The PSU turned very hot.
A lot of dust had accumulated!
But the fan did not spin anymore at all.
I had an idea!
Maybe dust would have made it's way between the magnet and the coils.
I tried to pour some white spirit into the fan, and soon, the fan again would spin freely.
After powering up, it works again!
I have inlined a 3W red LED as well, to reduce voltage.
Low speed is enough, since I do not use all available Watts.
Good work I think.
I tried the same method on a tiny GPU cooler as well. In a few minutes, it was spinning again!
Just a little white spirit did the trick.
https://www.youtube.com/watch?v=XPE8irEZUdc
Rated current]]>
The level of continuous DC current that can be passed through the inductor, this DC current level is based on a maximum temperature rise of the inductor at the maximum rated ambient temperature, the rated current is related to the inductor's ability to minimize the power losses in the winding by having a low DC resistance, it is also related to the inductor's ability to this power lost in the winding, thus, the rated current can be increased by reducing the DC resistance or increasing the inductor size, for low frequency current waveforms, the RMS current can be substitued for the DC rated current, the rated current is not related to the magnetic properties of the inductor.
DCR(DC resistance)]]>
The resistance of the inductor winding measured with no alternating current, the DCR is most often minimized in the design of an inductor, the unit of measure is ohms, and it is usually apecified as a maximum rating.
OHM
The unit of measurement for resistance and impedance, resistance is calculated by ohm's law.
Q value]]>
The Q value of an inductor is a measure of the relative losses in an inductor, the Q is also knows as the "quality factor" and is technically defined as the ratio of inductive reactance to effective resistance and is represented by Q=XL/R=2pifL/R
Since XL and R are functions of frequency, the test frequency must be given when specifying Q,XL typically increased with frequency at a faster rate than Re at lower frequencies, and vice versa at high frequencies, this results in a bell shaped curve for Q vs frequency, R is mainly comprised of the DC resistance of the wire, the core losses and skin effect of the wire, based on the above formula, it can be shown that the Q is zero at the self resonant frequency since the inductance is zero of the point.
Number turns]]>
The series impedance of a high frequency ferrite device can be increased by running two or more turns of the treated conductor through the ferrites core, magnetic theory predicts that the impedance of the device will increase with the square of the number of turns, however, due to the lossy and non-linear nature of EMI suppression ferrits,aferrite bead with wo turns will yield somewhat less than four times the impedance of an identical part wound with only one turn of the conductor.
DCR(DC resistance)
The resistance of the inductor winding measured with no alternating current, the DCR is most often minimized in the design of an iductor, the unit if measure is ohms,and it is usually specified as a maximum rating.
Standard inductance tolerances are typically designated by a tolerance letter, standard inductance tolerance letters include:
F:+/-1%
G:+/-2%
T:+/-3%
J:+/-5%
K:+/-10%
L:+/-15%
M:+/-20%
P:+/-25%
N:+/-30%
Test frequency
The frequency at which inductors are tested for either inductane or Q or both,some test fequencies used widely in the industry include:
1KHz power inductors( wide value range)
79.6K inductors(above 10000uH to 100000uH
252K inductors(above 1000uH to 10000uH)
796K inductors( above 100uH to 1000uH)
2.52M inductors(above 10uH to 100uH)
7.96M inductors(above 1uH to 10uH)
25.2M inductors(above 0.1uH to 1uH)
50M inductors(above 0.01uH to 0.1uH)
Most of these test frequencies have been designated by military specifications, however there are some conflicting frequency assignments among the military specifications, there is present trend to assign test frequencies that match the user frequencies, this is particularly true for evry low values, these user frequencies do not match those listed above. ]]>
Inductance]]>
The property of a circuit element which tends to oppose any change in the current flowing through it, the inductance for a given inductor is influenced by the core material, core shape and size, the turns count and shape of the coil,inductors most often have their inductances expressed in microhenries(uH),the following table can be used to convert units of inductance to microhenries, thus,47mH would equal 47,000uH
1H=10^6uH
1mH=10^3uH
1uH=10^3nH
Inductor]]>
A passive component designed to resist changes in current,inductors are often referred to as "AC Resistor",the ability to resist changes in current and the ability to store energy in its magnetic field,account for the bulk of the useful properties of inductors,current passing through an indutor will produce a mangetic field, a changing magnetic field induces a voltage which opposes the field-producing current,this property of impeding changes of current is known as inductance, the voltage induced across an inductor by a change of current is defined as: V=Ldi/dt, thus, the induced voltage is proportional to the inductance value and the rate of current change. .
There are 2 modules associated with the compare section the CCP module which well use for the compare function, and the Timer 1 module. The function of the compare module is to compare (obviously, as the name suggests) the value of the CCPR1 register against the Timer 1 value. The CCPR1 register is actually composed of two 8-bit registers that together form a 16-bit register. The high 8-bits the high byte make up the CCPR1H register and the low 8-bits the low byte make up the CCPR1L register.
For example, if the value of CCPR1H is 30 and the value of CCPR1L is 47, what is the value of the 16-bit CCPR1 register?
Solution:
CCPR1H = 30 = 0x1E
CCPR1L = 47 = 0x2F
So, the high byte = 0x1E. The low byte = 0x2F.
CCPR1 = 0x1E2F = 7727
Like CCPR1, TMR1 is a 16-bit register composed of two 8-bit registers TMR1H and TMR1L. If TMR1H = 30 and TMR1L = 47, TMR1 = 7727.
Back to the function of the compare mode. The 16-bit value of CCPR1 is compared against the 16-bit value of TMR1. RC2 pin is associated with the CCP module. Upon match of the CCPR1 register against TMR1 (when CCPR1 is equal to TMR1), one of the following can happen to RC2 (RC2 is the pin associated with the CCP module) depending on the setting of the CCP1CON register:
RC2 is driven high
RC2 is driven low
RC2 is unaffected
If RC2 is to be used, TRISC2 must be cleared to make RC2 a digital output.
Upon match of CCPR1 and TMR1 (when CCPR1 is equal to TMR1), CCP1IF (bit 2 of PIR1) is set. If the CCP interrupt is enabled, a CCP interrupt is generated. To enable the CCP interrupt, CCP1IE (bit 2 of PIE1) has to be set.
Another interesting setting of the CCP module is that it can be set up so that upon compare match of CCPR1 and TMR1, the following happen:
CCP1IF is set
CCP interrupt is generated if CCP1IE is set
RC2 is unaffected
Timer 1 is reset (TMR1 = 0)
If the ADC module is enabled, an A/D conversion is started
What happens when there is a compare match of CCPR1 and TMR1 depends on the setting of the CCP1CON register. Here is the CCP1CON register:
When CCP1CON = 8 (0x08), CCP1 module is configured for compare mode and is set up so that upon a compare match of CCPR1 and TMR1, RC2 is driven high.
When CCP1CON = 9 (0x09), CCP1 module is configured for compare mode and is set up so that upon a compare match of CCPR1 and TMR1, RC2 is driven low.
When CCP1CON = 10 (0x0A), CCP1 module is configured for compare mode and is set up so that upon a compare match of CCPR1 and TMR1, RC2 is unaffected.
When CCP1CON = 11 (0x0B), CCP1 module is configured for compare mode and is set up so that upon a compare match of CCPR1 and TMR1, RC2 is unaffected, TMR1 is reset and an A/D conversion is started if the ADC module is enabled.
Remember that in all compare modes, whenever there is a compare match of CCPR1 and TMR1, CCP1IF is set.
Now lets take a look at a code example to make things clear. Lets say were running a PIC16F877A with a crystal oscillator of frequency 20MHz. Lets use the compare mode to generate a 50Hz output with 50% duty cycle.
Solution:
Tcy = (1/20000000)*4*106 us = 0.2us
Since one TMR1 increment takes 1 instruction cycle (with prescaler 1:1), one TMR1 overflow takes 65536 instruction cycles. Thats 65536 * 0.2us = 13107.2us = 13.1072ms. For 50Hz, time period = 20ms. So with 50% duty cycle, on time is 10ms.
The number of instruction cycles in 10ms is equal to (10000/0.2) = 50000.
When TMR1 = 50000 after counting up from 0, 10ms will have elapsed. So, well assign 50000 to CCPR1 so that a compare match occurs every 10ms. Well use the CCP setting for trigger special event where TMR1 is reset upon compare match.
CCPR1 = 50000 = 0xC350
Therefore, CCPR1H = 0xC3 and CCPR1L = 0x50
For our required mode of operation, CCP1CON = 11. As CCP interrupt will be used, CCP1IE must be set. GIE and PEIE also have to be set.
Well use RC0 as our output pin. Every 10ms, the state of RC0 must be toggled. Well do the toggling in the interrupt service routine (ISR). We chose to use RC0. Since in our mode of operation, RC2 is unaffected, we can use any pin really and I arbitrarily chose to use RC0. If any of the pins that are multiplexed to the ADC module or the comparator module is to be used, remember to disable the analog circuitry and use that pin as digital output.
The code should be quite easy to understand. Here it is:
Code:
//Programmer: Syed Tahmid Mahbub
//Target Microcontroller: PIC16F877A
//Compiler: mikroC PRO for PIC
//Using Compare Section of PIC CCP Module - using PIC16F877A
void interrupt(){
if (CCP1IF_bit == 1){
RC0_bit = ~ RC0_bit; //Toggle RC0
CCP1IF_bit = 0;
}
}
void main() {
PORTC = 0;
TRISC = 0;
CCP1CON = 11;
CCP1IE_bit = 1;
GIE_bit = 1;
PEIE_bit = 1;
CCPR1H = 0xC3;
CCPR1L = 0x50;
//CCPR1 = 0xC350 = 50000
T1CON = 0x01; //Prescaler 1:1, start Timer 1
while(1){
//Do whatever else is required
}
}
Select All
Here is the output waveform:
Now lets take a look at another code example. Lets say were running a PIC16F877A with a crystal oscillator of frequency 20MHz. Lets use the compare mode to generate a 50Hz output with 75% duty cycle.
Solution:
Tcy = 0.2us
In the previous example, as duty cycle was 50%, the simple thing to do was to toggle the output pin. Here we cannot do that as duty cycle is 75%.
This time, well use RC2 as the output pin.
Time period = 20ms. With 75% duty cycle, on time is 15ms. Off time is 5ms.
Number of instruction cycles for 5ms = (5000/0.2) = 25000. Number of instruction cycles for 15ms = 75000. This is beyond the maximum for TMR1. So, well utilize the TMR1 prescaler. Well use a prescaler 1:2.
So, for 5ms, number of TMR1 increments = 12500. For 15ms, number of TMR1 increments = 37500.
Well change the CCP1CON setting from 8 to 9 to 8 to set the output high to low to high upon consecutive compare matches. Well need to alternately change CCPR1 between 12500 and 37500 as required.
12500 = 0x30D4
37500 = 0x927C
In the 2 modes of operation we chose to use, TMR1 is not cleared/reset upon compare match. So, in the interrupt, we must clear TMR1 manually to start TMR1 counting from 0.
If you understand how this code works, you should be clear about the compare module by now.
Here is the code:
Code:
//Programmer: Syed Tahmid Mahbub
//Target Microcontroller: PIC16F877A
//Compiler: mikroC PRO for PIC
//Using Compare Section of PIC CCP Module - using PIC16F877A
void interrupt(){
if (CCP1IF_bit == 1){
TMR1H = 0;
TMR1L = 0;
//TMR1 must be cleared manually
if (CCP1CON == 8){
CCP1CON = 9;
CCPR1H = 0x92;
CCPR1L = 0x7C;
//CCPR1 = 37500 --> 15ms
}
else{
CCP1CON = 8;
CCPR1H = 0x30;
CCPR1L = 0xD4;
//CCPR1 = 12500 --> 5ms
}
CCP1IF_bit = 0;
}
}
void main() {
TRISC = 0;
PORTC = 4;
CCP1CON = 9; //Clear RC2 on match
CCPR1H = 0x92;
CCPR1L = 0x7C;
//CCPR1 = 37500 --> 15ms
CCP1IE_bit = 1;
GIE_bit = 1;
PEIE_bit = 1;
T1CON = 0x11; //Prescaler 1:2, start Timer 1
while(1){
//Do whatever else is required
}
}
Select All
The compare section of the CCP module is very useful for timing-dependent applications. As shown above, you can use it for PWM at frequencies that are too low for the PWM section of the CCP module.
If clearly understood, it is very easy to use and I hope I could make the compare mode of the CCP module clear. Your comments and feedback are welcome. ]]>
Please read the introduction post.
Thanks!:)
1.9
Quote:
| For the figure shown in figure 1.10, with Vin=30V and R1=R2=10k, find (a) the output voltage with no load attached (the ope-circuit voltage); (b) the output voltage with a 10k load (treat as a voltage divider, with R2 and R1 combined into a single resistor); (c) the Thevenin equivalent circuit; (d) the same as in part b, but using the Thevenin equivalent circuit; (e) the power dissipated in each of the resistors |
1.10
Quote:
| Show that Rload=Rsource maximises the power in the load for a given source resistance. |
Quote:
| Determine the voltage and power ratios for a pair of signals with the following decibel ratios: (a) 3dB, (b) 6dB, (c) 10dB, (d) 20dB. |
Quote:
| Derive the formula for the capacitance of two capacitors in series. |
Quote:
| Show that the rise time (the time required to go from 10% to 90% of its final value) of this signal is 2.2RC. |
Quote:
| R1=R2=10k, and C=0.1μF in the circuit shown in Figure 1.34. Find V(t) and sketch it. |
1.15
Quote:
| A current of 1mA charges a 1μF capacitor. How long does it take the ramp to reach 10 volts? |
Quote:
| Use the preceding rules for the impedance of devices in parallel and in series to derive the formulas (Section 1.12) for the capacitance of two capacitors (a) in parallel and (b) in series. |
1.17
Quote:
| Show that if A=BC, then A=BC, where A, B and C are magnitudes. |
Quote:
| Prove that a circuit whose current is 90 out of phase with the driving voltage consumes no power, averaged over an entire cycle. |
Quote:
| Show that all the average power delivered to the preceding circuit winds up in the resistor. Do this by computing the value of VR2/R. What is that power, in watts, for a series circuit of 1μF capacitor and a 1.0k resistor placed across the 110 volt (rms), 60 Hz power line? |
1.20
Quote:
| Show that adding a series capacitor of value C=1/ω2L makes the power factor equal 1.0 in a series RL circuit. Now do the same thing, but with the word "series" changed to "parallel." |
Thanks for taking the time to help me with the questions to produce the world's first Art of Electronics Answer Guide.
Yay!
I intend to post questions up here. Any which need a schematic or formulae from the book will have that too.
In order to avoid people doing the same question, could you please reply and say which questions you want to do.
When you have worked each out, can you please send the workings and answer to me in a PM. Also, make it clear which question it is! :)
Thanks again. You guys are great.
Sparky ]]>
However, a MOSFET can be controlled as we want. Set the gate high (with a sufficient voltage) and current can flow from drain to source. Set the gate low and current can no longer flow. Convenient!
However, a MOSFET can only be used to control DC loads since it is a unidirectional switch - current flow can be controlled when it is flowing from drain to source, but can not be controlled from source to drain. So, certainly it can not be used to control AC loads. Right?
Well, directly, it can not be used to control AC loads the same way you'd control DC loads. But, with some clever circuitry, it can be used to control AC loads. And here's how:
Perhaps you don't see how it works now. But consider the two diagrams below, which show the flow of current during the two AC half cycles. I'm sure you'll get it better then.
As you can see, due to the bridge rectifier, the MOSFET always "sees" a DC voltage as the drain is always positive with respect to the source. Thus, with this combination of the bridge rectifier and MOSFET, by controlling a DC switch - the MOSFET, you can control the AC load.
The MOSFET must be turned on fully by driving it high by at least 8V above source level - 8V with respect to source.
So, you can turn the MOSFET on and off at any time and accordingly turn the load on and off as required. This makes driving the AC load so easy! ]]>
Previsouly I have built an AM transmitter but it never worked right, only very short distance.
cappels.org/dproj/FMXMTR/fmxmtr.htm
The design is somehow finicky, and does not work like shown. Adjustable resistor for the 330 Ohms, and 1K Ohms are essential.
As well, you need to tap the antenna with a gimmick capacitor.
-It only works at the very end of the FM scale
-There is still some noise contained. But with the indended frequency adjuster, it can be reduced.
Ideally if the frequency is set right, the circuit is sensible against tapping with a screwdriver. You will hear microphonic noise.
I have tested 10m distance so far!
www.youtube.com/watch?v=xMCjRE3jDCY - demo video