[Date: 2006- 1 1-27] Source: Electric Power Technology Application Author: Zhou [font: large, medium and small]
Abstract: the half-bridge inverter input circuit of electronic ballast determines the working frequency of the whole machine, which is an important factor affecting the inverter of switching power tube. Through the analysis of pulse transformer and ideal excitation current waveform, this paper discusses how to improve conversion efficiency and suppress oscillation, points out the existing design misunderstandings, and puts forward a new design scheme to reduce * * * state conduction and switching loss.
Keywords: conversion rate; Damping oscillation; * * * state conduction; Switching loss; New design
Introduction to 0
As we all know, the half-bridge inverter input circuit of electronic ballast is extremely important, which is directly related to the working frequency, switching loss, conversion efficiency and output power of the whole machine. At the same time, it also has a certain impact on EMC, THD, PF and other major technical indicators. At present, the practical technical scheme is that bipolar transistors are connected in half bridge, and the pulse transformer composed of magnetic rings generates self-excited oscillation by feedback, and outputs high-frequency pulse current for photoelectric conversion of fluorescent tubes. Due to the memory effect of the transistor base, the turn-off time is delayed; The capacitance of the collector junction makes the output pulse current have a bad influence on the input terminal; Especially for the pulse transformer in the input circuit, when the half-bridge inverter circuit works, the switching current will ring in its primary winding, and its positive peak value will work together with the positive charge stored in the base region, which will make the switching tube turn off late or turn on repeatedly, resulting in "* * * state conduction", which will increase the switching loss, make the transistor heat up and enter the secondary breakdown when it is turned on again, and any protection circuit can do nothing about it.
Therefore, it is necessary to study the ideal excitation conditions for the quick turn-on and complete turn-off of the switch tube, and design the best scheme of the electronic ballast half-bridge inverter input circuit composed of bipolar transistors.
1 Excitation condition of fast turn-on of switch tube
The current amplification factor β of bipolar transistor is not a constant, but decreases with the increase of signal frequency. When the step current Ib is injected into the base, the rise of collector current ic is nonlinear with time, that is
ic(t)=βoIb( 1-)( 1)
Where: βo is the β value at low frequency;
Ω o = 2π f β (f β is the cutoff frequency of the transistor emitter circuit).
If Tce= 1/ωo,
Then IC (t) = β OIB (1-) (2)
Equation (2) shows that when the step current Ib is injected into the base, the collector current ic gradually rises to the maximum value βoIb according to the exponential law. If the conduction time tk is the time required for the collector current ic to rise from zero to 90% of the maximum value βoIb, the above formula can be rewritten as follows
0.9βoIb=βoIb( 1-)(3)
Tk= 2.3Tce(4) of solution (3)
According to equation (4), figure 1 shows the influence of base step current Ib on ic.
(a) base phase current Ib (b) collector current ic
The influence of figure 1 Ib on ic
For example, transistors MJE 13005 commonly used in electronic ballasts, βo=20, fT=4MHz.
Then fβ=fT/β=4MHz/20=200kHz,
tce = 1/ωo = 1/2πfβ≈0.80μs，
Therefore, tk = 2.3tce ≈1.84μ s.
The single pulse width of the electronic ballast operating at about 30kHz is 16.7μs, and the on time accounts for 1 1%, which is relatively long. The longer the conduction time, the higher the switching loss. If the injected step current ib is increased to n times the critical saturation current Ib, the opening time tk can be shortened accordingly, that is
βoIb=βoNIb( 1-)(5)
Tk=Tceln(6) is used to solve equation (6).
Fig. 2 shows the influence of over-excitation on the opening time, from which it can be seen that: n >；; 2. The decrease of tk is not obvious. On the other hand, if the excitation is too large, it will cause the switch tube to be deeply saturated and consume too much excitation power. For the electronic ballast inverter circuit, it is appropriate to take N=2, that is,
ic≥(7)
Where: Icp is the peak value of collector current when the switch tube works.
(a) Influence of multiple n of excitation current (b)
Fig. 2 Influence of Overexcitation on Opening Time
Take MJE 13005 tube as an example. If the over-excitation multiple N=2, Tk = tceln = 0.8 μ s× LN2 ≈ 0.8 μ s× 0.7 = 0.56 μ s, which is similar to the curve shown in Figure 2, it is only 3/ 10 of the original breaking time.
2 switching tube fast cut-off excitation conditions
The physical process of switching tube from on to off is basically the same as that when it is on. Due to the base storage effect and collector junction capacitance, the collector current is kept at the saturation value Ic=βoIb when conducting. When the base injection current Ib suddenly drops to zero, ic does not suddenly drop to zero, but decreases exponentially:
ic=Ic(8)
If the falling time tx is defined as the time required for ic to fall to110 of βoIb, that is
0. 1βoIb=βoIb(9)
Tx of solution (9) ≈ 2.3tce (10)
The longer the falling time tx, the greater the switching loss. When the circuit works in the half-bridge inverter state, at the moment when one switch tube has not been completely turned off and the other switch tube has started to turn on, the DC loop is in a short circuit state, and the ic peak value is amazing. This "* * * state conduction" is an important reason for the secondary breakdown of the switch tube. When selecting the switch tube, the best scheme to shorten the falling time tx is to inject reverse current into the base and completely neutralize a large number of positive charges stored in the base region in a very short time, so as to achieve the purpose of fast turn-off. When the base reverse excitation current increases to -n'IB', the current ic decreases to-β on 'IB (this formula is only for illustration, because VCE >: 0, ic does not actually appear negative). Under the action of reverse excitation current, the time for ic to drop from Ic to zero is tx, then
0 =βoIb( 1+N’)( 1 1)
Tx of the solution = tceln (11).
According to the formula (12), the relationship between the back excitation multiple n' and the cut-off time in Figure 3 is obtained. As shown in Figure 3, when n' >:3 years later, the effect is not significant. In engineering, ib≥3Ib= is generally taken.
(a) Relationship curve of base injection reverse current (b) Cut-off time tx(N)
Fig. 3 Relationship between reverse excitation multiple and cut-off time
Take MJE 13005 as an example, when tx=2.3Tce≈ 1.84μs has no reverse overexcitation current, and n ′ = 3.
tx=Tceln=0.8μsln=0.224μs
Obviously, the falling time tx can be shortened from 1.84μs to 0.224μs by increasing Ib reverse overdrive by 3 times on the switch base. Therefore, due to the memory effect of bipolar transistor switching tube, the ideal excitation current waveform of base drive is shown in Figure 4.
In fig. 4, T 1-T2 is the base injection current ib at the moment when the switch tube is turned on, and the peak value is 2Ib, which is helpful to turn on quickly, shorten the transition period and reduce the switching loss.
Fig. 4 ideal excitation current waveform of base drive
T2-T3 is the time that the switch tube remains conductive. At this time, Ib should be as small as possible to avoid deep saturation, which is conducive to reducing the storage time of the switch tube.
T3-T4 is the moment when the switch tube is turned off, and its peak reverse current reaches 3 times of Ib value, which increases its base reverse current, thus reducing the storage time and falling time.
3 pulse transformer working state analysis
Electronic ballasts usually use ring pulse transformers with ferrite cores as driving elements. Because the excitation voltage pulse is a square wave, its flat top part contains many low-frequency components, while the leading edge and trailing edge of the pulse contain many high-frequency components. In this way, the requirements for pulse transformers are strict, requiring sufficient mutual inductance, small leakage inductance and small distributed capacitance. Therefore, R2K magnetic ring with rectangular hysteresis loop, low hysteresis loss and high saturation magnetic induction intensity Bs is the best, and its shape is φ 10 mm× 6 mm× 5 mm ... Its magnetic circuit is closed and its magnetic leakage is very small. In this process, parasitic parameters should be minimized.
Fig. 5 shows the structure and equivalent circuit of the pulse transformer. The equivalent circuit of four-terminal network can be obtained by Laplace transform method. In fig. 5, winding 1 is the primary winding and winding 2 is the secondary winding. Rs is the internal resistance of the signal source, LP is the leakage inductance, LM is the magnetizing inductance, C is the distributed capacitance, and RL' is the resistance value of the base circuit of the switch tube.
(a) structural diagram (b) equivalent circuit
Fig. 5 Pulse transformer
When choosing the lamp tube, ballast inductor and starting capacitor of electronic ballast, its oscillation frequency mainly depends on the base circuit of switch tube, the material, geometric size and number of turns of primary winding of pulse transformer. The engineering oscillation frequency f can be derived from equation (13).
f=( 13)
Where: Vs is the driving voltage of the primary winding;
N is the number of turns of the primary winding;
βs is the saturated magnetic flux density of the magnetic core;
S is the effective cross-sectional area of the magnetic ring;
K is the coefficient, and the rectangular wave is 4.0.
The effective cross-sectional area s of the magnetic ring described in this paper is
s = h =×5≈ 10 mm2 = 0. 1 cm2
Let Vs be 2.5V, βs=0.45T, and n take 4 turns and substitute it into the formula (13).
F = = = = 34.72kHz
The above calculated values are only for the reference of engineers and technicians when designing. In actual debugging, due to the influence of the ts value of the switch tube, the impedance of the base input loop and the compensation capacitor value connected in parallel with the switch tube, its working frequency is slightly deviated.
Fig. 5 shows the structure and equivalent circuit of the pulse transformer. The equivalent circuit of four-terminal network can be obtained by Laplace transform method. In fig. 5, winding 1 is the primary winding and winding 2 is the secondary winding. Rs is the internal resistance of the signal source, LP is the leakage inductance, LM is the magnetizing inductance, C is the distributed capacitance, and RL' is the resistance value of the base circuit of the switch tube.
(a) structural diagram (b) equivalent circuit
Fig. 5 Pulse transformer
When choosing the lamp tube, ballast inductor and starting capacitor of electronic ballast, its oscillation frequency mainly depends on the base circuit of switch tube, the material, geometric size and number of turns of primary winding of pulse transformer. The engineering oscillation frequency f can be derived from equation (13).
f=( 13)
Where: Vs is the driving voltage of the primary winding;
N is the number of turns of the primary winding;
βs is the saturated magnetic flux density of the magnetic core;
S is the effective cross-sectional area of the magnetic ring;
K is the coefficient, and the rectangular wave is 4.0.
The effective cross-sectional area s of the magnetic ring described in this paper is
s = h =×5≈ 10 mm2 = 0. 1 cm2
Let Vs be 2.5V, βs=0.45T, and n take 4 turns and substitute it into the formula (13).
F = = = = 34.72kHz
The above calculated values are only for the reference of engineers and technicians when designing. In actual debugging, due to the influence of the ts value of the switch tube, the impedance of the base input loop and the compensation capacitor value connected in parallel with the switch tube, its working frequency is slightly deviated.
It should be noted that the output voltage waveform of the half-bridge inverter circuit is a typical square wave, and the rising edge and falling edge of the current flowing through the pulse transformer winding will produce ringing phenomenon, and the waveform will be distorted, as shown in Figure 6.
Fig. 6 Ringing phenomenon caused by rising and falling edges of current
For electronic ballasts, the ringing current at the falling edge is large and harmful. The fundamental reason of ringing current is the overshoot of the rising edge and falling edge of rectangular pulse. Because the leading edge and trailing edge of the pulse contain rich high-frequency components, the higher the frequency, the greater the inductance ω LM value of LM, and when the equivalent impedance is large enough, oscillation will occur here. The intensity of oscillation is related to the equivalent impedance of the basic circuit. The damping coefficient can be expressed by equation (14).
δ=( 14)
According to equation (14), three kinds of damping curves shown in Figure 7 can be drawn.
Fig. 7 damping characteristics at different δ.
Take critical damping δ= 1.
When δ >; 1 is over-damped, and the transition between the upper and lower edges of the waveform is slow, which leads to the delay of the switch tube entering the amplification area, the increase of loss and the heating of the switch tube.
When δ
Some technicians reported that when the circuit loss was adjusted below110 of the power consumed by fluorescent tubes, the MJE 13005 radiator could not feel the temperature rise and worked normally all the time. It is puzzling that the lamp tube is replaced or the power supply voltage changes slightly, and it will be broken down at the moment of starting. This phenomenon changes from δ
Input circuit design
4. 1 Design of pulse transformer
In electronic ballast, pulse transformer, like the heart of human body, is the key to determine the working efficiency and reliability of the circuit.
The first is the choice of materials. In order to realize the ideal driving base current waveform, it is required that the initial magnetic permeability μi and saturation magnetic flux density BS of the magnetic core are high, while the smaller the remanence Br and coercivity HO, the better the current conversion. It has become the common knowledge of engineers and technicians that Curie temperature TC and reluctance Rm are high, and the circuit has good working stability and low loss. Choosing domestic RM2KD ferrite material can generally meet the requirements. For φ 10 mm× 6 mm× 5 mm magnetic ring, μi=2500, TC=220℃ and BS=0.45T are measured.
Secondly, the determination of primary winding of pulse transformer. Generally speaking, the inductance LM of the primary winding is calculated first.
LM = tuRL′/δ( 15)
Where: tu is the pulse duration;
In delta engineering, 0.8 is usually taken as the distortion coefficient of pulse top drop.
According to the formula (16), the number of turns n of the primary winding can be estimated.
N=( 16)
Where: L is the average magnetic circuit length of the magnetic ring;
μ△ is the magnetic permeability of the iron core;
S is the cross-sectional area of the magnetic ring.
In the design, RL ′ varies with the primary turns ratio, the current limiting resistor Rb in series with the base of the switch tube and the emitter resistance of the switch tube, and this equivalent impedance should be adjusted appropriately.
4.2 Design of Base Input Circuit
After the parameters of pulse transformer are determined, the design of switching tube input circuit is very important. The basis of design is to meet the ideal excitation current waveform as much as possible. According to the actual situation of market competition and process requirements, combined with the inherent characteristics of pulse transformer, it is required that the circuit structure is simple, the performance is stable, the consistency is good and the practicability is strong.
One scheme: reverse diode damping circuit, as shown in Figure 8.
Fig. 8 Reverse diode damping circuit
In fig. 8, the fast recovery diode D is connected in anti-parallel with the base current limiting resistor Rb of the switching tube, and the anti-vibration capacitor C is also connected in parallel between the base and the ground. For the pulse rising edge and flat top section, I b+ is slightly larger than Ib due to the negative connection of D and the current limiting effect of Rb. By properly adjusting the number of turns of the secondary winding, the switch tube can be quickly turned on without entering the supersaturated state. When the falling edge of the pulse comes, D is turned on with positive bias, and the negative overshoot current output by the pulse transformer is smoothly injected into the base of the switch tube, which quickly neutralizes the positive charge stored in the base region at the rate of DIB-/DT and instantly enters the off state. The function of C is to further eliminate the ringing current generated by the rising edge and falling edge of the pulse, so that the switch tube can work safely.
Scheme 2: RC parallel damping circuit, as shown in Figure 9.
Fig. 9 RC parallel damping circuit
In fig. 9, RC is connected in parallel between the base of the switch tube and the ground. The connection of damping resistor R makes the anti-vibration capacitor C have a discharge circuit, which enhances the damping and vibration reduction effect. At the same time, the resistance of R is generally close to the input impedance of the switch tube, about 33 ~100Ω, especially when the switch tube is turned off, the damping effect of R on the pulse transformer is obviously enhanced. Another advantage is that the access of R further improves the Vcer value of the switch tube, especially for the high-power electronic ballast, its reliability is significantly improved. The circuit has simple structure and low cost, but it is very practical. Used with pulse transformer, the debugging is normal and the work is stable and reliable. Designed in lamps and lanterns, even if there is no lamp open circuit protection circuit, it will not cause damage to electronic ballasts.
Scheme 3: On the basis of the above two schemes, a 20 ~ 50 μ h inductor is connected in series in the base circuit of the switch tube, which presents great impedance and attenuation to the high-frequency ringing current. The biggest advantage of this scheme is that it is simple to debug and the effect is twice the result with half the effort.
The author recently published the invention patent "electronic ballast with dual power factor correction and low peak ratio", which is equipped with dual power factor correction and low peak ratio circuits, high-efficiency filament preheating and abnormal state protection circuits, and its input circuit adopts RC parallel damping circuit. Therefore, the power factor reaches 0.99, the total harmonic distortion THD is less than or equal to 12%, and the crest ratio CF is less than or equal to 1.65. EMC technical indicators of electromagnetic compatibility meet the relevant provisions of IEC. It also has no power supply (PTC) preheating start, which further improves the circuit efficiency. The average service life of lamps and lanterns switch is more than 10000 times.
5 design misunderstanding
5. One of the design misunderstandings of1
In order to improve the switching characteristics of bipolar transistors, a small capacitor of 1000 ~ 3300 pf is often connected in series in the base drive circuit. Using the principle that the voltage across the capacitor can't change suddenly, it can provide a large driving current instantly, which not only accelerates the turn-on, but also accelerates the turn-off. However, it is not suitable to inject the ringing current caused by the rising or falling overshoot in the pulse transformer winding into the base of the switch tube at the same time. This circuit is only suitable for IC drive, not for pulse transformer drive.
5.2 The second design misunderstanding
In the common ballast circuit in China, a diode is connected in reverse parallel between the base and emitter of the switch tube, which is beneficial to conduction and eliminates part of the ringing current. But it also eliminates most of the reverse current, which is not conducive to the accelerated turn-off of the switch tube, and the ideal base driving current waveform can not be obtained.
The ideal design is to connect a number of forward and reverse diodes in series and parallel between the base and collector of the switch tube to form an "anti-saturation circuit", such as a "Becker clamp circuit". Although the deep saturation during conduction is eliminated and the storage time is shortened, the reverse excitation current is also reduced. But it needs 3 ~ 4 fast recovery diodes, and its circuit structure is complex, so it is rarely used in electronic ballasts.
6 conclusion
The difficulty of electronic ballast design lies in the current storage effect of bipolar transistor base and the overshoot ringing current of pulse transformer rising edge and falling edge. Thirdly, the negative resistance characteristic of gas discharge lamp makes the load circuit of ballast must be inductive, which makes the design worse. Through the analysis of the half-bridge inverter input circuit of electronic ballast, the reverse diode damping circuit and RC parallel damping circuit are practical and basically solve the above difficulties. The application of this technology in invention patents has been further verified. Due to the limited space, this paper does not involve the influence of zero-current switching compensation capacitor, inductive load in output circuit and discharge lamp on base circuit.