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AlbertHall
- Joined Jun 4, 2014
- 12,660
In series with the current that you want to measure. Connect the two leads from the 'scope one to each end of the resistor.
What is the main principle of charging a supercapacitor?
- Thread starter Taha Muhammed
- Start date
Ok will try it tomorrow and will soon get back to you thanks
Do you need to allow for the internal resistance of the capacitor? (or the AA battery) Is the resistance of the wiring and capacitor more than that of the resistor? What kind of switch are you using to discharge the capacitor? Does it have a resistance?
The super capacitor has a significant internal resistance. Even the huge ones that are rated at high discharge rates will have enough internal resistance, and equally significant at this short (bogus) charging cycle, inductance, which will limit the current to way less than 1000 amps at this applied voltage. In any event, the resistor will not burn at that short of a duty cycle.
The 500 microsecond charge specification is a bogus figure. To charge a super capacitor you will need seconds of charge time even for a relatively small MFD super capacitor. Example: A 1 farad super capacitor charged at 1 amp initial starting current will take 1 second to reach 63% of the charging voltage. . The 5 rc you state in your post reflects a resistance determined charge to around 97%, which would take around 5 seconds if the initial charge current was 1 amp. So in the example from a 5 volt charging voltage you would use a 5 ohm resistor to get the initial charging current to be 1 amp. The capacitor will then charge to 97% of the 5 volts in 5 time constants which would be 5 seconds.
You can play with the values in the example to determine any charging time, capacitance, resistance, and voltage you want by simply factoring the numbers.
The 500 microsecond charge specification is a bogus figure. To charge a super capacitor you will need seconds of charge time even for a relatively small MFD super capacitor. Example: A 1 farad super capacitor charged at 1 amp initial starting current will take 1 second to reach 63% of the charging voltage. . The 5 rc you state in your post reflects a resistance determined charge to around 97%, which would take around 5 seconds if the initial charge current was 1 amp. So in the example from a 5 volt charging voltage you would use a 5 ohm resistor to get the initial charging current to be 1 amp. The capacitor will then charge to 97% of the 5 volts in 5 time constants which would be 5 seconds.
You can play with the values in the example to determine any charging time, capacitance, resistance, and voltage you want by simply factoring the numbers.
Using a 'scope to measure voltage drop across the current sensing resistor, one termination of the resistor must be connected to circuit ground. Usually circuit ground is also ground of the 'scope which is the reference potential of one of the two connectors of the probe of the 'scope.
There are ways around this but, I agree, measuring voltage with a scope to calculate current seems a bit convoluted - especially if scope is grounded and circuit is charged from mains or a non-isolated transformer.
The current sensing resistor and 'scope is not bad if you want to see the waveform of the current, or when current is drawn relative to a voltage of the circuit. Of course a multi-meter will generally do very poorly at measuring a changing current.
In the past I remember finding that isolating the 'scope from earth ground would allow measuring voltage drop across a current sensing resistor that was not connected at one end to circuit ground. Do you think that not earthing the 'scope is bad practice? In my case this is the old style cathode ray 'scope.
How about the computer-based scope? With that type of scope, does the probe take a floating voltage or is one end of the probe grounded with that type of 'scope also?
This info I suppose might be helpful for the OP, otherwise I'll discontinue pursuing the topic here.
-Pete
AlbertHall
- Joined Jun 4, 2014
- 12,660
With a 2 channel 'scope you can usually use probes connected to both channels set to invert one channel and display the sum so you a differential input to the 'scope. Provided the voltage to ground isn't too great this gets around the grounding problem.
New to this topic here, but am I to understand that the theory the original poster (OP) is thinking this way?
5 volts, 1 ohm = 5 amps?
If so then 5 volts, 0.001 ohm should equal 500 amps. In theory, yes. However, someone asked before (don't remember the post) "can your power supply deliver that much current?" That question needs to be considered more seriously, as if the best your supply can give is 1 amp, no amount of place holders in your decimal equation will give you any more power than is available. By simply lowering the resistance you can't get "More Power".
Also needing consideration is the imperfections of conductors. Copper wire of a given size will have a given resistance over a given distance. Since I don't have the numbers in front of me I'll make something up just for the sake of clarification. Suppose a copper wire, 12 gauge, can handle 20 amps at 120 volts (AC) on a run of 100 feet. At the end of the run the useful voltage is 119 volts. That means in 100 feet you have a 1 volt drop. The calculated resistance would be 1 ÷ 20 = 0.05 (ohms). Divide that by 100 (feet) and you get 0.0005 ohms per foot. Keep in mind I'm using made up numbers, but the calculations are accurate (I hope). Adding an even lower resistance isn't going to lower the overall resistance. Instead, it will increase it.
The whole point I'm trying to make is that theory is fine on paper. But in the real world there are other things that need to be considered. Such as "How much resistance is my circuit experiencing just based on the wire being used?" Using gold wire will further reduce the resistance, but I imagine even gold wire has resistance of some sort, measurable or not. And your other components also have internal resistances to be considered. But the paramount concern is "How much current can your supply deliver?" If all it can muster is one amp, lowering the resistance ever closer to zero ohms is NOT going to give you "More Power". Not unless you discover "Negative Resistance", which would increase current as opposed to limiting it. Remember, resistance "RESISTS" current. Not voltage. Resistance is like kinking a garden hose. Unkinking it (zero resistance) means the hose can deliver as much as the spigot can (provided the hose itself is big enough). And even then, the longer the hose the more "resistance" the water flow (current) experiences.
Your SC (Super Capacitor) will take whatever charge it will take at whatever rate it will take it. You can't force it to take more - faster. You also need to understand (and I don't pretend to understand much about capacitors), but you need to understand that "Current preceeds voltage." That means when a capacitor is at zero volts and you apply a given current, the max current drawn by that capacitor will be the max the supply can give at the moment you begin charging it. Over time (5T) the capacitor will charge to full capacity. As the capacitor takes more and more charge the current drawn is less and less. At the end of 5T the capacitor should be drawing (theoretically) no current at all.
Charging a cap for 0.5 milli-seconds isn't going to result in any kind of useful power.
5 volts, 1 ohm = 5 amps?
If so then 5 volts, 0.001 ohm should equal 500 amps. In theory, yes. However, someone asked before (don't remember the post) "can your power supply deliver that much current?" That question needs to be considered more seriously, as if the best your supply can give is 1 amp, no amount of place holders in your decimal equation will give you any more power than is available. By simply lowering the resistance you can't get "More Power".
Also needing consideration is the imperfections of conductors. Copper wire of a given size will have a given resistance over a given distance. Since I don't have the numbers in front of me I'll make something up just for the sake of clarification. Suppose a copper wire, 12 gauge, can handle 20 amps at 120 volts (AC) on a run of 100 feet. At the end of the run the useful voltage is 119 volts. That means in 100 feet you have a 1 volt drop. The calculated resistance would be 1 ÷ 20 = 0.05 (ohms). Divide that by 100 (feet) and you get 0.0005 ohms per foot. Keep in mind I'm using made up numbers, but the calculations are accurate (I hope). Adding an even lower resistance isn't going to lower the overall resistance. Instead, it will increase it.
The whole point I'm trying to make is that theory is fine on paper. But in the real world there are other things that need to be considered. Such as "How much resistance is my circuit experiencing just based on the wire being used?" Using gold wire will further reduce the resistance, but I imagine even gold wire has resistance of some sort, measurable or not. And your other components also have internal resistances to be considered. But the paramount concern is "How much current can your supply deliver?" If all it can muster is one amp, lowering the resistance ever closer to zero ohms is NOT going to give you "More Power". Not unless you discover "Negative Resistance", which would increase current as opposed to limiting it. Remember, resistance "RESISTS" current. Not voltage. Resistance is like kinking a garden hose. Unkinking it (zero resistance) means the hose can deliver as much as the spigot can (provided the hose itself is big enough). And even then, the longer the hose the more "resistance" the water flow (current) experiences.
Your SC (Super Capacitor) will take whatever charge it will take at whatever rate it will take it. You can't force it to take more - faster. You also need to understand (and I don't pretend to understand much about capacitors), but you need to understand that "Current preceeds voltage." That means when a capacitor is at zero volts and you apply a given current, the max current drawn by that capacitor will be the max the supply can give at the moment you begin charging it. Over time (5T) the capacitor will charge to full capacity. As the capacitor takes more and more charge the current drawn is less and less. At the end of 5T the capacitor should be drawing (theoretically) no current at all.
Charging a cap for 0.5 milli-seconds isn't going to result in any kind of useful power.
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Thanks for letting us know about this procedure. On my dual trace 'scope, there are the controls that would allow me to use that method.
My current scope (DS1054) has been used to do just that in measuring voltage across a minimal resistance to compute current flow.
I use 4 X 10,000 uF group of caps to charge up to 10 volts. With a 1/2 amp current limited supply you can watch the voltage increase while it charges the caps. I then dump it across an 'inductor' of unknown inductance to determine the probable actual value. It's not uncommon to dump 80 amps for about 10 us measured across a .1 or .01 ohm resistance. If I get 100 mv across the .1 ohm resistor, it's easy compute.
I know many people who've fried scopes (especially multi channel models) by cross connecting grounds. I would not say it's a bad practice, if you're aware enough to be doing this stuff, you should be aware of what's ground and what isn't. You really had to watch this in the vacuum tube days. Most stuff is low voltage and the worst is that you fry the item you're working on.
This link is may be helpful to show the use of this technique determine 'unknowns', in this case the value of an unknown inductor. The schematic is simple and explained. I use a DSO so I don't need to keep firing it over and over, like an old scope needed. I use a very similar trigger circuit and cap storage. I also use a micro for a single desired pulse duration (mosfet 'on' time.)
http://ludens.cl/Electron/lmeter/lmeter.html
IMHO ;-}
I use 4 X 10,000 uF group of caps to charge up to 10 volts. With a 1/2 amp current limited supply you can watch the voltage increase while it charges the caps. I then dump it across an 'inductor' of unknown inductance to determine the probable actual value. It's not uncommon to dump 80 amps for about 10 us measured across a .1 or .01 ohm resistance. If I get 100 mv across the .1 ohm resistor, it's easy compute.
I know many people who've fried scopes (especially multi channel models) by cross connecting grounds. I would not say it's a bad practice, if you're aware enough to be doing this stuff, you should be aware of what's ground and what isn't. You really had to watch this in the vacuum tube days. Most stuff is low voltage and the worst is that you fry the item you're working on.
This link is may be helpful to show the use of this technique determine 'unknowns', in this case the value of an unknown inductor. The schematic is simple and explained. I use a DSO so I don't need to keep firing it over and over, like an old scope needed. I use a very similar trigger circuit and cap storage. I also use a micro for a single desired pulse duration (mosfet 'on' time.)
http://ludens.cl/Electron/lmeter/lmeter.html
IMHO ;-}
