Total Device Dissipation RF-5 Board (1)P
T A =25 °C
Derate above 25°C1.8mW/°C
Thermal Resistance, Junction to AmbientR
Total Device DissipationP
Alumina Substrate, (2) T A = 25°C
Derate above 25°C2.4mW/°C
Thermal Resistance, Junction to AmbientR
Junction and Storage TemperatureT J , T
– 50Vdc
– 50 Vdc
– 3.0Vdc
– 50mAdc
D
θJA
D
θJA
stg
225mW
556°C/W
300mW
417°C/W
–55 to +150
MMBT5087LT1
3
1
2
CASE 318–08, STYLE 6
SOT– 23 (TO–236AB)
°C
ELECTRICAL CHARACTERISTICS (T
= 25°C unless otherwise noted)
A
CharacteristicSymbol MinM axUnit
OFF CHARACTERISTICS
Collector–Emitter Breakdown VoltageV
(I C = –1.0 mAdc, I B = 0)
Collector–Base Breakdown VoltageV
(I C = –100 µAdc, I E = 0)
Collector Cutoff CurrentI
(V
= –10 Vdc, I E= 0)—–10
CB
(V
= –35 Vdc, I E= 0)—–50
CB
1. FR–5 = 1.0 x 0.75 x 0.062 in.
2. Alumina = 0.4 x 0.3 x 0.024 in. 99.5% alumina.
(BR)CEO
(BR)CBO
CBO
– 50—Vdc
– 50—Vdc
n Adc
M17–1/6
Page 2
LESHAN RADIO COMPANY, LTD.
MMBT5087LT1
ELECTRICAL CHARACTERISTICS (T
= 25°C unless otherwise noted) (Continued)
A
Characteristic SymbolMinMa x Unit
ON CHARACTERISTICS
DC Current Gain h
(I C = –100µAdc, V
(I C = –1.0 mAdc, V
(I C = –10 mAdc, V
= –5.0 Vdc)250800
CE
= –5.0 Vdc)250––
CE
= –5.0 Vdc)250––
CE
Collector–Emitter Saturation Voltage
(I C = –10 mAdc, I B = –1.0 mAdc)
Base–Emitter Saturation Voltage
(I
= –10 mAdc, I B = –1.0 mAdc)
C
V
V
FE
CE(sat)
BE(sat)
––– 0.3V dc
––– 0.85Vdc
SMALL–SIGNAL CHARACTERISTICS
Current–Gain — Bandwidth Product
(I C = –500 µAdc, V CE= –5.0 Vdc, f = 20 MHz)
Output Capacitance
(V CB= –5.0 Vdc, I E = 0, f = 1.0 MHz)
Small–Signal Current Gainh
(I C= –1.0mAdc, V
= –5.0Vdc, f = 1.0 kHz)
CE
Noise FigureN FdB
(I C = –20 mAdc, V CE= –5.0 Vdc,Rs=10kΩ, f = 1.0 kHz)— 2.0
(I C = –100µAdc, V CE= –5.0 Vdc,Rs=3.0kΩ, f = 1.0 kHz)— 2.0
f
T
C
obo
fe
40 —MHz
—4.0pF
250900—
––
M17–2/6
Page 3
TYPICAL NOISE CHARACTERISTICS
(V CE = – 5.0 Vdc, T A = 25°C)
10
BANDWIDTH = 1.0 Hz
7.0
IC=10 µA
5.0
30µA
3.0
1.0mA
2.0
, NOISE VOLTAGE (nV)
n
e
1.0
102050 100200500 1.0k2.0k5.0k 10k
100µA
300µA
R S 0
LESHAN RADIO COMPANY, LTD.
MMBT5087LT1
10.0
~
~
7.0
5.0
3.0
2.0
IC=1.0mA
300µA
1.0
0.7
0.5
0.3
, NOISE CURRENT (pA)
n
0.2
I
0.1
102050 100200500 1.0k 2.0k5.0k 10k
100µA
30µA
10µA
BANDWIDTH = 1.0 Hz
~
8
R
~
S
f, FREQUENCY (Hz)
Figure 1. Noise Voltage
1.0M
500k
200k
100k
50k
20k
0.5 dB
10k
5.0k
2.0k
1.0k
500
, SOURCE RESIST ANCE ( Ω )
S
R
200
100
1020 3050 70 100200 300 500 700 1.0K
1.0 dB
I C , COLLECTOR CURRENT (µA)
BANDWIDTH = 1.0 Hz
Figure 3. Narrow Band, 100 Hz
1.0M
500k
200k
100k
50k
20k
10k
5.0k
2.0k
1.0k
500
, SOURCE RESIST ANCE ( Ω )
S
200
R
100
1020 3050 70 100200 300 500 700 1.0K
0.5dB
10 Hz to 15.7kHz
I C , COLLECTOR CURRENT (µA)
Figure 5. Wideband
NOISE FIGURE CONTOURS
(V CE = – 5.0 Vdc, T A = 25°C)
1.0M
500k
200k
100k
50k
20k
10k
5.0k
2.0k
2.0dB
3.0 dB
5.0 dB
1.0dB
2.0dB
3.0 dB
5.0 dB
1.0k
500
, SOURCE RESIST ANCE ( Ω )
S
R
200
100
1020 3050 70 100200 300 500 700 1.0K
Noise Figure is Defined as:
NF = 20 log 10 (
e n= Noise Voltage of the Transistor referred to the input. (Figure 3)
I
= Noise Current of the Transistor referred to the input. (Figure 4)
n
K = Boltzman’s Constant (1.38 x 10
T = Temperature of the Source Resistance (°K)
R s= Source Resistance ( Ω )
DUTY CYCLE, D = t
D CURVES APPLY FOR POWER
PULSE TRAIN SHOWN
READ TIME AT t 1 (SEE AN–569)
= r(t) • R
Z
2
θJA(t)
T
J(pk)
– T A = P
1
θJA
(pk) Z θJA(t)
t, TIME (ms)
Figure 14. Thermal Response
/ t
C
ob
2
M17–5/6
Page 6
4
10
V
= 30 V
CC
3
10
I
2
10
1
10
CEO
I
CBO
AND
0
10
–1
, COLLECTOR CURRENT (nA)
10
C
I
–2
10
–4–20+20+40+60+80 +100 +120 +140 +160
I
@ V
BE(off)
= 3.0 V
CEX
T J , JUNCTION TEMPERATURE (°C)
Figure 15. Typical Collector Leakage Current
LESHAN RADIO COMPANY, LTD.
MMBT5087LT1
DESIGN NOTE: USE OF THERMAL RESPONSE DAT A
A train of periodical power pulses can be represented by
the model as shown in Figure 16. Using the model and the
device thermal response the normalized effective transient
thermal resistance of Figure 14 was calculated for various
duty cycles.
To find Z
the steady state value R
Example:
Dissipating 2.0 watts peak under the following conditions:
t 1 = 1.0 ms, t 2 = 5.0 ms. (D = 0.2)
Using Figure 16 at a pulse width of 1.0 ms and D = 0.2, the
reading of r(t) is 0.22.
The peak rise in junction temperature is therefore
∆T = r(t) x P
For more information, see AN–569.
, multiply the value obtained from Figure 14 by
θJA(t)
(pk)
x R
.
θJA
= 0.22 x 2.0 x 200 = 88°C.
θJA
M17–6/6
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