The circuit above takes 9mA all the time via the middle 1k resistor and the LED takes 8mA when it is illuminated.

The improved circuit takes 2 microamp when light falls on the Light Dependent Resistor and the first transistor is turned OFF. This makes it "disappear" from the circuit and all we have is two 10k resistors connected between the base and emitter of the PNP transistor and these turn the transistor OFF. So nothing else in the circuit is taking any current. A huge saving on current.

COMMON EMITTER AMPLIFIER
YouTube is filled with instructional videos explaining the Common-Emitter amplifier and similar related circuits.
Most of these demonstrations are quite impractical, as they use component-values that I have never seen in reality and none that I would recommend.
But more important, you have to "look outside the box" and understand that you cannot design a circuit without knowing the characteristics of the input signal and the impedance of the stage that is connected to the output.
You cannot specify a signal that comes from a sinewave generator as it is completely different to that found in a normal circuit.
In fact there is nothing you can specify or predict, as the results will be completely different to reality.
It is absolutely pointless going into the fine detail of including base-emitter voltage drop and using a value of 100 for the gain of a transistor as you are finishing up with a "dream" value for the stage.
Even learning the fundamentals of knowing how to find the gain of a stage is an unnecessary exercise as it will bear no resemblance to the actual value. 
I have never used a mathematical calculation in my life as I know it produces NOTHING.
University students are being led "up the garden path" with questions and computations that are just a waste of time.
But a University cannot charge $70,000 and tell the student to build the circuit and measure the results. So they have to lull the student into  false sense of achievement by complicating the issue with mathematics. It's a SCAM. And after the COVID lockdown, a lot of the University myths have been uncovered and debunked. Even Elon Musk has mentioned the ineffectiveness of University teachings. In some areas they are effective as no student has the technical equipment to carry out medical operations, but in many another areas he has said the only way to learn is on-the-job.
And the fastest way to learn is called REVERSE LEARNING. A medical student learns the fastest when he attend a dying heart patient and sees how the operation is carried out. An electronics person learns more by diagnosing a faulty device and seeing how to locate the fault.
 

To make a long story short . . .  In 50 years I have never used any of the internet mathematics to design a circuit. It is not relevant and does not apply. It is remote from what is needed and does not take into account external influences. It is a complete waste of time putting pen-to-paper. But telling the "Professors" is a waste of time.  Nothing has changed. I sent my concepts to all schools and colleges in Australia, over 40 years ago and got no reaction. I explained that the only way to learn electronics was through building projects.  Not one school or University bought a single kit in 40 years. The reason  . . . .  when the project did not work . . . because of student incompetence . . . . the instructor did not have the skills to fix it in front of the student.  It's as simple as that.  The blind leading the blind.   Talk is CHEAP.  Blabbering in front of  blackboard is easy.  Proving your blabberings is difficult.

2 LAMPS

PCB

555 IC MISTAKE

The whole idea of drawing a circuit is to present the components in the simplest way possible.
The original circuit has been improved but the placement of the 555 pins is not ideal. However the major mistake is connecting a 470u directly to pin7. This pin has a low-current transistor inside the IC that "shorts-to-ground" and the energy in a 470u can blow-up the transistor.

ELECTRONICS MAKER MAGAZINE 

India's second largest electronics magazine has ceased providing construction projects in the magazine.
Why do they keep the title ELECTRONICS MAKER if there is no "maker" in the magazine.
A few years ago I provided them with articles for 18 months and did not get one single reply from any of the 40,000 subscribers. That's not-one-in-720,000 views.
India is just fooling itself with slogans like: "Make in India."
I don't know where the fault lies, but the University lecturers on the web have very little understanding of basic electronics and what is needed to be taught.  They just fill the whiteboard with mathematics and equations and come with up with answers that are far more accurate than the data !!!
I have never seen them explain how a circuit works, and certainly no mention of how to test it. 
And now the magazine does not have any articles for the beginner or the constructor.
They are just allowing China to take over the field.  And they are. China is designing many electronics items and they show enormous skill in programming and buying microcontrollers that are only available to the Chinese market, for a price that is one-tenth of our purchase price.
The Western World needs electronics engineers to design products because the Chinese do not know what we want or how to design things aesthetically. 
Unfortunately everything is going "down-hill" and China is rising.
However the recent downturn in China-economics has seen millions laid-off and more than 10 million apartments either abandoned or unfinished with millions of purchasers unable to reside in their unfinished dwellings.
And this is only the beginning of the collapse.

UNDERSTANDING THE 555 IC
The circuit shows an IR transistor connected to pin 2 of a 555 IC. Technically the 555 will trigger when the voltage on pin2 is lower than 33% of the supply voltage.
But there is no resistor supplying a voltage to pin2. The 555 will react when the voltage on pin2  is lower than 33% of rail voltage as there are some resistors inside the chip that cause the chip to start-up in a "set-mode" but you cannot use them in your designs.
When the Infrared light is turned off, the chip requires a HIGH on pin6 to change state but the voltage on pin2 must be above 66% of rail voltage to this to occur, as pin2 controls the action over the voltage on pin6.

The solution is to place a high value resistor between pin2 and the supply rail.

The IR transistor creates a voltage divider with this resistor and we call this a HIGH IMPEDANCE or LOW IMPEDANCE SET-UP, depending on the value of the resistor.
If the resistor is 10k, we can say it is low-impedance, because this is a low-value for this type of set-up.
It will require a high level of IR light to change the state of the 555.
On the other hand, a 100k will produce a high-impedance input and the 555 will respond to a very low level of illumination. This is called sensitivity. The following circuit shows a sensitivity pot with a 10k "stop resistor" to prevent damaging the transistor. 
Pin6 needs to be connected to a resistor or switch to complete the circuit.

COMMON EMITTER STAGE

Too much time and effort has been put into describing and analysing the Common Emitter Amplifier.
The facts are these: The gain of a transistor can be anywhere from 150 to 450 and it is a waste of time producing a mathematical result for the gain of the stage.
The gain will depend on the impedance of the stage connected to the output and if this is unknown, nothing can be calculated.
However the mistake in the circuit above is the biasing.
You cannot connect the base to the supply rail. If you replace Rb with an adjustable resistor, you will find the collector voltage is equal to the supply voltage and as you decrease the resistance, it will immediately fall to nearly zero.
You have no opportunity to set the collector voltage to mid-rail and thus the output voltage will be "clipped" or "bottomed."
In other words it will be distorted and this type of circuit is useless for a demonstration. In fact it is useless for anything.


 
TIMER
We have covered lots of faulty circuits from Mr Mohan Kumar as he does not test any of his designs and most of them do not work.
The main problem with this circuit is the 10k resistor on the base of the T2.  This will allow very little current to flow and depending on the resistance of the relay it may not "pull-in." The resistor should be 1k to 2k2.

The circuit  has been simplified to a transistor,10k and relay.
Here is how to instantly work out what is happening.
The transistor will have a gain of 100. It might have a gain of 330, but when a small current or medium current flows, you can only allow a gain of 100.
The transistor divides the 10,000 ohm resistor by 100 and the transistor turns into a 10,000/100 = 100 ohm resistor.
The supply is 6v and we have a 100 ohm resistor in series with a 5v relay.  Suppose the relay needs 100mA to operate. Its resistance will be 5/.1  = 50 ohms.  We have 100 ohms in series with 50 ohms and 2v will appear across the relay.  Obviously, this will not work.  If the relay requires 50mA, the resistance will be 5/0.05 = 100 ohms and the relay will see 3v.
You can instantly see this circuit will not work reliably.

That's how you see all EMITTER FOLLOWER circuits.

Here is a voltage regulator circuit using an EMITTER FOLLOWER.  It is a Power Supply that takes a higher voltage and converts it to a lower voltage, depending on the value of the zener diode.

It is also called a SERIES REGULATOR (to distinguish it from a SHUNT REGULATOR arrangement).
The point we are making is this: How much current will the circuit pass? The only thing that is taken into account is this: The transistor will have a gain of 100 and it will convert into a 1,000/100 = 10 ohm resistor.
If you draw 100mA, the voltage across the transistor will be 0.1 x 10 = 1v.  If you draw 300mA, the voltage drop across the transistor will be 3v and if you draw 400mA, the drop will be 4v, but you don't have 4v available !!!

A new "club" was promoted on Jameco website, where beginners can join for $19,99 for a 12 month subscription.  But when you fill out the application, the charge is $99,00   A typical "bait and switch."

The site is filled with technical mistakes and diagrams:

The main fault with the diagram above is this:  A switch is placed on the POWER RAIL. Imagine if all the lights in your house were wired like this.  They would all still be "live" when switched off !!

A lot of the wording is not common terminology:  "Glowing a LED" 
Preset??  . . . . should be "mini trim-pot."

Continuity Test of of an LED !!!   (proof-reading needed.)
We don't talk about "Continuity of a LED." We talk about the characteristic voltage drop due to the colour.

AMERICAN KITS
 I was sent an email recently with a link to an American supplier of educational kits for beginners.
What I found was a total disaster.
The Printed Circuit Boards had no overlay, no tinned lands and it looks like they had been hand-made with tape, some 20 years ago. They were single-sided with lots of links,
What I am complaining about is this. In this "day and age," to deliver rubbish like this to hobbyists is unfathomable.
What are they trying to present?
I am not blaming the supplier. He does not have any intelligence or any understanding.
I blame the teachers for accepting this rubbish and not forcing the supplier to improve his product.
I go back 50 years when the Chinese produced rubbish like this and responded to complaints.
Over a period of time they improved and now stand as the worlds foremost producer of acceptable electronics.
This reinforces my statements of the educational standards in the USA.
I was on my own with saying their standards were the worst in the world, but I have been backed-up recently.
The cost of a degree or diploma or "Masters" is astronomical for the "product" it provides.
I have seen the courses, the content and was staggered by the lack of content and training.
Not one course covers the skill in soldering. Not one course mentions TIP-CLEANING PASTE and not one course explains how to remove a component.
Not one course explains how a transistor really works and not one course explain ANYTHING understandable to designing the simplest circuit.
This is because the "Professor: has never designed a circuit in his life and certainly never repaired 35,000 electronics devices, like I have.
One University graduate came to me for an application and said: "Oh, I don't do soldering."
And another said: "When I design a circuit, it will work."   I asked, "Where did you get that idea from?"   "That's what the University Professor told us."
It's called MESMERISM.
Fooling the student into thinking has has an understanding of electronics, just because he has paid $70,000.

Look at the links and the lack of component values

Another "Junk" board
What does it do?

Look at this quality kit from Aliexpress

I have included a number of circuits from the web to show that it does not matter how the circuit is constructed or the type of gate (NAND or NOR).
The point l want to make is this.
How does the capacitor get discharged?
No one has covered this point and it does not matter until you need to describe how the circuit works.
That’s the position l was placed in before l started my magazine TALKING ELECTRONICS.
Things don’t matter until they matter.
It seems such a trivial point.
But a capacitor has to discharge so that it can charged for the next cycle. Ideally, it discharges fully.
To discharge fully, both leads have to be at the same potential. They can both be at 12v or 0v. In other words, the capacitor can be discharged while “high” or discharged while at zero voltage.
And the thinking is this: The input of a CMOS gate has a very very high input impedance (resistance), so how does any capacitor connected to it, discharge?
Here’s the secret.
As soon as the voltage on the input rises above the supply voltage, a diode connected to the input, and also connected to the power rail, is activated and the voltage cannot rise above the supply voltage.
This diode is only active when the input line is higher than the supply. Otherwise is does not do anything.
Obviously you cannot put a high voltage on the input as it will immediately flow through this tiny diode and the current will destroy the diode. However the current from the discharging capacitor does not damage the diode.
This is how the capacitor discharges. One lead cannot rise above the supply voltage and if the other lead rises to this voltage, the capacitor sees equal voltage on both leads and discharges.
It’s as simple as that.

BEEPER CIRCUIT

Not only does the BEEPER circuit have a mistake, but the component values are incorrect for CMOS gates. The value of 1k and 470u are related to TTL gates where the input impedance was much lower and had an effect on the charging and discharging of the capacitor.
CMOS devices have input impedances that are 1,000 times higher and it is intelligent to use values that reflect this. The capacitor should be from 100n to 1u and the resistor from 100k to 1M
These values show you have an understanding of what you are presenting.
The corrected Beeper Circuit is shown below:

The problem is this:
The part of the cycle he is describing is the DISCHARGING of the capacitor. And it does not discharge via the resistor. But via the input of the first inverter.
But he knows nothing about the diodes on the input of the gate and thus he should not be writing on the web.
All the rest of his discussion is completely incorrect.
Also he does not understand the fact that a gate changes state when the input voltage is slightly higher and slightly lower than mid-rail voltage. It does not have to rise to 100%. Maybe something like 30%, 70%.

I will take up where he has started. The capacitor is fully charged and the output rises. As it rises, it delivers a voltage to the input that is higher than the supply. A protection diode on the input of the first gate will not allow the input to rise higher than the supply, so, as the output rises, the top of the capacitor "hits a ceiling" and will not rise. So it gets discharged via the protection diode. Now the output is HIGH and the input is HIGH.
It would stay like that all day if it were not for the resistor. The resistor CHARGES the capacitor, because the bottom lead of the capacitor is HIGH and the bottom of the resistor is LOW.
As the capacitor charges, the voltage on the input of the first gate reduces and when it fall to about 30-40% of rail voltage, point B goes HIGH. This makes the output go LOW and the whole capacitor drops by a voltage equal to rail voltage.
The input voltage had about 30% rail voltage on it and now the voltage drops by full rail voltage.
This means the input voltage will drop below 0v, but a protection diode on the input prevents it dropping below 0v, so the input just goes to 0v and the rest of the voltage on the capacitor is quickly removed.
Now the output of the first gate is HIGH and it begins to charge the capacitor via the resistor and when the voltages reaches about 60% of rail voltage, the two gates change state to start the next cycle.
Note: the capacitor has times when it is partially charged and discharged very quickly via the protection diodes.
This has never been mentioned in any discussion. Everyone thinks it is a "smooth cycle."

The "ACCEPTED" way to draw a diagram but the regulator has to be "flipped over."