ltc5598 Linear Technology Corporation, ltc5598 Datasheet - Page 5

• Manufacturer Part Number: ltc5598
• Description: 5mhz To 1600mhz High Linearity Direct Quadrature Modulator
• Manufacturer: Linear Technology Corporation

TYPICAL PERFORMANCE CHARACTERISTICS
0dBm single-ended, BBPI, BBMI, BBPQ, BBMQ common-mode DC voltage V
0.8V
f
NOISE
–20
–30
–40
–50
–60
–70
180
170
160
150
140
85
80
75
70
65
60
55
PP,DIFF
–40
= f
Supply Current vs Temperature
Output IP2 vs RF Frequency
Image Rejection vs LO Frequency
LO
, two-tone baseband input signal = 2MHz, 2.1MHz, 0.5V
– 6MHz; unless otherwise noted. (Note 11)
10
–15
10
5V, 25°C
5.25V, 25°C
4.5V, 25°C
5V, –40°C
5V, 85°C
LO FREQUENCY (MHz)
RF FREQUENCY (MHz)
TEMPERATURE (°C)
10
5.25V
4.5V
5.0V
100
100
35
5V, 25°C
5.25V, 25°C
4.5V, 25°C
5V, –40°C
5V, 85°C
60
1000
1000
5598 G01
5598 G04
5598 G07
85
–145
–150
–155
–160
–165
10
–1
–2
–3
–4
–5
Noise Floor vs RF Frequency
(No AC Baseband Input Signal)
8
6
4
2
0
Voltage Gain vs RF Frequency
Output 1dB Compression
vs RF Frequency
10
10
10
RF FREQUENCY (MHz)
RF FREQUENCY (MHz)
RF FREQUENCY (MHz)
PP,DIFF
100
100
5V, 25°C
5.25V, 25°C
4.5V, 25°C
5V, –40°C
5V, 85°C
100
each tone, I&Q 90° shifted (lower side-band selection);
5V, 25°C
5.25V, 25°C
4.5V, 25°C
5V, –40°C
5V, 85°C
(NOTE 3)
CMBB
5V, 25°C
5.25V, 25°C
4.5V, 25°C
5V, –40°C
5V, 85°C
1000
5598 G08
= 0.5V
1000
1000
V
5598 G02
5598 G05
CC
= 5V, EN = 5V, T
DC
, I&Q baseband input signal = 100kHz,
–10
–20
–30
–40
–50
–60
0
RF Two-Tone Power (Each Tone),
IM2 and IM3 vs RF Frequency
–40
–50
–60
–70
29
27
25
23
21
19
17
f
IM3
Output IP3 vs RF Frequency
LO Feedthrough to RF Output
vs LO Frequency
= f
10
LO
f
A
IM2
f
IM3
RF FREQUENCY (MHz)
– 2*f
= 25ºC, f
10
10
f
RF , EACH
= f
= f
5V, 25°C
5.25V, 25°C
4.5V, 25°C
5V, –40°C
5V, 85°C
BB1
LO
RF FREQUENCY (MHz)
LO FREQUENCY (MHz)
LO
– f
+ 2*f
+ f
100
= f
BB1
BB2
RF
LO
BB1
– f
LTC5598
– f
= f
100
100
BB2
+ f
BB1
LO
BB2
– f
1000
BB
5V, 25°C
5.25V, 25°C
4.5V, 25°C
5V, –40°C
5V, 85°C
5598 G09
, P
1000
1000
LO
5598 G03
5598 G06
–40
–50
–60
–70
–80
–90
–100
5
5598p
=