Here‘s a small article on the real logic behind angular bisection which no teacher would have told you

Link : http://iamsuhasm.wordpress.com/tutsproj/intuitive-math-angular-bisectors/

My Mini PCB Driller

I’ve written an article on this DIY PCB Driller I made sometime back. Here is the link to the article :http://iamsuhasm.wordpress.com/tutsproj/diy-pcb-driller/.

And if you haven’t read my PCB making tutorial , here is the link to that too :

PS : Dont forget to leave comments if you like the post

Prove that maximum power is transferred to the load only when source impedance is equal to load impedance.

A.K.A. the maximum power transfer theorem.

Instead of blindly googling the proof , i decided to find the proof myself. Thankfully , AOE gave a hint that the proof involved calculus. Otherwise , it is doubtful if my math challenged mind would have been able to work this out.

Not finding any ‘noob – understandable’ proofs for the MPTT on the internet , i have posted the proof here.

The problem involves finding when maximum power is delivered to load L.

It is pretty clear that power delivered to the load will be zero when both Rl is 0 and infinity. So , the maximum power should be transferred somewhere in between.

At the point where the power transfer peaks , the tangent to that point will be horizontal. Therefore , the derivative at that point will be zero.

So , we can find the point of maximum power transfer by finding the maxima of the derivative of the power function.

First step : Find the Power Function

The power transferred to the load will be equal to the voltage across it times the current in it.

Rl and Rs form a voltage divider , hence the voltage across Rl will be

The current through Rl will be

So , the power dissipated by Rl will be the product of the above two equations bcoz P=IV.

Second Step : Differentiate the Power Function.

if we try to differentiate this entire equation , it will result in a reasonably complex derivative for my noob brain due to the presence of the variables V and Rs.

Instead , to simply the process , i have taken V and Rs as two constants and then have found the derivative.

I have taken V = 5V and  Rs = 7Ω.

So , we get the Power as :

When this is differentiated , you get the derivative as :

Third Step : Find the Maxima

Setting the derivative to zero to find the maxima :

The solution to this equation is 7 !! Which is precisely the same as the value of Rs !!!

If you find the solution to the equation by taking Rl and V differently , you will still find that Rl = Rs each and everytime!!

So , it is indeed true that maximum power is transferred when source resistance = load resistance.

The cotton candy machine

While digging back through all my old photos , i found this picture of a cotton candy machine (alright , its not exactly a ‘machine’ ) that i had built 2 years back. It was made from a toy motor and a badam milk tin can punched with holes. And it really did work!!. You can see the candy that it produced on a glass rod in the pic.

Atmega 8 , the most popular AVR uC

Programming AVR MCUs in C

I am starting to write a series of tutorials on programming AVR uCs in C.  I plan to write articles covering Interrupts , ADC , timers , internal EEPROM, PWM and USART. This is the first in those series and it covers basic I/O operations. Here is the link to the page.

URL : http://iamsuhasm.wordpress.com/avr-gcc-tutorial/

So , first we need the function for a circle. Every point on the circle is equidistant from the origin. Hence the equation is clearly x^2 + y^2 = r^2 , where r is the required radius.

Rearranging to get a proper function ,

Now , if i take this function and plot it , it should ideally give me a circle. And it does. Unfortunately , for reasons unknown , Microsoft math is seeming to clip off the negative half of the circle. I dont know why. But that really doesnt matter  as i will only be taking the part of the circle in the 1st quarant and will be finding its area. Then we can multiply the area by 4 to get the area of the whole circle.

Now since i have my function , i can find the area of the circle by integrating the function between o and r. This integration will give me the area of a quarter of a circle. We can multiply that area by 4 to get the full area.

Ok. To integrate , we need to find the antiderivative of the circle function. Microsoft math tells me that it is :

The Anti derivative

Now , let me call this function F(x).

Finding F(r) – F(0) should give me the area.

F(0) is very clearly 0 , as both the terms in the equation become zero.

F(r) is

as arctan of  (x/sqrt(0)) is apparently 90 = π/2 radians. (I dont know why. Maybe by taking the limit of the function?)

and there you have it folks , ( πr^2 /2) , which is (πr^2/4)  !!

Which is precisely the area of a quarter of a circle.

So , now if we multiply it by 4 , we get the area of the whole circle = πr^2 !!!

Using LCDs

I’ve just written an article on using HD44780 based LCD s. Click here if you want to read it.

link : http://iamsuhasm.wordpress.com/using-lcds/

Its a Vartech 5020 Dual channel 20Mhz Cathode ray o’scope. I’ll post a few waveforms being displayed after i figure out how to use all those buttons and knobs.

And one more thing , can anyone tell me what the handle like thing on the top is? It doesnt seem to be movable. I’ve been racking my brains from the past few hours trying to figure out what that is.

What is an EEPROM?

To cut it short , EEPROM is a memory device , similar to the memory stick in your cell phone or camera.

The Pin layout of an EEPROM is very similar  for most Static RAMS. So once you know how to program an EEPROM , you also know how to program RAM.

So lets get started.

Before knowing how to use an EEPROM , you have to know how it organizes its data.

The EEPROM is divided into 8912 “houses” with eight rooms in each house. You are allowed to store either a 1 or 0 in each room.

When you wish to write some data to the EEPROM , you have to specify the address of the “house” where you want to store your data.

(Did you ask yourself why the number of  “houses” is something as arbitary as 8912?  Its because 2^13 is 8912. And there are 13 Address choose pins on the IC.)

Then , using 8 input pins , you have to give 8 values which will be stored in the eight “rooms” of selected “house”.

The procedure is very similar when you want to read the data back.

Again , you will have to select the address of the “house” and the 8 values in the eight rooms will appear through the output pins.

If add up the total number of pins : 8 for input +8 for output + 2 for power + 13 for  address + 3 control pins

= 34 pins!!!

So to reduce pin count , Manufacturers have combined the Input and Output pins and made them into a single set of 8 I/O pins. This makes it a bit tougher to use the IC as both the functions have to be done by the same set of pins.

The  Procedure :

EEPROMs have  3 control pins : CE  (Chip Enable) , WE (Write Enable) , OE  (Output Enable)

This truth table should make it pretty clear how the 3 pins work :

Here’s the explanation :

The Chip can Read/Write only when CE is low. If CE is held high , then the chip is disabled and the I/O pins are tristated.

When CE is LOW, OE is  low and WE is held high , the data in the memory location chosen by the address pins , appears at the I/O pins.

On a new EEPROM which has never been programmed before , all the memory locations are programmed with 1′s by default. So , while reading from a new EEPROM , you will see only 1′s in every memory location.

To write , input the 8 bits that you wish to write at the I/O pins while holding OE high and WE low. Then raise WE to high and then back to low again. Then , the data that you have given on the I/O pins  will be written on the  rising edge of the pulse to the address that is present on the address pins at the rising edge.

The waveform will be like this :

So , if you have not understood from the above waveform , let me make it very clear.

This is the way to manually write to an EEPROM :

1) Hold OE and CE low and WE high – Read Mode – Whatever is in the location determined by the address pins will appear at the I/O pins.

2)Hold CE Low, OE high and WE high – I/O pins are tristated – disconnect the output leds (or whatever) from the I/O ports

3)Hold CE low , OE high and WE low – Write Mode – Give the data to be written at the I/O port and the address where you want it       to be written at the address pins.

4)Hold Ce low ,OE high and WE High – I/O pins are tristated AND data is written – Disconnect the Switches or wires that you used to give data to the I/O port.

5) Hold CE and OE Low and WE high – Read Mode – You  are back to step 1. Whatever is in the location determined by the address pins will appear at the I/O pins.

The procedure is same for RAMs as well. The only difference between EEPROMs and RAMs is that RAMs lose whatever has been written when the power is turned off , but EEPROMs dont.

There are many EEPROM and RAM ICs available. The EEPROM that i generally use is AT28C64B from Atmel. It is available at around 100 INR (~2$) here in bangalore.  The cheapest RAM IC is 6264 which is available for 35 INR (0.7$)

The writing procedure that i have described above is rather crude. There are better ways of writing to the same EEPROM. But , this is best way if you are using the EEPROM by giving CE ,WE, OE signals by hand.

Set 2  – 74150 to 74300

Zip file which contains trimmed datasheets (74150-74300)

All the trimmed datasheets combined into a single pdf (74150-74300)

Set 1 7400 to 74150

Zip file which contains trimmed datasheets (7400-74150)

All the trimmed datasheets combined into a single PDF (7400-74150)

Here “trimmed” means that I’ve edited the pdfs and have removed all the crap like ” IC Physical Dimensions “, which we never use.

I have the unedited pdfs with me. I couldnt upload them as 110mb.com (where i store my files) has a cap of 8mb per file.

If anyone wants the unedited collection , please email me at iamsuhasm@yahoo.co.in or leave a comment. I will be more than happy to email it to you.