Can buck converter increase current if voltage step down ...
2 answers
Jan 17, 2018 - 2 Answers. So before I thought I was getting 3 amps from my previous buck converter. But I was using the multimeter to read the amps. And I forgot that because the multimeter has very low resistance therefore the amps will be very high. So it was a faulty reading!!A buck converter can output almost all the power it's getting from its input (less typically 5-15% losses), which means if the output voltage is significantly lower than the input, the output current can be significantly higher.
This new DC-DC buck converter has a digital read out for amps - and so I only get a reading of 2 amps max (and that is at great resistance pedaling on the bicycle). So I can get only a trickle charge of 1 amp. It's better than nothing but it's hardly anything at all. Here a DC-DC Buck Converter on a bicycle DC motor is getting only 200 to 300 milliamps charge output - vid
So when I pedal faster than I do get an increase in amps but also it's much harder to pedal. I thought maybe if the battery is not as charged up then it should have lower resistance. So I ran the battery down to in the 20 percentile range. It made hardly any noticeable difference.
One thing that really bothers me is your use of the battery as a load. If you have a battery of 12V and a resistance of 0.1 ohms at a point where the circuit is at a potential of 14.4V, you only have a voltage of 2.4V across that resistor. So you're going to get a maximum of 24A pulled through the load resistor.https://forum.allaboutcircuits.com/threads/buck-converter-steps-down-voltage-but-does-not-step-up-current.115519/page-2
interesting!
https://www.quora.com/How-exactly-does-a-DC-buck-converter-produce-a-low-voltage-high-current
So a DC-DC Buck converter tutorial vid
Note the fact that the inductor is an element that:
a) stores energy in the form of magnetic field
b) “doesn’t like” the change in current - it tries to oppose it by producing a voltage to counteract this change
One way to imagine an inductor is to imagine a water turbine, assuming the pressure is the voltage and the water flow is current.
There are really two switches here - one is presented just as a switch (in reality it is a transistor, controlled by a controller) and one is a diode, which conducts only when voltage at the anode is greater from voltage at the cathode (so when the current through the inductor is decreasing, as you’ll see later). I’ll refer to them as S and D.
- At the beginning, S is open (not conducting). No current flows and voltage across the load is zero.
- S is closed (begins conducting). Current flowing through inductor starts to increase and inductor responds to this change by producing voltage of the opposite polarity, so the voltage across the load is equal to the difference of input voltage and voltage produced by the inductor because of change in current - therefore, it’s smaller than supply voltage. Diode is not conducting at this moment. We’re also ignoring the voltage drop across the switch for simplicity.
- S is open (stops conducting). The current (from the source, that’s important) rapidly drops to zero, but there’s still magnetic field accumulated in the inductor that has to go somewhere. Remember that the inductor responds to change of current by generating a voltage. This voltage will have opposite polarity from the voltage generated when the current was increasing (the minus will be on the left, the plus on the right). Diode starts conducting and now the inductor powers the load through the diode with a decreasing current. Note that the input source is now disconnected and current flows only from the inductor, through the diode, to the load. Because of that, the average current flowing through the load is greater than the average current drawn from the source.
Note that even if the current is drawn from the source in pulses, it is delivered to the load even in the absence of current being drawn from the source (when the switch stops conducting), and I’m deliberately avoiding the word “continuously” here, because it’s not always the case. The average current through the load is larger from the average current drawn from the source.
Another way to think about it is that because energy has to be conserved and no energy has been wasted here (we’re ignoring losses, because even if they limit the final efficiency, they’re not relevant to the theory of operation, as opposed to linear regulators, in which power loss is their principle of operation), the same energy with the lower voltage delivered is able to sustain a bigger current - hence increase of average current.
OK so basically because I'm using a battery as a load - that has lower resistance - then the DC-DC buck converter is losing too many amps due to the back current being too strong. I'm not sure what is happening.
But at least with this new device I could get a better reading of my amps.
So then I decided to use my battery generator - the Chafon - to power the Treadmill DC motor. I hooked up a DC 3 amp 12 volt output from the Chafon to the Treadmill motor. So then I knew what 12 volts was. So then I hooked up the bicycle poly V-belt so that it the bicycle wheel was not powered by the Treadmill Motor via the Chafon - and not the other way around! This was a lot of fun - and so the 12 volt speed is as slow as I can go. 1.5 HP DC Motor to 48V Electric Generator DIY - Amazing Idea 2019 vid shows a lot of torque at 12 volts!
So then I hooked up the Multimeter again - and when I pedal - I can not get a steady output speed from the bicycle - I guess since the torque is higher at slow speed. So then I did an amp reading - and literally when I spin the bicycle wheel by HAND then I get the highest amp reading - over 2 amps and almost 3 amps. So the irony being that a Treadmill motor is designed for higher amps at walking speed. Only to charge a battery then I lose amps by increasing the voltage.
So that is why a DC-DC Buck converter does not increase amps much via a treadmill motor.
For the new 12-volt TV, the specs say a 3.17 amp draw maxSo he powers a 12 volt 7 inch tv by walking on the treadmill - vid
So this person has the SAME set up that I did - and pedals at just over 12 volts and says he gets about 2 amps (but he was not real clear on the amps!)
So this person says - don't pedal over 13 volts or the exact energy is just wasted in heat sinks, etc.
Here someone is getting 1 amp max for steady output - vid
The PS2 requires 30 Watts of power through an AC inverter. The purpose of the inverter is to convert 14.8 DC (Direct current) volts from the pedal power bike generator into 110 Volts AC (Alternate Current) which is what you use when you plug something into the wall outlet in your Home. So when you pedal you will be supplying a total of 30 Watts and about 2.0 Amps. Note: the way you calculate Watts is Current AMPS X Volts. In this case: 2.0 Amps x 14.8Volts DC = 29.6 Watts. If you want to see voltage current Amps and Watts waveform graph data from this application go to http://pedalpowergenerator.com
So he recommends the charge controller so that the generator doesn't blow up...
So it is good that I am using the DC-DC buck converter for that reason.
https://www.askaprepper.com/diy-bicycle-generator/
alternator runs off RPM, not HP. A 100 amp alternator must turn at 1200 RPM to produce 100 amps. At 600 RPM it will generate half that or 50 amps.So based on my previous DC motor post - I would need a huge amount of rpms -
So to get 1 amp is actually not so bad based on the motor rated at over 6000 rpm for 18 amps I think.
6700 RPM 18 AMP MAGNET
The lower the RPM of a motor, the more torque it will have – torque being the measure of a tendency to cause rotation; in other words, the power to turn. This allows the motor to last longer. I strongly recommend a motor with an RPM rating of 4000 or lower, but never more than 5000.https://blog.johnsonfitness.com/blog/treadmill_drive_motors_and_the_question_of_horsepower/
This is crucial because this is where a lot of manufacturers begin to play with, and boost, the ratings above 2.5 in to the 3.5 and 4 HP range. In a lot of those cases, if we examine the plate stamped on the treadmill, you will find RPM ratings in the 7000 to 8000 range. This is how they can get away with these bloated HP rating numbers.
And second, generally speaking they only generate the rated HP at their top speeds, which in my experience has been anywhere from 4,500 to 7,000 rpm.
here's another treadmill motor spec:
1.75 HP Continuous Duty @ 90 VDC
Amps 15
4200 RPM @ 90 VDC
1.75 hp = 1330W (hp x 760)
4200 rpm = 439.8 radians/sec (2 x PI x rpm/60)
torque =1330/439.8
=3.02Nm
DC motors produce maximum torque at zero RPM.
an Australian history movie, ( John Flynns Outback ) and they had radios in very remote areas mainly used to call the “Flying Doctor” that were peddle operated, and this was in the 1930’s. I would think if peddle power was ever going to be used as a main power supply, it would of happened before now, but still its usablehttps://www.instructables.com/id/Best-DIY-Bike-Trainer-Generator/
So he can get over 4000 rpm at 15 mph!!
Oh he is converting to AC power so when he says 400 watts he means 3.3 amps. But he blew the circuit at 300 watts. So he's back down to ... 2 amps. Not bad.
A treadmill motor from a pro form treadmill.
2.65 hp (1.5 hp cont. Duty at 95 vdc). 21.4 amps. INS. class H. 25c ambient. Open const. External fan
The treadmill motor is rated at 95 volts so if you use 48 volts
It will only put out about 0.4hp to .66hp but should last a long time.
In Brazil they have inmates on bikes chargeing batteries to power street lamps ....they bike 8 hours a day! At a moderate pace!
I had a little time last weekend to tinker again with the treadmill.
i hooked the multimeter to the pos and neg leads to the motor to see what the motor runs at. What i found is that for every mph you set the treadmill for, the voltage changes accordingly: 1 mph = 10volts; 2 mph = 20volts; all the way to 10 mph = 95volts
then i connected the motor to a 12 volt battery. the motor spun (silently) but had almost no tourque. at 10volts from the treadmill it spins (noisily) with loads of tourque.
Your math for Ohm's law for solving resistance is correct, but it has flawed data for current. Specifically, the 21.4 amps is not the stall current. You have to take into account the motor's back EMF when it is turning. He would have to lock down the shaft and apply low voltage to the motor then read the current that flows to the motor and measure the voltage at the motor terminals to get as accurate a result as possible.
Exactly!
http://www.diygokarts.com/vb/archive/index.php/t-4728.html
As I also blogged:
https://elixirfield.blogspot.com/2019/02/eb-aka-back-emf-aka-counter.html
So now I'm challenging this person's claim of getting 300 watts - because he is not taking into account the RPMs needed for the amps - and also what the Ohms are based on the rpms!
https://www.thedoityourselfworld.com/articles/article.php?id=11488
If it's a 90V motor (I have a Baldor treadmill motor rated at that) you're already seeing less than half the rated voltage, and thus half (or less) the HP. Actually, you're probably down to more like a sixth or even just an eighth the HP- the power curve on those lightweight, high speed motors falls off real fast as the RPM drops.https://www.garagejournal.com/forum/showthread.php?t=186594
in the case of a typical cheap treadmill motor, results in the HP falling by the square of the speed- as in, at half RPM, HP is a quarter rating, at a quarter speed it's a sixteenth, etc.
If you lower the voltage the current will be lower.So that's 44 watts at 4.44 ohms and 14 volts!!
So in order to calculate the actual power we need to find the R or resistance of the motor. If we use ohms law V = I x R and rearrange it to V/I = R then 95Volts/21.4amps= 4.44 ohms.
The formula for Power in watts is Vsqrd/R or
48v x 48v / 4.44ohms = 519 watts
At full voltage 95v x 95v / 4.44 = 2033 watts
V = volts, I = current, R = resistance in ohms.
The Kv rating of a brushless motor is the ratio of the motor's unloaded rotational speed (measured in RPM) to the peak (not RMS) voltage on the wires connected to the coils (the back EMF).
9.1 amps at 130vdc is more like 1.58 hp theoretical, probably 1hp actual shaft hp.
amps squared is proportional to watts lost in the rotor.
https://electronics.stackexchange.com/questions/162187/does-a-motors-stall-current-change-with-applied-voltage
Once the rotor starts to rotate it will induce a back emf so the voltage 'seen' across the rotor will be reduced. The net current through the motor will be reduced.So the Chafon is 97 volts and 31 amps - so the Ohms is ~ 3.2
That means at maximum solar input the panel can deliver its rated power into a 18 volt / 5.55 amps = 3.24 ohms.Ah - now it's starting to make sense! So again that's the "LOAD" of the Chafon battery.... as it charges up, the resistance goes up. https://bbs.homeshopmachinist.net/forum/general/13538-dc-motor-question-inline-choke
The motors may not have sufficient internal inductance to smooth out the current flow. Current is what produces torque, so if current flow is uneven, the torque is too.
If you have access to a DMM with an inductance meter in it (Some HP handhelds, IIRC), you can measure directly. Otherwise you need an inductance bridge, or a signal generator, scope, and a capacitor or two to get at the value.
If the speed control delivers more than the motor can immediately use, the excess is converted and stored in the form of a magnetic field around the coil in the core. As the motor catches up, the magnetic field collapses, converting back into electrical energy to be dumped, along with the supply current, into the motor.
Speed is proportional to voltage and torque is proportional to current. ...The DC resistance of the DC motor is quite low. It does NOT take high voltage to get high torque at low speed, it only takes the voltage required to push the required current thru the low resistance. And motors run on torque.
Most of the voltage driving a DC motor is used to overcome back EMF, or "self generation". The base speed at any voltage produces a certain back EMF. Back EMF is dependent on RPM.
The DIFFERENCE between the back EMF and the applied voltage is SMALL, but it is what controls the torque to turn the motor at that speed against the load.
The "base speed" is the speed at which the difference between back EMF and applied voltage produces just enough current through the DC resistance of the motor to produce the required torque to keep it going at that speed.
1) THEREFORE, at a low speed, with resulting LOW back EMF, it requires only a SMALL voltage to produce the same torque.
At low speeds a DC motor is generally constant torque, variable power.
DC motors slow down with increasing load. The motor load can be sensed with a resistor in series with the motor. The sensed voltage is amplified and causes the voltage to the motor to increase with load. This increased motor voltage cancels the RPM drop with load.
The technique is called 'IR compensation' and it works satisfactorily down to about 100 RPM.
When the motor is loaded, it slows a bit, and draws more current. It MUST slow in order to draw that added current through the circuit resistances. That causes the resistances to have a (possibly) considerable effect on the speed regulation.
The IR compensation provides a small (usually adjustable) boost of voltage that is proportional to current draw. The proportionality is important.
What that does is to partly negate the effect of the resistances so they have less influence on speed. It makes them "appear to be" much smaller. The motor need not slow as much to draw the same increased current, because a slight current increase also boost the drive output voltage. The same current increase is obtained with only a slight slow down, less than otherwise required. In that way speed regulation under load is improved.
The reason the treadmill motors are said to be over-rated in power is that folks try to run them slow. They are high speed, low torque motors, that must turn fast to produce rated power. they have a high "base speed". When people try to run them slow, they notice the low torque, and tend to equate that with an overall lack of power, but it is really just a misuse of the motor that is causing a problem.https://www.motioncontroltips.com/faq-difference-between-torque-back-emf-motor-constant/
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