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The whole purpose of Talking Electronics website is to teach electronics without introducing any technical details. It's pointless knowing how much doping is required to produce a zener diode, if you don't know how to connect it to a circuit.
I have seen so many faulty designs in an Indian magazine: Electronics For You, from writers who have a list of qualifications after their name, that it makes their degree in electronics questionable.
I know no-one likes to be criticized but when you are teaching how to design a circuit, it is important the circuit not only works, but is designed correctly.
It is not good enough for a circuit to work, it must be designed correctly so it takes the least current and requires the least number of components. Every component has a "technically correct" value for each position in a circuit and that's what we teach. When the component-values are outside this range, it is obvious the engineer is lacking the skills in circuit-design. However if the circuit specifically requires a certain value, mention it on the diagram.
That's what a circuit is for. It MUST show all the values of all the components. It's a stupid idea to have a List Of Materials.  Only an IDIOT would draw a circuit with R1, R2 etc and provide a list of values. He obviously have absolutely no electronics ability AT ALL.  I can work out what a circuit is doing by looking at the component values. That's why this website is so important. It is teaching ELECTRONICS UNDERSTANDING.
The traffic to our site is increasing all the time and it has just reached 7,000 visitors per day.
This is spread over hundreds of articles, but the eBooks are getting most of the readers while those that are really interested in electronics are reading through the 14 pages of Spot the Mistake.  
This type of teaching has never been done before, and I wish it was available when I stated in electronics.
You learn more from other's mistakes than a whole book of successful circuits.  


Electronics For You magazine is filled with mistakes and these will keep us busy for years to come.
Here is a DICE project with a number of mistakes:

The capacitor on pin 5 of the 555 is needed to keep the oscillations of the chip reliable. In this circuit we are requiring unreliability so the 10n capacitor is not needed.
A single resistor on the common line of the display will make the segments fade when a lot of segments are illuminated. To keep the segments equal-brightness, a resistor is needed on each drive-line. 
The diode on the reset line serves no purpose. In fact it is creating a high impedance on the line and can cause a reset at any time. It should be removed. 

Here we have a CMOS/TTL LOGIC PROBE from the Laboratories of Electronics For You Magazine:
It has lots of design-faults.

We start the discussion with Inverter N3. This is commonly called an Inverter or NOT-gate.
The input is sitting HIGH so the output will be LOW.
Only a HIGH-to-LOW transition on the input will change the state of the gate. This means the output will go HIGH.
It does not matter what voltage is on the anode of Diode D1, any rise in voltage on the cathode is not passed through the diode. That's not how a diode works. The diode is around the wrong way.
Now let's look at Diode D3. The input of N5 is LOW so the output is HIGH. No current can flow through the diode to the input of N4 and thus this inverter sits with an unknown voltage on the input.
Now let's look at the yellow LED. To turn on the pulse LED (yellow), the output of N6 must go HIGH. The input to N6 must go LOW and this means the input to N5 must go HIGH.
For the time-delay circuit (made up of C4 and R10) to work, Pin2 must be HIGH and C4 must be charged. Pin 2 goes LOW very briefly and the fully charged capacitor gets instantly discharged via pin 2 and the input protection diode on the input of N5. Pin 2 then goes HIGH and brings C4 HIGH and puts a HIGH on pin 9. This illuminates the yellow LED.
The capacitor gradually charges via the 4M7 resistor and this creates the delay.
But for this to happen, pin 3 of N4 must be LOW and if D1 is turned around the other way, pin 3 will be taken HIGH when a pulse is detected.  D2 and C3 are not needed. A 4M7 is needed between pin3 and 0v rail and D3 is not needed.
This circuit is certainly a mess.

This circuit has a number of technical mistakes.
I have received a reply from Electronics For You, and the replies are in
red. The reply was unsigned and the author of the circuit is unknown. Here is the datasheet for BT8031-xx that clearly shows the -02 version is Ding Dong. The sheet also shows the need for a buffer resistor on the output.

The inputs to the NOT-gates are high impedance and the TOUCH PLATE could be connected to one of the inverters (NOT-gates) and this would eliminate the need for the 741 op-amp.
There are many ways to design the circuit to get the same output. This circuit shows the use of inverter and 741 op-amp - but a circuit should be as simple as possible - not as complex as possible.
The need for the 10k resistor (R1) is unknown. It serves NO purpose.

The need for the 10k resistor (R4) is unknown as the op-amp pulls low.
You will not know unless you wire and check yourself.  What sort of answer is this?
Gates N2 and N3 are not needed as they simply produce the same result as the output of N1.
Not in this case. N2 and N3 gates are needed, there are reasons behind it.  What sort of answer is this?
The 22E (22R) resistor R5 is far too low. The CD4049 can source 10mA and to deliver this current to the 2v2 zener requires a 560R resistor.
You have no idea why we use these values. Your doubt is based on theoretical observations. The reason why the resistor had to be so low was due to the absence of the buffer resistor on the output. One mistake led to another.
The output of the BT8031 requires a buffer resistor on the output of about 1k. The IC should not be connected directly to the base of a transistor. See the data sheet.
The output current of BT8031 is less than 2mA. You don't need a buffer resistor at its output pin in this design. We have the datasheet of BT8031. The data sheet clearly shows a 5k to 10k resistor on the output.
The BT8031 is actually BT8031-02 for the Ding Dong version.
OK you say so. We have not seen or used BT8031-02
The BD678 is actually a PNP Darlington transistor and should be drawn as a Darlington.
Normally we used transistors symbols either npn or pnp irrespective of Darlington or not. This SAVES time and space. But your point is noted.
The BC547 transistor is not needed as a PNP BD679 transistor could be used as the buffer transistor.
The BT8031-02 can be supplied with 3v and it may be necessary to deliver 3v to drive a Darlington transistor.
There is no current limiting resistor on the collector of the BC547. Without this resistor a very large current will flow through the junctions of the two transistors when driving the speaker and this is WASTED CURRENT.
The value of R6 is technically too low. It should be 10k.
Your doubt is based on theoretical observation. The design is working fine and no changes are required. The circuit may be "working fine" but it has so many technical faults that it is obvious the designer of the circuit has no technical ability AT ALL. This would be ok if he just produced the circuit for himself, but it is being presented in an electronics magazine for thousands of readers to study and observe and it should be technically correct.

This is one of the worst-designed Delay circuits:

The circuit has been designed to discharge the delay capacitor when the circuit is turned off. This is a very unusual way to design, however if the connections to the gates of T1 and T2 are reversed, the discharging components will turn ON for a brief period and discharge the electro before the timing begins. A low-value resistor will need to be added in series with the 2,200 to generate the required time-delay.
The delay is made up of a 10k multi-turn pot and 10,000u electrolytic. Normally a delay is made up of a high resistance and low value capacitor. The mid-point of these two components is then taken to a high impedance detecting circuit. In this case it is the gate of a FET. The 10k should be increase to 100k or higher and then the capacitor can be reduced to 100u to 220u.
A muli-turn pot is the worst choice of pot as you don't know where the wiper is positioned. 10-turn pots are NOT designed to be constantly adjusted.
The 1N4007 diodes can be changed to 1N4004 to show that 1,000v diodes are not needed.
The value of R1 and R2 should be increased to 47k to show the surrounding circuit is high impedance.
The 78L05 is not a LDO (Low Drop Out) Regulator.
The designer of the circuit says "resistors of high values are less durable  What does this mean?  If he means they are less reliable, it is not true. A 100k or 470k resistor is just as reliable as 10k.
The designer suggests a 3v3 regulator. The 2N7000 requires up to 3v to turn ON when 75mA is needed. If the relay needs 100mA or more, the gate may require slightly more than 3v and the circuit will not work.
The circuit is over-designed and uses expensive components.

Here is another over-designed circuit:

All that is required is two transistors:

The Darlington transistor provides a gain of more than 10,000 and the 1k resistor on the base will turn the TIP122 on hard enough to illuminate the 21watt globe.  Don't forget, the globe will require 5 or more amps to start glowing as the resistance of the filament is very low when the lamp is cold and that's why it takes up to 6 times more current.

Here's another over-designed circuit with mistakes:

1. What is the purpose of D3?   It is not needed. The battery cannot feed-back into the power supply as diodes D1 and D2 will block the voltage.
2.  D1, D2 and D4 should be identified as 1N4002 to indicate the reverse-voltage only needs to be 100v. A 1N4007 is a 1,000v diode.
3.  The output of a 9v - 0v -9v transformer will be a lot higher than 9v on NO LOAD due to the poor regulation of these types of transformer. The output voltage is designed to fall to 9v on full load. This means supply rail will be a lot higher than 12v and the 9v battery will be overcharged via the 68 ohm resistor. 
4. R3 R4 and R6 can be converted to a single resistor.
5. You cannot drive a diode through a 10M resistor and expect to get a result. Any voltage-change will be partially or wholly absorbed by the 10M resistor.
6. What does C2 and R8 do? There is nothing at the join of these two components and they perform no function when connected in series. This type of arrangement is only used when the circuit is oscillating at a fairly high frequency. When it oscillates, the 10k and 22k resistors produce a voltage-divider that is detected by the inverted input of the op-amp. This feature does not apply when the change is very slow.
7. The value of 10p for C3 will have no effect on the circuit.
8. The maximum output current for a CA3140 is 40mA for a 15v supply.  The supply will possibly be about 15v due to the point mentioned above and the output will be about 10v due to the arrangement of the 10M resistors and the fact that the output is connected directly to the inverting input. It is unknown what current will be delivered to the LED but the 100R on the LED will make the output very low impedance and the maximum output voltage is unknown.
9. The op-amp will have UNITY GAIN in this arrangement and that is one of the big mistakes of the circuit. 
10. The value of R11 is too low. When the pot is turned fully clockwise, up to 30mA may flow through the zener and base-emitter of the transistor. It's just a bad design. Many of the values are far from "design values." Design Values are those you would find in a properly designed circuit. 
That's 10 mistakes in a single circuit.

I put the circuit on an electronics forum and got a number readers detecting some of the faults. But none saw the circuit as a whole by identifying all the faults.

One reader: "amspire" produced a list of answers that were entirely incorrect. I will go though his list:

You have not identified a single fault. Using a 1N4007 instead of a 1N4001 is very common - the 1N4007is often the same price and sometimes cheaper. 1N4001's can be hard to find at times and really only exists for historical reasons that are now irrelevant.
Component values should be shown for a reason. A beginner to electronics may wonder why a 1,000v diode is needed. That's why the circuit should show 1N4002.
You do need a compensation cap on that opamp otherwise it is not stable at unity gain - it is 100% guaranteed to oscillate without the 10p cap.
A 10p capacitor will have NO EFFECT on reducing the instability. There is no mention of the need for a compensation capacitor in the data sheet.
The 10Meg resistor does not swamp out the diode. At low temperatures, it is the diode that swamps out the 10M resistor.
This is entirely UNTRUE. The 10M resistor in series with the diode will reduce the voltage-change (from the diode) entering the op-amp by a ratio equal to the 10M to the input-resistance of the op-amp. The input impedance of the op-amp is less than 5M and thus less than 35% of the voltage-change will be detected.   
C2 and R8 increase the sensitivity to a fast rise in temperature.
The voltage-increase from the diode is very slow in terms of op-amp speed and the capacitor will have no effect.   
The maximum that can flow into the transistor base is probably 15mA with the pot at full (and it will not be at full after adjusting) and 15mA will not damage the transistor or zener so that is absolutely OK.
The circuit is so badly designed that you cannot find a starting-point to discuss this section.
However the zener diode is not needed as the voltage-drop across it is NOT 3v1. It will start to leak at a much lower voltage and the transistor will turn on. The zener serves NO purpose.
There are no "design values" for a transistor that you should adhere to - you can do absolutely anything you like with a transistor as long as you do not exceed limits and blow it up. Even Maximum ratings can be exceeded if you know what you are doing - eg using a transistor in avalanche mode for a fast pulse generator.  I can go through the whole list, but you have not identified a single fault.

It is lucky we have websites like Talking Electronics to show you how to design a circuit. Of course there are "design values" like 470R feeding a LED and 4k7 to 10k feeding the base of a transistor.

This circuit was sent to Electronics For You via their request in the February issue:

I pointed out 10 mistakes in the above circuit and got a reply from D.Mohankumar:

Mr Mitchell is a Critic of all circuits published in EFY and he is tarnishing the Indian publications and authors. So I am not interested to reply to his queries.

I then said I would analyse circuits before publication in  Electronics For You because so many of the circuits had faults and this was giving faulty information to newcomers to electronics.
The reply from the "Technical Editor" was:
mind your own bussiness, (he can't spell business) we have enough people to look into the mistakes and take corrective action.

Where are these technical personnel????   Why didn't he send my corrections to one of these "technical people"?
Obviously the magazine does not have any technical personnel as it has been publishing circuits with hideous mistakes for over 10 years. Nearly every one of their circuits has technical faults and they don't even want anyone to correct them.
I call this sabotage. They are sabotaging electronics enthusiasts by feeding them with faulty designs and making their chances of successful employment almost zero. 
It's people like this and magazines like  Electronics For You that should not be in the publishing business. They are just ruining the name of electronics.

Here's another circuit with a couple of mistakes. You can see who designed the circuit as it looks just like some of the previous circuits we have looked at:

If the 10k pot is turned fully clockwise, photo-transistor T1 will blow up.
When the voltage on the base of T2 is 0.55v, the transistor is not turned on. When it is 0.7v, the transistor is turned on. The 470u C2 is only used for this small voltage range.
By placing a resistor between C2 and T2, (on the base) the voltage-range will be increased and its effectiveness will be increased.

Here's another D.Mohankumar circuit.  None of his circuits work.  You must not build them.

Transistor T2 will never turn off.  The voltage across diode D1 is 0.7v.  The voltage across the collector-emitter of T1 is 0.2v.  This is 0.9v. The voltage on the base of T2 must be less than 0.6v to turn it OFF.
  Here's another D.Mohankumar circuit:

You cannot get a 7805 to work from the output of a 7806.   A 7805 requires 8v input.

Here's another D.Mohankumar circuit.  It is a battery charger:

Connecting a slightly flat battery to the circuit will blow up the LED. When the circuit is redrawn, you can see why:

There is a direct connection between the 15v supply and the 12v battery, via a diode-drop of 0.7v which is the base-emitter junction of the transistor. There is no current-limiting resistor and the LED will blow up.

Here's another D.Mohankumar circuit. It is an emergency lamp:

A 1watt LED takes 300mA.   The voltage drop across a 100R when 300mA flows is 33v!! The 100R can only deliver 25mA.  
What is pulling the base LOW when the power fails? Another untested circuit.

Here's another D.Mohankumar circuit. It is a battery charger:

The circuit is supposed to turn the current off when the battery reaches 13.5v.  But the "turn-off" circuit consists of an 8v2 zener, the base-emitter junction of T1 and the voltage drop across a red LED.  This is 8v2 + 0.7v + 1.7v = 10.6v    The zener needs to be 11v.

This circuit toggles the relay each time the switch is pressed.

Pressing S1 will blow up transistor T1. A 4k7 resistor should be connected in series with S1.
The operation of this circuit is very clever and quite complex. See the full explanation of how the circuit works in "101-200 Transistor Circuits". Toggle A Relay

Here's another circuit from Electronics For You.  It has not been checked by a qualified design-engineer. The circuit possibly works but when you find out how it operates, you will be HORRIFIED. 

It is absolutely essential to draw a circuit so it is instantly understandable.
The connection of the first three transistors looks unusual, so the circuit was drawn in an understandable way. 

You can see the 3v supply goes through the emitter-base junction of the 3rd transistor and collector-emitter junction of the second transistor, without any current-limiting resistor.
When the first transistor is NOT turned on, the second transistor is turned on via the 22k resistor and this pulls the base of the third transistor towards the 0v rail. The emitter will have about 0.9v on it and the current through the two transistors will be very high.
This is an extremely BAD design.
The voltage on the collector will be about 0.7v. The transistor driving the relay will turn on but the LED on the base will not illuminate and the Piezo Buzzer will not produce a sound.
When the first transistor is turned ON, the second transistor is turned OFF and the third transistor is turned OFF.
What a terrible design.
Electronics For You informed me they do not have a technical that understands electronics technology and simply publish the circuit because "it works."
Now you can see how damaging this simplistic approach is. 40,000 young electronics enthusiasts are a getting the wrong idea on circuit design because the magazine does not know how to test a circuit.

Here's another disaster from Electronics For You magazine.

It's an FM Bug that connects across the phone line. All Talking Electronics Phone Bugs connect to one of the lines by cutting the wire and inserting the circuit. A bridge on the circuit automatically adjusts to either polarity and you don't have to worry about the high voltages of the phone line.
The following circuit is a disaster.  The output transistor is a 20v device and it will zener (actually break-down) when connected to the line and that's why the "laboratory" at Electronics For You had to add the 22k. Instead of working out why the circuit engaged the phone when connected to the line, they added the resistor.
But the major faults are:
The 47k resistor is not connected to the top rail and the oscillator transistor i s not turned ON.
The coil in the collector does not have a capacitor across it to make a TANK CIRCUIT and the circuit will not work.  
C1 should be connected to the base of the transistor.
C2 is not needed.
That's 5 faults in a single circuit.
Another terrible design.



Here's an over-designed circuit, with mistakes. If the 10k is turned clockwise, pin 7 will be damaged.


Here's the circuit with corrections and simplified:

A 6v battery must be charged from a supply that is no more than 6.8v.
This circuit delivers 7.5v and the battery will gradually DRY OUT. The circuit needs to be redesigned with a current-limiting resistor or a power diode on the emitter of T1 to reduce the output of the regulator to 6.8v.

This circuit detects heat via the thermistor. As it gets to 100C, its resistance decreases to about 1k.
The diode, R2 and electrolytic are not needed. By adjusting the value of R8, the voltage on the base of T1 can be sent so the circuit turns on at an elevated temperature.
As the voltage on the base of T1 increases, the transistor turns on and the emitter rises. This means the voltage on the base must rise higher to turn the transistor on more. This is not a problem in this circuit but resistor R3 (470R) should be in the collector of T1 so that a rise in base voltage will be passed to the output.
The worst feature of the circuit is the current-drain. The circuit takes 10mA and most fire alarms take only a few microamps so the battery lasts for a year or more.
By connecting pin4 and 8 of the 555 to the collector of T2, the chip will not be powered during the waiting period and the circuit will take much less current.

Here's another D.Mohankumar faulty circuit. It is a solar battery charger:

Here is a reply from a reader:

I assembled the Solar Charger circuit as per your design. The solar panel when not connected in circuit shows 22.5VDC. But when I connect it in the circuit the output shows 0.79VDC.

D.Mohankumar could not answer the question.
A zener diode is not designed to be used as shown above. A zener diode is very "leaky" and any voltage on the cathode is transferred to the other lead and it will appear on the base of the transistor.
That's what is happening in this circuit and the output only has to rise to 0.9v and the transistor is already turned on and taking the adj terminal of the regulator to 0v rail.
This circuit does not work. Another untested circuit by D.Mohankumar.
Why use a 12v panel to charge a 6v battery? This makes the circuit less than 50% efficient.
Why have any electronics AT ALL? A 4.5aHr battery can be constantly supplied with 300mA and it will not be overcharged. Use a 6v 300mA - 500mA solar panel with a single diode to prevent current flowing back into the panel when it is not illuminated. That's all the circuitry you need as the panel will be supplying energy for no more than 8 hours per day and the battery will not get overcharged.  
You have to think "outside the square" and ask: "Is a complex circuit necessary?"


There are a couple of technical faults with the circuit:
The 10k base resistor to the first transistor is far too low. It should be at least 100k.
A low-value resistor will reduce the incoming waveform from the microphone by about 50%.
The electret microphone will produce a waveform of about 2mV to 20mV. The first transistor will have a gain of about 100. The result will be a maximum of 1,000mV
The second transistor is an emitter-follower and will not increase the amplitude of the waveform.
The third transistor has a fixed gain of 10 due to the ratio of the 10k collector resistor to the 1k emitter resistor.
The circuit will produce sufficient waveform (about 1,000mV x 10)  to turn on the fourth transistor but the 10u feeding the diodes is around the wrong way and the 4u7 will have a greater effect if it is away from the base by adding a 4k7 resistor.

If the pot is turned fully clockwise, the BD140 will take more current and this will flow through the base-emitter junction of the BD139. This current will be more than required by the BD139 and it will be damaged. 
The current taken by the 1watt LED will be 12v - 4v = 8v / 33 = 240mA. The wattage of the 33R should be 8 x .24 = 2 watts.

Transistors are not added to a circuit to make it more complex. These transistors are NOT NEEDED.

Remove the transistors and the circuit will work exactly the same as before:

Here we have a circuit from Swagatam Majumdar.  It has a number of faults:

I have only identified the resistors as this is where the faults are located.
1. The value for R1 is 0.12 ohms for 10 amp load due to the fact that diode D3 has been placed on the emitter of transistor T1. This diode is not needed. Removing the diode allows R1 to be 0.06 ohms and 6 watts instead of 12 watts.  A huge difference.
2. D1 is not needed. It serves NO purpose.
3. The value of R4 is far too high. It will NEVER keep T1 in conduction. R2 needs to be increased to 10k and the circuit will work.
Obviously the circuit has never been tested.  Two unnecessary parts and two incorrect-value resistors.

Here's another untested circuit from Swagatam Majumdar:

The voltage on the collector of the first transistor will be 0.7v plus 0.2v  = 0.9v when the transistor is conducting and this will not turn off the second transistor. The diode must be removed. It serves NO purpose. He adds a diode to the emitter in all his circuits and this is completely unnecessary and creates faults circuits. He claims to be an "electronics engineer" and yet most of his circuits have not been tested and DO NOT WORK.

Here's another untested circuit from Swagatam Majumdar.
I have only identified the 7805 and 4017 as this is where the fault lies:

The voltage on the collector of the PNP transistor is about 12v. The voltage on the supply-rail of the 4017 is 5v.
The output of the 4017 cannot rise higher than 4.5v. To turn off the PNP transistor (T2) the base voltage must rise  to about 11.5v  
In this circuit the transistor will NEVER TURN OFF.
That's why you use an NPN transistor on the output in this type of circuit. You can then use different-voltage supplies. 

Another circuit from Swagatam Majumdar with a few minor mistakes. The circuit detects when a car is hit or bumped and sounds the horn for a short period of time.

The 2k7 load resistor for the microphone is too low. It should be 47k or higher. You only want to detect a strong hit or bump.
The 100 ohm resistor on the base is not needed. It serves no purpose.
The "coil" on the base of T3 is an unknown component and should be removed.
The two 39k resistors on the input to the "Darlington" arrangement should be 100k so the 220u can be reduced to 100u.

Here we have a capacitor-fed power supply with a slow turn-on feature:

When the voltage is applied, the output of the bridge is effectively 330v with a current of 70mA. The 330v will be directly across the first 10k resistor and it will start to charge the 470u. The voltage will then pass through the second 10k and start to charge the second 470u.
Suppose we remove the LEDs and see what happens.
The result is 330v on the output of the bridge, 172v on the first electro and 15v on the second electro. The transistor now turns on and the 25 white LEDs produce a voltage-drop of 75v. The voltage on the first electro drops to 45v, but it will blow up. The 100k must be reduced to 10k.
Now we come to the next problem.
The 1u capacitor will deliver 70mA and in this type of circuit the 70mA current is being delivered to the load and it must be able to accept (dissipate) the wattage. The wattage will depend on the voltage across the load and in this case we have two paths. The 10k plus 10k plus 15v zener path takes 3mA because the supply is 75v.
The remaining 63mA will flow though the white LEDs and BLOW THEM UP it they are ordinary LEDs. They have to be high-wattage LEDs and although 1watt LEDs require 300mA for bright illumination, 63mA will produce a very good output. 
As you can see, the physics behind this simple circuit is much more complex than you think.

Here's another untested circuit from Swagatam Majumdar.

The circuit will not work because the author failed to realise the Darlington TIP127 transistor has inbuilt resistors. To pull the base down by a voltage equal to 0.7v plus 0.7v, resistor R3 must be less than 60k. At 60k, the transistor is just at the point of turning on.
Now we come to the next problem.
A car horn requires about 4 amp to produce an output. But the coil inside the horn is only being activated for about half the time and this means the starting current to activate the electromagnet is 8 amps. This point has never been mentioned before and is one of the hidden problems with driving a horn.
To fully drive the transistor (fully saturate it), the base current must be increased and the base resistor will have to be reduced to 33k.

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