Current Logic Group, Ltd

Trusted manufacturer of
DC-DC converters | AC-DC Power supply | Latest News

Current Page: Index
Losses associated with the filter capacitor
1nput and output filter capacitors are not the main loss sources of switching power supplies, although they have a significant impact on the operating life of the power supply. If the 1nput capacitor is not chosen correctly, the power supply will not work as efficiently as it should.
Each capacitor has a small resistance and inductance in series with the capacitance. The equivalent series resistance (ESR) and the equivalent series inductance (ESL) are parasitic components caused by the structure of the capacitor, both of which prevent an external signal fr0m being applied to the internal capacitance. Capacitors therefore perform best when operating at DC, but perform much worse at the switching frequency of the power supply.
The 1nput and output capacitors are the only sources (or reservoirs) of high frequency currents generated by the power switches or output rectifiers, so the current flowing through the ESR of these capacitors can be reasonably determined by observing these current waveforms. This current inevitably generates heat in the capacitor. The main task of designing filter capacitors is to ensure that the internal heat of the capacitor is low enough to ensure the life of the product. Equation (4) gives the calculation of the power loss due to the ESR of the capacitor.
Not only does the resistive part of the capacitor model cause problems, but if the parallel capacitor leads are not symmetrical, the lead inductance will cause uneven internal heating of the capacitors, thus shortening the life of the capacitor with the highest temperature.
Additional loss
The additional losses are associated with all of the funct1onal components required to operate the power circuit, including the circuitry associated with the control IC and the feedback circuitry. These losses are generally small compared to other losses in the power supply, but some analysis can be done to see if improvements are possible.
The first is to start the circuit. The starting circuit obtains direct current fr0m the 1nput voltage, so that the control IC and the driving circuit have enough energy to start the power supply. If this startup circuit fails to cut off the current after the power supply is started, the circuit will have a continuous loss of up to 3 W, depending on the 1nput voltage.
The second major aspect is the power switch driver circuit. If a bipolar power transistor is used for the power switch, the base drive current must be greater than the transistor's peak collector current divided by the gain (hFE). The typical gain of a power transistor is between 5 and 15, which means that for a peak current of 10 A, a base current of 0.66 to 2 A is required. There is a 0.7 V drop between the base and the emitter, and if the base current is not taken fr0m a voltage very close to 0.7 V, there will be significant losses.


To improve the efficiency of DC-DC converter
To improve the efficiency of switching power supplies, it is necessary to identify and roughly estimate the various losses. The internal loss of switching power supply can be roughly divided into four aspects: switching loss, conduction loss, additional loss and resistance loss. These losses often occur simultaneously in lossy components and are discussed separately below.
Losses associated with power switching
Power switch is one of the two most important loss sources in a typical switching power supply. Losses can basically be divided into two parts: conduction losses and switching losses. The conduction loss is the loss when the power switch is in the conduction state after the power device has been turned on and the driving and switching waveforms have been stabilized; Switching losses are losses that occur when a power switch is driven into a new operating state and the drive and switching waveforms are in transition.
The conduction loss is measured as the product of the voltage and current waveforms across the switch. These waveforms are approximately linear, and the power loss during conduction is given by Equation (1).
A typical way to control this loss is to minimize the voltage drop during the conduction of the power switch. To achieve this, the designer must operate the switch in saturation. These conditions are given by Eqs. (2a) and (2b), which ensure that the collector or drain current is controlled by an external component rather than by the power switch itself, via a base or gate overcurrent drive.
Switching losses during power switching are more complex, both due to their own factors and the impact of related components. The waveform related to the loss can only be observed by an oscilloscope with a voltage probe connected to the drain and source (collector and emitter), and an AC current probe can measure the drain or collector current. When measuring the loss at each switching instant, a short lead probe with shielding must be used because any length of unshielded wire may introduce noise fr0m other power supplies, which may not accurately display the true waveform. Once you have a good waveform, you can roughly calculate the area enclosed by the two curves using a simple piecewise summation of triangles and rectangles
This result is only the loss value during the turn-on period of the power switch, and the total loss value during the switching period can be obtained by adding the turn-off and conduction losses. Losses associated with the output rectifier
Of the total losses within a typical non-synchronous rectifier switching power supply, the output rectifier losses account for 40% -65% of the total losses. So it's very important to understand this section. \n Rectifier losses can also be divided into three parts: turn-on losses, conduction losses, and turn-off losses.


The internal loss of DC-DC converter
To improve the efficiency of switching power supplies, it is necessary to identify and roughly estimate the various losses. The internal loss of switching power supply can be roughly divided into four aspects: switching loss, conduction loss, additional loss and resistance loss. These losses often occur simultaneously in lossy components and are discussed separately below.
Losses associated with power switching
Power switch is one of the two most important loss sources in a typical switching power supply. Losses can basically be divided into two parts: conduction losses and switching losses. The conduction loss is the loss when the power switch is in the conduction state after the power device has been turned on and the driving and switching waveforms have been stabilized; Switching losses are losses that occur when a power switch is driven into a new operating state and the drive and switching waveforms are in transition.
The conduction loss is measured as the product of the voltage and current waveforms across the switch. These waveforms are approximately linear, and the power loss during conduction is given by Equation (1).
A typical way to control this loss is to minimize the voltage drop during the conduction of the power switch. To achieve this, the designer must operate the switch in saturation. These conditions are given by Eqs. (2a) and (2b), which ensure that the collector or drain current is controlled by an external component rather than by the power switch itself, via a base or gate overcurrent drive.
Switching losses during power switching are more complex, both due to their own factors and the impact of related components. The waveform related to the loss can only be observed by an oscilloscope with a voltage probe connected to the drain and source (collector and emitter), and an AC current probe can measure the drain or collector current. When measuring the loss at each switching instant, a short lead probe with shielding must be used because any length of unshielded wire may introduce noise fr0m other power supplies, which may not accurately display the true waveform. Once you have a good waveform, you can roughly calculate the area enclosed by the two curves using a simple piecewise summation of triangles and rectangles
This result is only the loss value during the turn-on period of the power switch, and the total loss value during the switching period can be obtained by adding the turn-off and conduction losses. Losses associated with the output rectifier
Of the total losses within a typical non-synchronous rectifier switching power supply, the output rectifier losses account for 40% -65% of the total losses. So it's very important to understand this section. \n Rectifier losses can also be divided into three parts: turn-on losses, conduction losses, and turn-off losses.


surge protection module
ZLG's SP00S12 surge protection module can be used in various signal transmission systems to suppress harmful signals such as lightning strikes, surges and overvoltages, and to protect the signal ports of equipment. Isolated CAN transceiver with ZLG's fully isolated CTM or SC series, as shown below. It can greatly improve the integration level of products and greatly reduce the development cycle at the same time.
The Necessity of Resistance-capacitance Circuit Grounding
The principle of grounding after bus isolation and the recommended circuit have been described above, which must be very clear to everyone. In the field, many customers will mention why RC grounding is needed after bus isolation? Here is a brief dession:
1. Capacitor: fr0m the perspective of EMS (Electromagnetic Immunity), this capacitor is used to reduce the possible impact (the impact of high-frequency interference signals on the circuit with the ground level as the reference) on the premise that PE is well connected to the ground, in order to suppress the transient common-mode voltage difference between the circuit and the interference source. In fact, the direct connection of GND to PE is the best, but the direct connection may not be operable or safe. fr0m the point of view of EMI (electromagnetic interference), if there is a metal shell connected with PE, this high-frequency path can also avoid the radiation of high-frequency signals.
2. 1m resistor: This is used for ESD (electrostatic discharge) test. Because of this system (floating system) that connects PE and GND with capacitors, during ESD test, the charge injected into the circuit under test has no place to release, and will gradually accumulate, raising or lowering the level of GND relative to PE, to a certain extent, exceeding the voltage range that can be withstood by the weakest insulation between PE and circuit, GND and PE will discharge. In a few nanoseconds, tens to hundreds of amperes of current are generated on the PCB, which is enough to shut down any circuit due to EMP (electromagnetic pulse), or to damage the components of the signal connection at the weakest point of insulation between PE and circuit. But sometimes it is impossible to connect PE and GND directly, so a resistor of 1 ~ 2m is used to slowly release the charge to eliminate the voltage difference between them. Of course, the value of 1 ~ 2m is se1ected according to the ESD test standard, because IEC61000 stipulates that the maximum number of repetitions is only 10 times per second. If you make a non-standard ESD discharge of 1000 times per second, then I don't think the resistance of 1 ~ 2m can release the accumulated charge.


Equipment control test suspension
Equipment control test suspension
In this state, the reference ground of the control side of the device is suspended and has no connection with the PE, and the bus side has access to the protection ground (PE).
The analysis follows:
When the bus-side interface is subjected to electrostatic discharge, the electrostatic energy is discharged to the PE through the internal bus-side device of the isolated interface module. However, if the ESD energy exceeds the ESD immunity of the internal bus-side device of the interface module, the bus interface may be damaged.
When the control side interface is subjected to electrostatic discharge, because the control side is suspended, the energy can only be discharged through the equivalent capacitor Ciso of the isolation barrier. Because Ciso is very small, the voltage Viso at both ends will be very high, and all the voltage will be applied to the isolation barrier of the isolation interface module. If the voltage exceeds the voltage tolerance of the isolation barrier, it will cause damage to the internal isolation barrier.
3. Improvement measures
In view of the above two situations, the isolation interface module needs effective electrostatic protection. It is recommended to increase Cp, Rp and TVS when designing the isolation interface to improve the ESD immunity of the isolation interface.
The funct1on of the capacitor Cp is to reduce the pressure on the isolation barrier and provide a low-impedance path for the electrostatic energy. Most of the electrostatic energy is discharged through this capacitor. In order to achieve good results, the value of Cp should be much larger than Ciso, and it is recommended to be between 100pF and 1000pF.
funct1on of TVS tube: for the static electricity on the bus side, the static energy will be discharged through the protection device. Note: its conduction voltage must be less than the maximum voltage that the isolation interface can bear and greater than the signal voltage; When the communication rate is high or the number of nodes is large, it is also necessary to se1ect devices with small equivalent capacitance as far as possible to avoid affecting the normal communication of the bus.
Note: If the product has no safety requirements, it can be connected with Cp in parallel with a large resistance discharge resistor, such as 1m, to prevent static accumulation; If there is a safety requirement, it is generally necessary to remove the bleeder resistor and se1ect the safety capacitor. \n Perfect bus interface protection circuit
The mechanism of ESD has been analyzed before, but as industrial products have higher and higher requirements for the EMC level of communication interfaces. Many applications require IEC61000-4-2 electrostatic discharge level 4 and IEC61000-4-5 surge immunity level 4. The protection level of general transceiver ESD and surge is relatively low


How is the bus grounded after isolation?
CAN and 485 are field buses commonly used in industrial communication. Engineers must be very familiar with the bus isolation scheme, but they may encounter the situation that the communication is still abnormal after the bus isolation scheme is adopted. This article will take you to discuss how to ground the bus after isolation?
In order to ensure the communication stability of the bus network, the communication interface is usually isolated. The main purpose of isolation is: \n Safety considerations: protect equipment and personal safety, and isolate potential high voltage hazards;
Improve the stability of communication: eliminate the influence of earth potential difference;
Improving the reliability of the device: eliminating the influence of the ground loop;
Low coupling: improve compatibility between systems.
At present, there are two ways to achieve bus isolation: using discrete components to build or using integrated modules.
Principle of isolated grounding
Although adding isolation to the bus can ensure stable and reliable communication of the bus, the equipment with isolated communication interface will show completely different ESD characteristics under complex environment or installation state. Only by understanding the influence mechanism of ESD on the interface, can we add protection devices pertinently and improve the ESD capability of the isolated interface. Taking the communication interface with isolated CAN or RS-485 as an example, the action mechanism of ESD under common equipment conditions is analyzed, and the corresponding improvement measures are proposed.
1. The bus side is suspended
In this state, the control side of the device is connected to the protection earth (PE), and the reference earth of the bus side is suspended without any connection to the PE
The analysis follows:
Assuming that sufficient protection measures are taken at the control side, when the interface at the control side is subjected to electrostatic discharge, the energy is discharged to the PE through the protector at the control sides
When the bus interface is subjected to electrostatic discharge, because the bus side is suspended, the energy can only be discharged through the equivalent capacitor Ciso of the isolation barrier, because Ciso is very small, only a few picofarads to more than ten picofarads, Ciso is rapidly charged, the voltage Viso at both ends is very high, almost equal to the discharge voltage, and all the voltage is applied to the isolation barrier of the isolation interface module. If the voltage exceeds the voltage range of the isolation barrier, the internal isolation barrier will be damaged
Note that for a typical isolated interface module, the isolation barrier can withstand an electrostatic discharge voltage of only 4 kV, which is very vulnerable to higher levels of static electricity of 6 kV or 8 kV, and is very vulnerable to damage.


voltage dividing network of the EN pin is not considered to be set in the "appropriate" range
Because of the improper setting of the divider resistor network in the circuit, the threshold voltage of EN is reached when the 1nput voltage is very low, resulting in premature enabling of the output of the power supply chip. This is an example in which only the voltage of the EN pin of the power supply chip is considered to be set below the withstand voltage value in the design process, but the voltage dividing network of the EN pin is not considered to be set in the "appropriate" range.
So where is the appropriate location for the voltage divider network of the en pin?
When the 1nput voltage is low, the enable threshold of VEN is reached, and the output of the chip is enabled. At this time, the output is affected by the 1nput fluctuation and the power-on is slow, which affects the working stability of the subsequent circuit.
When the 1nput voltage VIN rises to 70% -80%, VEN reaches the enable threshold. At this time, the output of the chip eliminates the unstable stage of the 1nput power supply. The power supply is fast and the output is stable, which reduces the impact of 1nput voltage fluctuations.
At the same time, 20% ~ 30% margin is reserved to avoid the problem of output shutdown caused by the fluctuation of 1nput power supply;
Therefore, it is reasonable to set the EN threshold voltage of the power supply chip at 70% ~ 80% × VIN through the voltage divider network, and the EN threshold can be found in the chip manual. According to the known EN threshold and 1nput voltage, the appropriate divider resistor ratio can be obtained.
Network resistance is assigned based on a known EN threshold
The output waveform after adjusting the resistance of the divider resistor of the en pin shows that the output voltage fluctuation has been significantly improved. A more stable output waveform can be obtain by continuously adjusting that resistance value of the voltage divide resistor, and the method simply and effectively solves the problem of unstable output mentioned above.
It can be seen that improper setting of a small EN pin can also cause a lot of trouble, so it is also very important to stabilize the 1nput voltage of EN within the "appropriate" range according to the actual situation under the premise of meeting the EN withstand voltage value. Have you learned this little skill?
2. Smart use of EN funct1on to realize power-on sequence
In circuit design, chips or modules often need a variety of working power supplies, and the power-on sequence of these power supplies is also required. Failure to meet these power-up timing requirements can result in bus conflicts, device latch-up, and other faults.


Detailed explanation of the principle of voltage doubler circuit
Detailed explanation of the principle of voltage doubler circuit
Note: To understand a voltage doubler circuit, first think of the charged capacitor as a power source. It can be connected in series with the power supply, just like the principle of connecting ordinary batteries in series.
DC half-wave rectification voltage circuit 1) In the negative half cycle, i.e. A is negative and B is positive, D1 is turned on and D2 is turned off. The power supply charges the capacitor C1 through D1. Under ideal conditions, D1 can be regarded as a short circuit in this half cycle, and the capacitor C1 is charged to Vm. In the positive half cycle of its current path and the polarity of the capacitor C1, i.e. A is positive and B is negative, D1 is turned off and D2 is turned on. At this time, the voltage of the power supply in series with C1 is 2 Vm, so C2 is charged to the maximum value of 2 Vm.
It should be noted that: (1) In fact, the voltage of C2 cannot be charged to 2 Vm in a half cycle, and it must gradually approach 2 Vm after several cycles. For the convenience of explanation, the following circuit dession also assumes this.
(2) If the half-wave voltage doubler is used in a power supply without a transformer, we must connect C1 in series with a current limiting resistor to protect the diode fr0m the inrush current at the beginning of charging of the power supply.
(3) If there is a load connected in parallel to the output of the voltage doubler, as generally expected, the voltage on capacitor C2 will drop during the negative half cycle (at the 1nput) and then be recharged to 2 Vm during the positive half cycle as shown in the following figure. Therefore, the voltage waveform on the capacitor C2 is a half-wave signal filtered by the capacitor filter, and the voltage doubler circuit is called a half-wave voltage circuit.
(4) In the positive half cycle, the maximum reverse voltage borne by the diode D1 is 2 Vm, and in the negative half cycle, the maximum reverse voltage borne by the diode D2 is also 2 Vm. Therefore, a diode with PIV > 2 Vm should be se1ected in the circuit.
Imple DC voltage double circuit
n At the moment when the 1.5V is switched on, the 1.5V DC voltage charges C2 through the energy storage inductor coil L and R1. Since the voltage at both ends of the capacitor cannot change suddenly, the base voltage of VT1 is almost zero, so VT1 is turned on, so that VT2 is saturated and turned on. At this time, the current of L will gradually increase fr0m small to large, and L will convert electric energy into magnetic energy and store it. In this process, VD2 is cut off, Vo = 0V, and the voltage stabilizing circuit composed of VT3, R2, VD3 and VD1 does not work. ② When the current in L does not change any more, the base potential of VT1 also increases to the maximum.


Influence of Power Chip EN Pin on Motor Control Board
Influence of Power Chip EN Pin on Motor Control Board
Embedded hardware design will become the core technology of microelectronics in the 21st century. Three key technologies in SoC design and some research fields of mutual integration are described in detail, and the challenges and development trends of SoC design are prospected.
A motor control panel has a power recovery funct1on, and when there is no power battery, the rotation of the motor can continue to supply power to the control panel. However, the uneven rotation of the motor will produce rapidly fluctuating voltage, which will cause the power supply chip to output extremely unstable voltage, so that the post-stage equipment will be powered up and down frequently in a very short time, resulting in frequent loss of firmware or even burnout of the Bluetooth module on the board, and reducing the product performance. Later, the problem was solved perfectly by adjusting the relevant configuration of the EN pin of the power supply chip. Do you want to know what was done to him? What kind of great wisdom does the small earth contain?
I. Overview
EN means "enable". Different chips have different names, such as EA, RUN, etc. Their funct1ons are basically the same, that is, only when the pin is activated, the chip or module can output normally. For this funct1on, we can add some simple peripheral circuits to realize the funct1on of stabilizing the chip or outputting the power-on sequence. The en pin of some advanced power supply chips usually has hysteresis characteristics.
II. Application Skills
1. Using Dividing Resistor to Realize Stable Output of Power Supply Chip
For the power supply chip, we usually use a divider resistor to connect the EN signal to the 1nput pin of the power supply to prevent the voltage at the EN terminal fr0m exceeding its withstand voltage value. Under the condition of meeting the withstand voltage value, the voltage of EN pin should also be set in the "appropriate" range.
For example, as mentioned at the beginning of the article, the 24V power supply of a motor control board not only supplies power to the motor, but also outputs 12V to other circuits through DC/DC: MP2451. When there is no booster battery, the motor generates electricity to supply power to the control panel, but the rotation of the motor is not uniform, resulting in a large fluctuation of voltage, as shown in Figure 1 below. The yellow line is the reverse generation voltage of the motor, and the green line is the output voltage of MP2451.
The DC/DC output is enabled when the generated voltage (DC/DC 1nput voltage) VIN of the motor is about 6.2 V, and the 1nput voltage is less than the set 12 V output voltage


Working Principle of Push-pull Transformer Switching Power Supply with Rectifier Output
Working Principle of Push-pull Transformer Switching Power Supply with Rectifier Output
Working principle: The typical circuit of push-pull switching power supply is shown in Figure 1. It belongs to the double-ended conversion circuit, and the magnetic core of the high-frequency transformer works on both sides of the hysteresis loop. The circuit uses two switching tubes VT1 and VT2, and the two switching tubes are alternately turned on and off under the control of an external excitation square wave signal, so that a square wave voltage is obtained at the secondary side of a transformer T, and the square wave voltage is rectified and filtered to become a required direct current voltage. \n The advantage of this circuit is that the two switches are easy to drive, and the main disadvantage is that the withstand voltage of the switches must reach twice the peak voltage of the circuit. The output power of the circuit is large, generally in the range of 100-500 W.
As two switch tubes work alternately, that rectify output push-pull transformer switching power supply is equivalent to that two switch power supplies output power at the same time, and the output power is about twice of the output power of a single switching power supply. Therefore, the output power of the push-pull transformer switching power supply is very large, the working efficiency is very high, and after bridge rectification or full-wave rectification, the output voltage ripple of the push-pull transformer switching power supply can be very small only by a very small filter inductor and capacitor.
Except for the rectifier and filter circuit, the working principle of other circuits is basically the same as that in Fig. 1-27. The bridge rectifier circuit consists of D1, D2, D3 and D4, wherein C is an energy storage filter capacitor, R is a load resistor, Uo is a DC output voltage, and Io is a current flowing through the load resistor.
Similarly, except for the rectifier and filter circuit, the working principle of the rest of the circuit is basically the same as that of Figure 1-27 and Figure 1-30. However, the secondary of the switching transformer needs one more winding, and the two windings N31 and N32 output voltage in turn; The full-wave rectification circuit is composed of D1 and D2, C is an energy storage filter capacitor, R is a load resistor, Uo is a DC output voltage, and Io is a current flowing through the load resistor.
The push-pull transformer switching power supply with the bridge rectifier output uses two more rectifier diodes than the push-pull transformer switching power supply with the full-wave rectifier output, but the switching transformer with the full-wave rectifier output has one more set of secondary coils than the switching transformer with the bridge rectifier output. Therefore, the bridge rectifier output push-pull transformer switching power supply shown in Figure 1-30 is more suitable for the case of relatively small output current


reliability and safety of DC-DC converter
Nowadays, the circuit is more and more in pursuit of reliability and safety. Many circuits are equipped with overvoltage and overcurrent detection circuits to protect the circuit. The general control method for circuit overcurrent protection is shutdown or current limiting. The overcurrent circuit generally uses a fuse for current limiting protection or adopts a sampling resistor to obtain a circuit signal. When the circuit is too large, the subsequent circuit is turned off or the current is limited to a specific value. When the current is normal, the circuit works normally.
When a short circuit occurs in a line, one of the important characteristics is that the current in the line increases sharply, which requires a corresponding protection device to act in response to the current increase when the current flows through a predetermined value, which is called overcurrent protection.
The protection funct1ons of the power supply are mainly overvoltage and overcurrent protection.
The relationship between the two is:
When any kind of power supply fails, the output voltage or output current may be out of control. In order to prevent the user's load fr0m being damaged, the power supply of our company is generally equipped with overvoltage and overcurrent protection. For some loads, such as resistive loads, when the power supply fails, the voltage on the load may rise sharply, and the current increase may not exceed the overcurrent protection value. In this case, over-voltage protection should be used. For example, when working at 50 V, the voltage protection value can be adjusted to 55 V. If the power supply fails, the power supply will automatically cut off the voltage output as long as the voltage rises to 55 V. When some loads are capacitive loads, because the electrolytic capacitors with large capacity are connected in parallel, when the power supply fails, the current may rise sharply, but the voltage rise is not obvious. At this time, the overcurrent protection component inside the power supply will start first, and the power supply will automatically cut off the output.
The overvoltage protection value has a potentiometer on the panel, which can be set manually. The overcurrent protection value can not be set manually, and it has been fixed in the machine, generally 1.2 to 1.5 times of the rated current. It should be noted that the overvoltage protection will start immediately and quickly, while the overcurrent protection has a delay of about one second. This is because when the power supply works normally, if the load of the power supply is suddenly short-circuited, the instantaneous current output by the power supply at this time is several times or tens of times the rated current value, which can be considered as a current impact, far exceeding the value of overcurrent protection, but at this time it is not expected that the overcurrent protection will work. It is hoped that after the short circuit is removed, the voltage will automatically return to normal. Therefore, in the design of overcurrent protection, it is necessary to avoid the current impact of sudden short circuit, and only consider that the duration of output overcurrent reaches a certain value before starting overcurrent protection.


Next Page