Yes they are synonyms. CMOS stands for Complementary MOS because it uses both NMOS and PMOS.

You are quite correct, of course. Pardon my brevity.

 

Yes, sorry for my brevity. As the schematic shows, enhancement mode MOSFETs.

I honestly didn't know. I checked Google and YouTube, and couldn't find an answer fast enough, so I figured I'd ask you guys. lol. I kinda figured it, but just wanted to make sure.

Yes they are synonyms. CMOS stands for Complementary MOS because it uses both NMOS and PMOS.

Yum.. I love synonym toast crunch in the morning !! ...

So many people calling them as NMOS or PMOS, it's not even funny on YouTube. I like n-Channel & p-Channel better

Thanks for the help guys, much apreciated as always.

 

R x C x 5 = T
So how much capacitance do you need to discharge? How quickly? Automatically, through a transistor, or manually putting in a jumper?

I know this is a little late.. but.. I'm making a hand crank generator, and don't want to crank that hand, take small little breaks in between. So I'm trying to squash this capacitor thing. Yes, they're difficult for me. The hand crank generator will be hooked up to a diode, and also hooked up to either a buck converter, or linear voltage regulator, like an LM7805. 5 volts output, no matter how fast I crank the motor.

I need to discharge to a load. The load for starters will be a 3.3 Vf, 0.02A If LED. So it's going to discharge through a resistor, and into an LED.

So the resistor will be "for simplicity" 85Ω.

I want to take a 6 second break, or slow down cranking for 6 seconds.

T = R x C x 5

Can you help me.. I don't know how to figure out the capacitor value. Or where to plug in the 6 seconds.

 

You need to know one more parameter - the voltage that your cranking would charge the capacitor up to. That value is also important for protecting the 7805. It has a maximum operating input voltage and you don't want to exceed that.

 

Ok, so if I stop cranking the hand generator, and the buck converter stops providing 5V to the resistor and LED.

The resistor is 90Ω. Do I need to know the resistance of the LED? If so it's 165Ω. But I don't think it's needed. I'll do the calculations for the led too, if that's what I need to do in order to keep it lit. But I don't see why I'd need to include the resistance of the LED, regardless if it's in series with the resistor, etc.

So.. It's 5 x R x C.

So 1uF is 0.000001 farads. 1,000uF is technically, 1mF, but again, Mouser and people don't like that, so it's 1,000uF, which is 0.001F

5 x 90Ω = 450 let's keep that value, that's 5 time constants multiplied by 90 ohm's.

450 x 0.001 or 1,000uF = 0.45 seconds.

Let's shoot for 6 seconds now. How..?
What's 0.45 seconds relationship to 6 seconds?
6s / 0.45s = 13.3 times

So we need to increase the capacitance by 14.
14 x 0.001F = 0.014F x 450 = 6.3 seconds !!!

Correct me if I'm wrong, but that's a 14,000uF capacitor?

30s / 0.45s = 66.6 let's just say 67 times more.

67 x 0.001 = 0.067F, so we need a 67,000uF capacitor?

450 x 0.067 = 30.15 seconds !!!!!

Am I on the right path here?

 

You need to know one more parameter - the voltage that your cranking would charge the capacitor up to. That value is also important for protecting the 7805. It has a maximum operating input voltage and you don't want to exceed that.

Ok, so if I stop cranking the hand generator, and the buck converter stops providing 5V to the resistor and LED.

The resistor is 90Ω. Do I need to know the resistance of the LED? If so it's 165Ω. But I don't think it's needed. I'll do the calculations for the led too, if that's what I need to do in order to keep it lit. But I don't see why I'd need to include the resistance of the LED, regardless if it's in series with the resistor, etc.

So.. It's 5 x R x C.

So 1uF is 0.000001 farads. 1,000uF is technically, 1mF, but again, Mouser and people don't like that, so it's 1,000uF, which is 0.001F

5 x 90Ω = 450 let's keep that value, that's 5 time constants multiplied by 90 ohm's.

450 x 0.001 or 1,000uF = 0.45 seconds.

Let's shoot for 6 seconds now. How..?
What's 0.45 seconds relationship to 6 seconds?
6s / 0.45s = 13.3 times

So we need to increase the capacitance by 14.
14 x 0.001F = 0.014F x 450 = 6.3 seconds !!!

Correct me if I'm wrong, but that's a 14,000uF capacitor?

30s / 0.45s = 66.6 let's just say 67 times more.

67 x 0.001 = 0.067F, so we need a 67,000uF capacitor?

450 x 0.067 = 30.15 seconds !!!!!

Am I on the right path here?

 

You need to know one more parameter - the voltage that your cranking would charge the capacitor up to. That value is also important for protecting the 7805. It has a maximum operating input voltage and you don't want to exceed that.

Can you please check my math, in the above post?

The capacitor would go after the LM7805 or buck converter, which accepts a voltage of much higher than 5V. But also, keeping in mind, ...

Actually, I lied. I own a LDO Linear Voltage Regulator (SPX2940U)
Which it's max input voltage is 16V.. Gonna go with a buck converter instead.

Like this guy is doing. He's also using a 10,000uF cap.


Vin - Vout x Iout = Watts. (Linear voltage regulator).

So 12V - 5V = 7V x 1A = 7W dissipated, but buck converters, are more efficient. As you probably already know, and as you can clearly see in the video.

 

Can you please check my math, in the above post?

If I get some time.

The capacitor would go after the LM7805 or buck converter, which accepts a voltage of much higher than 5V.

That limits the energy stored in the capacitor. If you store energy before the regulator or converter, you can store it at a higher voltage, a critical parameter which you still haven't specified. Most caps will be rated to voltages all above 5V, so you'd be "wasting" the energy capacity if you use them at only 5V. If your cap is rated to 35V, for a given capacitance rating, it'll store much more energy when taken to ~30V than when taken to 5V.

If you look at capacitor cost, if follows roughly with the amount of energy the capacitor can store. That is, price goes up with both voltage rating and with capacitance. To get the biggest bang for your capacitor buck, you'll want a voltage rating that is high enough to cover all situations but not, say, >2X more.

Another problem with storing energy after a regulator is that a regulator such as 7805 cannot tolerate a voltage on the output higher than the input. So if the input stops, and you have a big capacitor on the output, the 7805 will be damaged as the capacitor discharges through it. Many designs for the 7805 include a diode reverse biased across the 7805, to drain off any voltage at the output so it doesn't drain backwards through the 7805.

Which it's max input voltage is 16V.. Gonna go with a buck converter instead.

Excellent choice. You need to know the maximum input voltage so you can specify one.

 

If I get some time.
That limits the energy stored in the capacitor. If you store energy before the regulator or converter, you can store it at a higher voltage, a critical parameter which you still haven't specified. Most caps will be rated to voltages all above 5V, so you'd be "wasting" the energy capacity if you use them at only 5V. If your cap is rated to 35V, for a given capacitance rating, it'll store much more energy when taken to ~30V than when taken to 5V.

If you look at capacitor cost, if follows roughly with the amount of energy the capacitor can store. That is, price goes up with both voltage rating and with capacitance. To get the biggest bang for your capacitor buck, you'll want a voltage rating that is high enough to cover all situations but not, say, >2X more.

Another problem with storing energy after a regulator is that a regulator such as 7805 cannot tolerate a voltage on the output higher than the input. So if the input stops, and you have a big capacitor on the output, the 7805 will be damaged as the capacitor discharges through it. Many designs for the 7805 include a diode reverse biased across the 7805, to drain off any voltage at the output so it doesn't drain backwards through the 7805.
Excellent choice. You need to know the maximum input voltage so you can specify one.

It's a 12VDC motor (12V x 2 = 24V)
The Capacitor : 598-SLPX153M025C7P3
The Buck Converter : 2pcs Re DC-DC 3A Buck Converter Adjustable Step-Down Power Supply Module LM2596S

The capacitor will go before the buck converter, and after the motor and diode.

 

I would like to understand the equation for discharging a capacitor in X seconds. I want it discharged when I say so (X).
Extra, but not the subject. I would also like to understand the equation for charging a capacitor in X seconds. I want it charged when I say so (X).

Can you please help?

Here's what I learned from MrAl in a previous thread a while back. But those do not teach me how to discharge, and charge a capacitor when I say so. So if I wanted it discharged or charged COMPLETELY in X seconds.. How would I do that?

Since the charge/discharge curve is exponential, you can never (theoretically) reach the source or sink voltage.

For charging, It's like the float valve in an old-timey toilet tank, where the closer it gets to the end the less water it allows to flow in to raise the valve, so the longer and longer it takes to fill the tank.

Is there a simpler equation for doing this..?

Yes, but you have to set realistic limits.

 

It's a 12VDC motor (12V x 2 = 24V)

The voltage you can make running a motor in reverse is often a lot less than the voltage it is meant to operate on. But the output is proportional to the RPM. So if you go nuts on it, it might produce even more than the operating voltage. Before you attach the regulator, you should just take a look at the voltage on the capacitor, after the diode. Do you know if the motor produces DC? If it's an alternator making AC, you should use a bridge rectifier to get more out of it. If it's really DC, the single diode will be fine.

 

The voltage you can make running a motor in reverse is often a lot less than the voltage it is meant to operate on. But the output is proportional to the RPM. So if you go nuts on it, it might produce even more than the operating voltage. Before you attach the regulator, you should just take a look at the voltage on the capacitor, after the diode. Do you know if the motor produces DC? If it's an alternator making AC, you should use a bridge rectifier to get more out of it. If it's really DC, the single diode will be fine.

It's a $40 12VDC HD Premium Planetary Gear Motor.
I'll check the voltage on the capacitor after the motor and diode.

The motor will have either a 12" handle, or a 6" handle. Maybe just a 6 inch one. Yes, I've done the calculations on how much force will be needed to turn the motor, under 5V @ 2A. I also have a good understanding how fast it needs to be turned. Yes even after the diode.

If I need to, I'll just drop the whole project. Because hand crank generators are stupid, and pointless to make. Better to set a Anker Solar charger.

But I honestly would like to make a hand crank generator. Just because it's free energy, and also it's powered by YOU. It can be cranked in the shower, in the rain, underground, above ground, in a train, in a plane, under a table, under the covers, standing on your head, etc. Let's face it.. Hand crank generators might be a pain in the ass to crank, but it's free energy to charge devices. Oh yeah, and hand crank generators, if the voltage is high enough, can be hooked up to an inverter, and spit out 120VAC.

 

It's a $40 12VDC HD Premium Planetary Gear Motor.

That's a brushed DC motor, so it should behave as a DC generator. You won't need a bridge, just a single diode.

It has a very high 71:1 gear reduction. Are you sure you'll be able to start it turning?

 

That's a brushed DC motor, so it should behave as a DC generator. You won't need a bridge, just a single diode.

It has a very high 71:1 gear reduction. Are you sure you'll be able to start it turning?

Idk.. That's the only part I'm worried about. Kinda.. Because it's 118 RPM.
I actually made an app to do all the calculations for me in Windows OS.

But I'll show the calculations manually.

Voltage : 12VDC
Stall Current @ 12VDC : 20A
RPM : 118 (no-load)
Stall Torque @ 12VDC : 958.2 oz-in

Alright.. I might have missed something here. Including the software I wrote and made.
958.2 ounces is 59.8905 lbs. Midas well just say 60 lbs. (I'll also update my DC Motor generator post)

60 lbs / 12 inches = 5 lbs

So even with a 1 foot crank handle, that's 5 lbs @ 20A @ 12VDC.

Looks like I'm going back and looking at the motors again.
Not to mention the software has to be re-written.

It's honestly not that big of a deal.. 5lbs really isn't much, but this thing has to fit in a bag for travel, camping, etc.

Ok..680.5 oz-in 165 rpm 12vdc 20A motor

That's not very much smaller, but it is smaller.

680.5 ounces is 42.53125 lbs or.. in-lbs

That's with a 1 inch crank handle.
Let's just look at the maximum stall current for 12VDC. Which is 42.53 lbs @ 20A @ 12VDC.

The torque will decrease if the crank handle arm length is increased.
42.53 lbs / 12" = 3.5441 lbs

What a royal pain in the A$$.

If it's not the 4 lbs or force required, then it's the 165 turns per minute that's gonna hurt.

Start it turning? Yes.
Keep going for 4 hours?

I probably could do that for 4 hours. For $40? Then the capacitor. The buck converter. Handle parts?
Even if I do buy them from the same place where I get the motor.

It seems like a waste of time and money.
To actually beat the competitors.. To make and own a 12VDC hand crank generator.. Seems nice.

I REALLY want/wanted to make a hand crank generator. I really do/did.

 

I probably could do that for 4 hours.

Probably? No offense, but I doubt it. To do any repetitive motion for that long you need very good ergonomics and a lot of training for that activity. Otherwise you'll get blisters on your hands and you'll strain whatever the weakest link in your body is. I don't know what that will be but you'll know after 30 minutes or less. I'm a decent bike rider but I consider a 2-hour ride 'significant'. A four-hour trip would take some training and planning before I'd attempt it.

Have you ruled out a bicycle-type arrangement? You can generate a lot more power with your lower body than with your arms. And you can use off-the-shelf parts. Of course it gets bigger and heavier.

If you're willing to do all this effort for some electricity, just carry a battery. Charge it at your leisure at home. Powering anything with human energy just converts expensive food into electricity, with piss-poor efficiency. It's fine if you're trying to burn off calories, but that's not usually the scenario out in the field.

 

Fixed my software, and my math.

 

I don't know what that will be but you'll know after 30 minutes or less.

You and everyone else on the other post, is correct. With these freaking motors, I'll be sick and tired of cranking those HD Premium planetary gear motors. I saw the 20A stall current, and was like.. I'M GETTING ONE... or.. THERES MY MOTORS !!


I can't fit a bike or any kind in my bookbag, or whatever.

I mean.. I have the entire internet at my fingertips, and I can't seem to finish this project, without sacrificing current output.

Ok, so I see these.. (Economy Spur Gear Motors)
All of these have straight cut spur gears. So they should be able to be turned by hand. I'm sure about 99.99%

MUCH BETTER, but much lower quality (imho, I think) than the HD Premium Planetary Gears.

I'm looking at either this one (98 RPM Econ Gear Motor)


This one (124 RPM Econ Gear Motor)


Still too high you think?


Carry another battery? Tried that. These cellphones just gobble up batteries. My $13 flip phone puts my Samsung Galaxy Note 4 to shame as far as standby time. I don't have to charge it for a little less than a month. The galaxy note 4 if it's not on airplane mode 24/7 will eat the battery, without Wifi in less than 2 days.

I totally hear what your saying, and I'm taking it to mind. I just wanted to make a hand crank generator, do a FHD video on YouTube about it, and have a hand crank generator that I can be proud of.. Not just for fun, but because I made something, and It's something I made, and the 2nd thing I've made that I could use everyday if I wanted to. The 1st being my unregulated box mod.

 

Ok.. So these are very nice too. But less torque.

NeveRest 40 Gear Motor


NeveRest 60 Gear Motor



Welp.. I'm pretty sure I know which ones I should get. But I want to hear it from you guys. NeveRest 60. Because the torque isn't much, and also it only takes 44 RPM for 5VDC.

 

Also.. FT-LBS x RPM = X / 5252 = HP x 745.699872W = Watts.
Also.. 1 oz-in = 0.00520833333 ft-lbs

I'll use the NeveRest 60 as an example.
593 oz-in x 0.00520833333 = 3.0885 ft-lbs

3.0855 ft-lbs x 105 rpm = 323.9775
323.9775 / 5252 = 0.0616865003808073115003808073115 HP
0.0616865003808073115003808073115 HP x 745.699872W = 45.999 Watts
I = P/E

45.999W / 12V = 3.83A

So it's capable of 12V, 46W, 3.83A, 105RPM, 593 OZ-IN or 3.08 FT-LBS

I have a question..
If line voltage 120VAC is powering a 60W light bulb, it draws 0.5A

What about if an inverter was powering the light bulb.. Wouldn't the current be more from the 12V battery?
60W / 12V = 5A

Wouldn't that be the same for the motor?
The motor is capable of outputting 46W @ 12V right?

So what if it was a 5V load?
46W / 5V = 9.2A ?

Or is it.. 3.83A / 12V = 0.319166A per volt x 5V = 1.59A

Which one is correct? Can you tell me please?

Also made software to do the calculations for me to convert torque & RPM to HP and Watts.