuit, or neglect to include any fusing/current protection!
- Neither this capacitor bank (or any battery) should EVER be placed directly on the output with any power supply that has a "crowbar" voltage protect circuit (such as is present on many power supplies such as the Astron "RS" and "RM" lines) as a power line or transient on the output could cause the crowbar to trigger and short the output of the power supply - including the capacitor bank or whatever battery you might have on the output. If this happens, expect significant damage to the power supply and possibly the connecting wiring!
- This sort of device must be housed within a rugged enclosure/container. The picture shows the prototype being built into a Hammond 1590 series die-cast aluminum box that is both very rugged and also provides heat sinking for Q1 and Q2. In the unlikely event of a catastrophic failure due to a wiring problem or a capacitor going bad this enclosure will not melt and is likely to contain the "mess" - even if it were to get very hot!
- If used in a vehicle, this circuit should be disconnected/disabled when the vehicle is started since these capacitors can supply "car starting" current on their own and it is possible to blow fuses, pop breakers, damage switches, circuits, relays and other mischief if the current from the capacitor bank were to "back-feed" through vehicle wiring that was not designed for that sort of current!
- It is the nature of power FETs such as Q1 and Q2 to have an intrinsic "reverse diode" - even when turned off. Be aware that if it so-happens that the bottom of capacitor bank is more negative than the "ground" (Battery -) side, these diodes - particularly the one in Q2 - will conduct!
Possible uses:
While it is possible to use this to allow an HF rig to be powered from a set of D cells, it is more practical to use smaller lead-acid or lithium-ion packs as the primary power source and use the capacitor bank as the "ballast" to supply the peak currents.
I'm certain that this device could be improved, but it seems to function as it is.
Conclusions:
How usable is it, in the real world?
It depends a lot on how - and with which radio - you plan to use it. For example, many older-vintage Kenwood HF transceivers will fail to function properly (e.g. operate with distorted audio, "FMing" of the signal, etc.) much below 12-12.5 volts while more modern, compact HF radios like the Yaseu FT-100 and FT-857 will happily run at 10-11 volts - perhaps at reduced output power - but fine otherwise. The upshot is that if you are considering a radio to be operated from "marginal" power sources be certain that you have done your research and consider how your candidate radios operate at low supply voltage and how/if they degrade "gracefully" - or not!
How about running an HF rig from alkaline "D" cells? As it turns out I can happily transmit 100 watt (peak) SSB using my old Yaesu FT-100 with the device described on this page using 10 "D" cells in series.
To do this effectively one must minimize the contact resistance of the battery contacts which pretty much rules out using cheap, spring-loaded plastic battery holders which, by themselves, can have almost as much resistance as the cells themselves. Aside from spot-welding tabs onto the alkaline cells (the heat of soldering would likely cause some damage and slight loss of capacity) the best holders are aluminum with heavy bus bars and the fewest number of springs and contacts (e.g. multiple cells directly in series) such as the Keystone four-cell holders, models 158 (two of them) along with one two-cell holder, a Keystone model 186.
One cannot "key down" with a carrier without significant voltage sag, but the FT-100 seems to work OK on typical SSB with voice peaks - but under such heavy loads don't expect to get much longevity before the alkaline cells' internal resistance increases: As noted previously not all radios (such as older Kenwood mobiles) behave so well at lower voltages, so do your homework!
As detailed in the article, a more practical use of this sort of device is as a "power conditioner" to help compensate for voltage sags due to resistance in the interconnecting cables, somewhat underrated power sources, aging battery packs and/or "small"-ish batteries.
* * *
How about using this with a solar power source?
Assuming that voltage from the panel is regulated to a safe value (15 volts or below) the capacitor bank could, in theory, maintain voltage if the solar array provided at least the average current, but considering that solar illumination can vary wildly due to time of day, sun angle, clouds and shadows it would be recommended that additional storage capacity (remember that 53-1/2 Farads at 12 volts has about the same theoretical storage capacity of a single "AA" alkaline cell!) be used as well, such as a 7-20 amp-hour lead-acid or lithium-ion based pack - but this sort of system could well be the basis of another article.
[End]
This page stolen from "ka7oei.blogspot.com".
Excellent article, thank you.
ReplyDeleteOne comment:
"The capacitor bank/power conditioner with 53-1/3 Farads,
in a package slightly larger than a 7 amp-hour, 12 volt
lead-acid battery. How much energy is actually
contained in 53.33F at 13 volts? Theoretically, about
the same as just one alkaline AA cell."
I disagree with that:
The bank E = CV^2/2 = 4500 J
AA cell E = V*mA*h = (typ) 1.6V * 6 Ah = 10 J
Two orders of magnitude if favour of your bank.
Thanks for the note!
DeleteIn my calculations I come up with an AA alkaline having a bit more than the capacitor bank. For example, an AA alkaline cell, a fairly low discharge rate, has approximately 2.2 amp-hours of capacity, so assuming an average cell voltage of around 1.25 volts one obtains around 2.75 watt-hours. Considering that there are 3600 Joules in a watt-hour, this would imply approximately 9900 Joules from the AA alkaline cell.
An interesting web page that shows similar comparisons of various battery chemistry and cell sizes (and includes some capacitor examples) is here: http://www.allaboutbatteries.com/Energy-tables.html
Practically speaking, there is a bit of an "apples and oranges" problem in comparing a capacitor and an alkaline cell:
- Only the "top" few volts of the capacitor bank are actually useful: The radio gear will stop functioning below 10-11 volts, leaving the remaining charge unused unless some sort of switching converter were employed. (There are problems with "constant power" converters with current increasing as voltage drops, but that's another discussion altogether!)
- The alkaline cell's stated capacity (2.2 amp/hours) is only usable under fairly low-drain conditions: It rapidly diminishes to a fraction of this at currents that even remotely approach the "C" (capacity) rating of the cell which makes it much less useful for high-current applications - hence the utility of a low-impedance current source such as the capacitor bank to minimize the losses during transient, high-current conditions. Drawing even a moderate amount of current would certainly reduce the amount of useful energy significantly from the aforementioned "9900 Joule" number!
Again, thanks for reading!
Sorry, I did not multiply by 3600 to get J. I could not believe that such a setup would store so little energy compared to a venerable AA cell thus when my maths confirmed that I did not double checked ;-(
ReplyDeleteThank you for clearing away my misbelief.
On the positive side it means that the supercaps can be charged quite quickly. I am now afraid to show my maths but it seems to me that charging a 1F supercap with, say, constant 100 mA current only will fully charge it in just 26 seconds. You do have a charging current limiter with a 5A fuse; do you know the value of the initial charging current and how long it takes before the capacitors are considered fully charged and Q2 connect their negative terminal to the ground? If some "average" charging current is available, one can calculate the efficiency of the charger and try to trade off charging time with efficiency which might be important for portable equipment.
It is unfortunate that at this time that the storage capacity of capacitors is several order of magnitudes worse than than of "conventional" batteries, volumetrically.
DeleteYou are correct in noting that capacitors can exceed most cell technologies in endurance (e.g. charge/discharge cycles) and charge/discharge current which is why they are finding increasing use in power-limited systems where high surge powers are required for only limited periodes (e.g. traction motors on vehicles, etc.)
The circuit depicted limits the charge current to something in the area of 2.5 amps - varying 20-30% with temperature - and this current is maintained from when the charging starts (capacitors discharged) to a very short time before equalibrium. The capacitors specified are rated for 350 amps which means that without such current limiting, the 5 amp fuse would be triggered and the charge rate would be limited a few 10's to 100's of milliamps (the "holding" current for the resettable fuse) or, without it, the likely capacity of the power supply, Q1, or the conductors themselves - whichever blew up first.
Had I wanted to spend more time on the charging circuit I would have implemented a buck-type power converter to cause a constant-current load on the source supply to produce varying current while the capacitors were charging, greatly reducing charging inefficiencies. While this would have improved efficiency while charging - and added some "smarts" to minimize the likelihood of some sort of - I decided that in the time I had to build this, and the added complexity - wasn't really worth the extra efficiency that would be gained only during the charge cycle.
Of course, how much energy is wasted in the charge cycle depends on how often the charge cycle is invoked, the supply voltage, and the capacitance of the bank.
I don't recall the time to charge the capacitor bank with the values depicted in the diagram from a full discharge, but I believe that it was somewhere between 60-120 seconds: I keep meaning to check...
Great Work and I really like the balance circuit with the Caps. But I have a few questions and maybe I am looking at it wrong. The part of the schematics where you have the leds. It shows the resistor then led then diode, starting from the Positive side. But looking at the picture, it looks like you actually did diode first with even the diode backwards compared to the schematics. ( I am 300% a newbie and I am in no way an expert), reason I am asking. I plan to follow the Picture. Last quick Q. Those caps that you used. Do they have a positive and negative bigger top pins, and then 2 soldering Pin only. It look like they have 4 pins on the top, 2 larger and 2 smaller. Great Work!!!Has helped me allot that is forsure!!!
ReplyDeleteThink I am just looking at it wrong. I drew it out on paper and then compared it to the picture and it looks great! So only the Zener diode has the blocking side on the positive. Zener goes from the negative to positive (cap facing the Positive). Then the Led part has the diode and led with the Cap facing the negative, followed by a 5.1ohm resistor. I like the led there, even if its only lighting when charged. I actually made a board with Easy EDA that takes Maxwell D C caps(350F 2.7V). It takes 6 of them in series, with your circuit used as a simple balance. Than I made the board where you can stack them with Std off or M8 bolts to parallel another board of 6 or 2. It's basically storage for a Capacitor Battery Tab welder I am making. I just wanted to double check everything before I built. Easyeda sent me the boards in 3 days. I also made sure to use 2oz copper for the tracks. Let me know if you think I may have something wrong. Thanks for posting this project! Its helped me BIG TIME!!!
ReplyDelete"njfulwider5" : )