业界超薄超级电容-CapX

更新时间:2023-05-10 14:36:22 阅读: 评论:0

HW207 / HW107 SUPERCAPACITOR
Datasheet Rev 2.0
Features
High capacitance  (450mF @ DC)
Low ESR  (100m Ω @ step change in current);  High peak current    High puld power
Thin form factor
Typical Applications
High power LED Flash
Improved audio performance  Automatic Meter Reading
PC Cards, Compact Flash Cards & USB  Load levelling for PDAs & cell phones
Power support during battery contact bounce
Electrical Specifications
Table 1: Nominal Characteristics
Device Nominal Capacitance 1
Nominal ESR 2 Tolerance about nominal value
Footprint Height Weight HW207 450mF 100m Ω ±20% 28mm x 17.5mm    3.0mm    1.5 gm HW107 900mF 50m Ω
±20%
28mm x 17.5mm
1.45mm
0.753 gm
1
At 23ºC DC. 2
Measured using a 0.5A step in current @ 23ºC. 3
To the nearest 50mg
Table 2: Absolute Maximum Ratings
Parameter Name Conditions Min Max Units Terminal
Voltage
Vc      5.8 V
Temperature T  -40 +85 °C
Table 3: Electrical Characteristics
Parameter Name Conditions Min Typical Max Units Terminal
Voltage Vc    5.5 V
Leakage Current 3
I L    5.5V, 23°C
72hrs    1.7 5 µA RMS
Current 4 I RMS 23°C  3.6 A
Peak
Current 5 I P 23°C  57 A
3After 72hrs @ 4.5V at 23°C. 4Continuous charge and discharge for 2min operation. 5
Single pul, non repetitive current.
Definition of Terms
In its simplest form, the Equivalent Series Resistance (ESR) of a capacitor is the real part of the complex impedance. In the time domain it can be found by applying a step discharge current to a charged capacitor as in figure 1. In this figure the supercapacitor is pre-charged and then discharged with a constant current pul (I ). The ESR is found by dividing the instantaneous voltage step (∆V after 50µc from start of current pul) by I. The instantaneous capacitance (C i ) can be found by taking the inver of the derivative of the voltage and multiplying it by I . The effective capacitance (C e ) is found by dividing the total charge removed from the capacitor (∆Q n ) by the voltage lost by the capacitor (∆V n ). Note that ∆V , or IR drop, is not included becau this is the voltage drop due to ESR. C e  shows the time respon of the capacitor and it is uful for predicting circuit behaviour in puld applications.
In the example of Fig 1, using an HW207, ΔV = 5.48V – 5.34V = 0.14V, I = 2.05A, so ESR = 0.14V/2.05A = 68.3m Ω. Similarly for C effective  at 100µc ΔV n  = 5.34V – 5.32V = 0.02V, Δt n  = 100µc, and I = 1.97A. Therefore, C = 1.97A X 100*10-6s/0.02V = 9.85mF.
55.1
5.2
5.35.45.5
5.6
-100
-50
50100
150
200
Time (µc)
V o l t a g e  (V )
-0.20.5
1.2
1.9
2.6
3.3
4
C u r r e n t  (A )
Figure 1: Definitions for Effective Capacitance, Instantaneous Capacitance and ESR
DC Capacitance
CAP-XX measures DC capacitance by charging the supercapacitor to 5.5V then disconnecting the supercapacitor from the source, and applying a constant current discharge of 100mA. At Cap-XX, capacitance is measure using the time taken for the voltage to drop from 3V to 1V, so C = 100mA x (time taken to drop from 3V to 1V) / 2V.
In the example of Fig 2, for a ∆V n  = 3.0V – 1.0V = 2V, the corresponding ∆t c  = 9.25 – 1.69s = 7.65s. C = I X ∆t c /∆V c  where I = 0.105A, therefore C = 0.105x7.56s /2.0V = 402mF.
Figure 2: Measurement of Capacitance
ESR Measurement
CAP-XX measures ESR by measuring the voltage drop across the supercapacitor when a current st
ep is applied to a supercapacitor. The supercapacitor is first charged to 5.5V then disconnected from the source, and finally the current step applied and the voltage drop after 50µc is measured. The 50µc delay allows time for the current pul to ttle before the measurement is made.
In the example shown in Fig 3 below ∆V = 5.47V – 5.28V = 190mV and ∆I = 2.05A (load pul), therefore ESR = ∆V/I = 92.7m Ω
.
55.15.2
5.35.45.55.6
-100
-50050100150
200
Time (µc)
V o l t a g e  (V )
-0.20.5
1.2
1.9
2.6
3.3
4
C u r r e n t  (A )
Figure 3: Measurement of ESR
Effective Capacitance
Figure 4 shows the Effective Capacitance for the HW207 @ 23°C. The supercapacitor was charged to and held at 5.5V until the current drawn by the supercapacitor dropped to less than 1mA. The supercapacitor was then disconnected from the source and a constant current discharge of 100mA was applied. The capacitance was measured at different times during the discharge.
20
40
60
80
100
1201E-04
1E-031E-02
1E-011E+001E+01
N o r m a l i s e d  C a p a c i t a n c e  % o f  D C  C a p a c i t a n c e
Time (c)
Normalid Capacitance vs Time @ 23°C
Figure 4: Effective capacitance at different times during the discharge.
Pul Respon
Figure 5 shows the voltage ripple for a class 10 GPRS pul. A HW207 provides a 1.8A load pul of
1.15ms duration @ 25% duty cycle and the source current is limited to 600mA, though there is some source current overshoot evident in the first 100μs. The low supercapacitor ESR and high effective capacitance result in the load eing a voltage ripple of only 330mV. The 1.8A load current would consist of 0.6A current from the supply and the remaining 1.2A from the supercapacitor.
GPRS Pul at 23C
0123
456
-1
1
2
3
45
6
7
8
9
Time (mc)
V o l t a g e  (V )
0.4
0.8
1.2
1.6
2
2.4
C u r r e n t  (A
)
Figure 5: Class 10 GPRS pul at 23C
Capacitance and ESR with temperature
Figure 6 and 7 below show normalized ESR and Capacitance respectively at different operating temperatures ranging from -40C to 85C.
Figure 6: Normalid ESR at different temperatures
Figure 7: Normalid capacitance at different temperature

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