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KH600AI Datasheet(PDF) 9 Page - Cadeka Microcircuits LLC. |
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KH600AI Datasheet(HTML) 9 Page - Cadeka Microcircuits LLC. |
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9 / 13 page ![]() KH600 DATA SHEET REV. 2 January 2004 9 Single-to-Differential Operation The KH600 is specifically designed for differential-to- differential operation. However, the KH600 can be used in a single-to-differential configuration with some performance degradation. The unused input should be terminated into 50 Ω. When driven single-ended, there will be a slight imbalance in the differential output voltages, see Figure 5. This imbalance is approximately 2.88dB. To compensate for this imbalance, attenuate the higher gain output. (If the signal is applied to IN1, attenuate OUT1.) Unused Inputs and/or Outputs For optimal performance, terminate any unused inputs and/or outputs with 50 Ω. Adjusting Supply Current The KH600 operates class A, so maximum output current is directly proportional to supply current. Adjusting the voltages on +Vb1 and +Vb2 in opposition to -Vb controls supply current. The default supply current of the KH600 has been optimized for best bandwidth and distortion performance. The main reason for adjusting supply current is to either reduce power or increase maximum output current. Adjusting the supply current will not significantly improve bandwidth or distortion and may actually degrade them. To adjust the supply current, apply voltages of equal magnitude, but opposite polarity, to the bias voltage pins. For example, setting +Vb1, +Vb2 to +5VDC and -Vb to -5VDC (as shown in Figure 3) results in the standard supply current condition. Setting +Vb1, +Vb2 to +5.5V and -Vb to -5.5V results in an approximate 10% increase in supply current. Figure 10 shows the how the total supply current of the KH600 is effected by changes in the bias voltages (Vb = +Vb1 = +Vb2 = |-Vb|). Figure 10: Total Supply Current vs. Vb Supply current is relatively independent of the voltages on +Vs and -Vs as shown in Figure 11. Figure 11: Total Supply Current vs. Vs Figure 12: -3dB Bandwidth vs. Is Figure 13: Harmonic Distortion vs. Total Is 160 140 120 100 80 0 Vb (V) 02 6 8 4 60 40 20 ±V s = ±5V pply Current vs. ± 100 70 60 50 40 0 Supply Voltage ( ±V) 02 6 8 4 30 20 10 90 80 ±V b = ±5V 1100 900 800 500 Total Supply Current (mA) 40 60 100 140 80 700 600 1000 120 -40 -50 -60 -70 -80 -90 -100 Total Supply Current (mA) 40 60 100 140 80 120 3rd 2nd 2Vpp @ 50MHz |
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