詳細信息
NTC熱敏電阻器系由特殊配置的金屬氧化物陶瓷材料制成，它可用來抑制高的突波電流。相對于受保護電路,熱敏電阻器具有較高的電阻。因此會抑制突波電流約1~2秒，在這一段時間內熱敏電阻的電阻將因溫度升高而下降，直至熱敏電阻兩端壓降到可被忽略的電阻值為止。如圖A以電源供應器為例, 在電源開的瞬間，電容器一般阻抗極低，橋式整流器通常承受很大的電流，熱敏電阻器特別使用于保護電源供應器。
NTC Thermistor devices are made of a specially formulated metal oxide ceramic material which is capable of suppressing high inrush current surges.
Thermistor devices, being of relatively high resistance, shall limit the inrush current for 1~2 seconds during which time the device decreased in resistance substantially to a point where its voltage drop is negligible .The devices are especially useful in power supplies (see Fig A) because of the extremely low impedance of the capacitor being charged, of which the bridge is ususlly subjected to an exceedingly high current surge at turnon point.
特征
●有效抑止突波電流。
●穩(wěn)定狀態(tài)下功率損耗極小(通常僅有1W或小于50W)。
●熱及電特性穩(wěn)定性高。
●寬廣的電性規(guī)格可供選擇。
FEATURES
●High inrush current restriction effect
●Small power loss in stationary state
●(Normally less than 50W power)
●High thermal and electrical stability.
●Wide selection of electrical performances
應用概述
如圖B所示，將一NTC熱敏電阻與一白熱燈絲串聯(lián)時，可以消除突波電流。若一只NTC熱敏電阻無法提供足夠之突流限制功能時，二只或更多的熱敏電阻可用于串聯(lián)電路上或供應電路的各個分路上(如圖A)。但要注意的是NTC熱敏電阻，不可并聯(lián)于電路上，因為其中一只NTC就可能會傳導幾乎所有的電流。熱敏電阻最好用于圖A所示AC電路的A1或A2處，或是DC電路D1或D2處。
在設計上，當電路剛被打開的瞬間，NTC熱敏電阻的阻值高于電路上所有白熱燈絲的總電阻值。當電流開始通過時，熱敏電阻隨時產(chǎn)生「自然」現(xiàn)象，并在1到2秒內，阻值會降到幾可忽略。以同樣的構想來看電動馬達的突波電流，亦可以被抑制到最低限度。圖C表示應用熱敏電阻保護直流馬達前后突波電流波形的差異。
APPLICATION
As shown in Fig.B, the current surge can be eliminated by placing a NTC thermistor in series with a filament string. Yet, if the resistance of one NTC thermistor does not provide sufficient inrush current limiting functions for your applicantion, 2 or more may be used in series or in separate legs of the supply circuit (Fig.A). Be noticed, the thermistor can not be used in parallel since one unit will tend to conduct nearly all the current available. Thus, thermistor may be used in the AC (point A1or A2)or the DC (point D1or D2) locations in the circuit. (see Fig.A)
The resistance of NTC thermistor is designed higher than the total resistance of NIC thermistor is designed higher than the total resistance of filaments when thermistor shall immediately self-heat. Then, in 1~2 seconds, its resistance will be reduced to a minimum and become insignificant to the total resistance of a circuit. With the same concept, current surges in electric motors can be held to minimum. Fig.C shows a typical DC motor s turn on surge before and after the application of a thermistor to the circuit
選用原則
1.熱敏電阻器的最大工作電流＞實際電源回路的工作電流
2.熱敏電阻器的標稱電阻值
R≥ 式中E為線路電壓 l m為浪涌電流
對于轉換電源、逆變電源、開關電源、
UPS電源l m=100倍工作電流
對于燈絲、加熱器等回路l m=30倍工作電流
1.Maximum operating current >Actual operating current in the power loop
2.Reted zero power resistance at 25℃
of which, E: loop voltage, lm: Surge current.
For conversion power, reversion power, switch power,
UPS power, lm=100 times. operating current
For filament, heater, lm=30 times operating current
電壓電流特性
當NTCR熱敏電阻在小電流下工作時（如圖F），由于功率太底，其電阻保持固定而表現(xiàn)線性關系（符合歐姆定律V/R=1）。如果電流增加，NTC熱敏電阻就會產(chǎn)生焦耳效應(P =V×1)而使自己發(fā)熱，其電阻值隨即減小表現(xiàn)「電流增加,電壓下降」的狀態(tài)。
When operating low current (see fig.F), due to very low power is unable to make the NTC thermistor self-heated, so its resistance value is thus maintained constant and displayed with a linear curve (in conformity with ohm-law V/R=1).If the current is increased, the NTC thermistor will follow Joule-efficiency(P= V×1) and make itself self-heated that results in a resistace value decreasing and thus displays with a status of voltage descending while current increased.
(如圖F)，說明NTC元件與環(huán)境達成熱平衡所需的時間，主要決定于材料熱容量（G）及散熱系數(shù)（δ）。當元件溫度由T1降到T0，則可得到下列平衡式：
－HdT=δ(T－T0)dt 其中－HdT=元件熱損失
δ(T－T0)dt:元件散熱量
積分后可得溫度與時間關系式T－T1= (T0－T1)×e－t/t
其中τ=H/δ
As shown in fig.G which explains the time needed to reach thermal equilibrium of NTC components with the enviroment. This characteristic depends on two important parameters. If a step change in temperature is applied to a component e.g. form high (T1) to low (T0) temperature, the energy lost (δ(T0－T1)dt) by the component (－HdT) is equal to the energy dissipated by it.
－HdT=δ(T－T0)dt
This equation yields: T－T1= (T0－T1)×e－t/t
τ=H/δ
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