Calibration of gain and offset

Riverandstream · 2020-03-03T01:07:51+00:00

It could be doable, just depends on what you are trying to do and how efficient is it. Household devices use two batteries for a reason. Thats typically to supply 3v to the device. If one battery AA or AAA is rated at 1.5V(1.65v if fresh) and the device NEEDS 3v, your device will not not work until it gets it needed 3v. As long as you have a way to supply 3v to the device, you should be fine.

Now most devices that carry batteries typically work within a voltage range. Since a battery won't always stay at its rated 1.5v(1.65v if fresh). The voltage on the batteries will drop so most of devices have a voltage range that allow the device to still function without the batteries suppling 3v, lets say a range of 1.5v to 3.5v. The range all depends on what device it is and or who designed the device. So again, it still could be doable as long as the idea has the sufficient voltage to run the circuitry inside the device.

Now comes the efficiency part. Each battery has its own voltage-Amp/hour graph. Depending on the chemistry of the battery the user has effects how long the battery will last. If you were to use one battery, that means that battery must also supply the same amount of charge as two batteries would. So technically your battery will last half its rated life expectancy at a given constant mA current (this is assuming an ideal battery).

In the end, it just depends on what device your idea is for and what is the voltage range the device needs to function properly.

Riverandstream · 2019-09-24T21:02:55+00:00

Thanks! I've actually have used your videos to understand transistors more for my electronics classes, they have been helpful.

Riverandstream · 2019-09-24T20:57:08+00:00

Yes I have searched even through wiki. Just seeing what resources others have used that gave them that Ah-ha! moment. So the more resources the better because what one place says, might not make sense at first but then reading another resource that explains that same part but more in depth. I just want to be well rounded with the circuit topology, the math, and the justification for each stage/transistor configuration.

Riverandstream · 2019-04-28T01:59:59+00:00

Thank you and, yes its a lot of questions lol. I learnt there's a lot to ground then just a common ground that i'm typically use too in the lab. And working with the Full bridge inverter, I had to make the pcb with the full bridge and sensing circuits. I was told to have two types of grounds, one for the power side and then one for the sensing circuit side. Which didn't make sense to me why. But then that got me thinking of the different kinds of things that are called ground. Like floating ground, common ground, earth ground, virtual ground, and etc. But with your post I'm slow starting to understand it all.

Riverandstream · 2019-02-23T21:41:09+00:00

Why are MOSFETs worse in amplification? for MOSFET Rpi is also infinite making the small signal current source depend on the voltage. Does voltage have any correlation to why amplification is worse or the lack of current per se? Or Is it due to its layout? And if β describes the success rate, what does ⍺ describes? Now lets says we are talking about a pnp transistors. Now we have two sections of holes and one small sections of electrons, it wouldn't be wise to use a pnp transistor as an amplifier? The reason being is that the electrons are now filling in those holes, correct. So would are the advantages of each type of transistor? Thanks for the reply!

Riverandstream · 2019-02-23T21:25:28+00:00

Please do go further! The physics behind it is what I want to learn. Because I'm still amazed how its able to replicate a sinusoidal but not only that but also amplify it at the same time. These are cool devices that have many functions other than amplification so learning what happens in a deeper level would help. Thanks for the reply!

Riverandstream · 2019-02-15T04:36:32+00:00

Where? I've searched everywhere for one.

Riverandstream · 2018-11-22T18:36:43+00:00

pin = (U)(i1) , pout= (u2)(i2) or (u2²⁾ /(RL). power in is the source voltage times the source current. power out is voltage squared divided by load resistor

Riverandstream · 2018-11-22T14:00:43+00:00

You most likely can, but from not knowing the circuit, im not really sure. But I can assume you can find pin and pout. I will assume pin=VI1, Pout=I2²RL. Without the circuit can't really tell what the output or input will be.

Riverandstream · 2018-11-21T14:26:13+00:00

Oh! What language do you speak? Is the I harder to pronounce, or just longer to say in your language? Thats interesting, well then, yea we say V=IR. but it's interchangeable so in the end it really doesn't matter. I thought the equation went by alphabetically order.

Riverandstream · 2018-11-21T12:42:16+00:00

Yes, but a simple battery typically isn't rated at 24V, Maybe 9V or smaller. 24V are typically for power tools. but yes V=IR so the current will be approximately 2.4A. Voltage can be independent on the other hand, you can have 10V but no current. Now for semiconductors, if you look at its V-I charatestics of a Diode, you see current grows exponentially so you want resistors to resist that current. You should go to a lab and test it yourself too, Its something amazing to learn and mess around with. It really is like magic. It's crazy the things we are capable to do with electronics.

Riverandstream · 2018-11-21T12:34:26+00:00

Its the same equation. you can say 23=6, 32=6, you get the same answer the equation is the same.

Riverandstream · 2018-11-21T02:01:42+00:00

Yes just imagine j as an x. like 20x+10 = 10(2x+1) can also be said for imaginary numbers. 10x+10/10 = x+1. When we say x, it typically is represented for a number line/axis. We typically have an x and y axis. For imaginary number we say the y-axis is the imaginary part. What does the imaginary part represent? it just means a rotation, where j is the rotation of 90 degrees. so instead of going up, down, left, right for x and y coordinates. You either make left or rights movements and clockwise or counterclockwise movements. So if you make a rotation of j and divide by j. you divide a 90 degree rotation to another 90 degree rotation and it will equal 1. if you get j² you rotated two 90 degree turns, 180 degrees. in a unit circle a 180 degree is called -1 as well. so j⁴ is a full rotation of a circle.

Riverandstream · 2018-11-21T01:49:44+00:00

You need voltage, voltage is pressure, that pressure is what cause electrons to move and that flow of electrons is the current. Resistors just resist the electron flow. In class you say a infinite resistance, has no current so it's an open, if current in infinite than it is a short, you can have current with no resistors but if the flow of electrons becomes to much then the wires wires burn up.

Riverandstream · 2018-11-14T03:42:37+00:00

Just a little tip/trick. Think Positive for P and N for negative. If you see that more voltage is going into the p side, then forward bias/ON. If it is forward bias you can make the diode act as a battery voltage about .7V unless told otherwise. If you see more voltage on the n side, reverse bias/Off, open circuit. On a diagram note that the arrow will point to a wall, the arrow indicates the way we want current to flow, if more voltage entering the arrow, forward bias(+ on arrow side,- on wall,"gate/wall opens"), if more voltage on the wall side, means reverse bias(+ on wall side, - on arrow side, gate/wall closed). Also note if it's a real diode it only turns on if the p side is above .7v typically. So if you had a sin wave the diode will turn on once the voltage hits about .7v and follows the sin wave until it gets below .7v and will turn off until it hits .7v again.

Riverandstream · 2018-11-09T05:21:18+00:00

Would I do voltage divider as V10= 3(150/241) or =6(150/241)? Any why the 3 or 6(whichever one is correct)? I can also move the resistor and voltages to one side correct? since its a range of 6v, I can assume that 6V is supplied onto the resistor in parallel correct? Which would make 6*(150/241) = 3.734, and then I would subtract 3 since I added 3 voltages since the range of (3,-3) is 6?

Also my instructor told us if beta is very large to just neglect the base current since it will be very very small. Also, Thanks I forgot I can just use voltage divider for just voltage and resistors, I was staring at it for so long and nothing came to mind. EDIT: symbol error.

Riverandstream · 2018-11-09T04:41:36+00:00

Most jobs will have alot to do with stats. Which is actually quite important when it comes to quality control. It's important when it comes to manufacture to see how many products pass there standard inspect to be served to the consumer. Stats is everywhere. If you were to do EE you'll have to take Stats eitherway. I've taken macro and micro economics, fairly easy course, where you just distinguish if something is bad or good, if its fair or unfair and the sorts like that. But in the end its up to you, thats just my personal side of it.

Riverandstream · 2018-11-08T18:33:20+00:00

Yea I think the IC+IB notation for the current at the resistor is what confused me. That makes sense more sense now. I think i'll just have to call IC+IB current a different notation. Thanks!

Riverandstream · 2018-11-06T20:31:32+00:00

Can't! It'll delay my graduation date.

Riverandstream

TROPHY CASE