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参数资料
| 型号： | BZ05FB682ZSB |
| 厂商： | AVX Corporation |
| 文件页数： | 21/26页 |
| 文件大小： | 0K |
| 描述： | CAP SUPER 6.8MF 15V BZ05 SMD |
| 标准包装： | 100 |
| 系列： | BestCap® BZ |
| 电容： | 6.8mF |
| 电压 - 额定： | 15V |
| 容差： | -20%，+80% |
| ESR（等效串联电阻）： | 500 毫欧 |
| 寿命@温度： | 70°C 时为 1000 小时 |
| 安装类型： | 表面贴装 |
| 封装/外壳： | BZ05，3 引线 |
| 包装： | 散装 |
| 尺寸/尺寸： | 0.787" L x 0.591" W（20.00mm x 15.00mm） |
| 高度 - 座高（最大）： | 0.228"（5.80mm） |
| 工作温度： | -20°C ~ 70°C |
| 其它名称： | 478-6277 |
BestCap Ultra-low ESR
?
High Power Pulse Supercapacitors
4
3.5
3
Cutoff Voltage Limits
PULSE CAPACITOR APPLICATIONS
As mentioned earlier, the voltage drop in a circuit is critical
as the circuit will not operate below a certain cut-off voltage.
There are two sources of voltage drop ( Δ V) which occur, the
first Δ V ESR is because of the equivalent series resistance
(ESR) and the second, called the capacitive drop, is Δ V C .
From Ohm’s law,
Cutoff Voltage
% Increase
voltage = current x resistance or V = IR
2.5
2
3.4 Volts
3.5 Volts
3.6 Volts
28%
73%
300%
Let us say that the instantaneous starting voltage is Vo, or
voltage for the circuit from where the voltage drops. If the
capacitor has an ESR of 100 milliOhms and the current is 1
amp,
0
100
200
Time (Minutes)
300
400
Δ V ESR = 1 amp x (0.100) ohms = 0.1 volts or 100 milli-volts.
Battery with Pulse Capacitor
Battery Alone
On demand, during the discharge mode, the voltage V = Vo
- Δ V ESR = (Vo - 0.1) volts
GSM Pulse @ 2 Amps
Figure 6a. Li-ION Battery at +25°C
The second voltage drop is because of the capacitance.
This is shown in the equation as a linear function because of
simplicity. Simply put,
4
3.5
LI-ION Battery
Q (charge) = C (capacitance) x V (voltage)
The derivative, dQ/dt = I (current, in amps) = C x dV/dt
Hence, Δ V C (dV, the voltage drop because of capacitance) =
I x dt/C. This formula states that the larger the capacitance
value the lower the voltage drop. Compared to a Ta capacitor
3
2.5
Cutoff Voltage
3.4 Volts
3.5 Volts
3.6 Volts
% Increase
28%
100%
300%
this Δ V C is reduced by a factor of about 10 to 100. So,
BestCap ? has an advantage where higher capacitance is
needed. If the current pulse itself is 1 amp, the current pulse
width is 1 second, and the capacitance is 10 millifarads, the
Δ V C = 1A x 1Sec/0.01F, or a 100 volts; such an application
is out of the range of BestCap ? . However, if the pulse width
2
0
100
200 300
Time (Minutes)
400
500
becomes narrower, say 10 milli-seconds, and the capaci-
tance is 100 millifarads, the Δ V C = 1 x (10/1000)/(100/1000)
Battery with Pulse Capacitor
Battery Alone
= 0.1 volt or 100 milli-volts. This shows the advantage of the
large capacitance and hence the term “pulse” capacitor.
GSM Pulse @ 2 Amps 0°C
Figure 6b. Li-ION Battery at +0°C
The specific power – specific energy graphs are used in the
battery industry to compare competitive products. As the dt
becomes smaller i.e.100 milliseconds, 10 milliseconds and
then 1 millisecond, our estimates show that the specific
power for the BestCap ? is the highest as compared to our
competitors because of our choice of internal materials
chemistry.
Conclusion: we now clearly show that BestCap ? has an
advantage over competitors for short current pulse whose
widths are smaller than a few hundred milliseconds.
20
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