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I'm using the LTC5598 IC for modulation. I am confused as to how to properly feed my BaseBand analog signal into the chip.

According to its datasheet:

The circuit is optimized for a common mode voltage of 0.5V which should be externally applied. The baseband pins should not be left floating because the internal PNP’s base current will pull the common mode voltage higher than the 0.6V limit. This condition may damage the part. In shut-down mode, it is recommended to have a termination to ground or to a 0.5V source with a value lower than 1kΩ. The PNP’s base current is about –68μA in normal operation.

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If I understand this correctly, if I want to feed it an instantaneous signal of 0.4V do I set:

BBPI = 0.9V
BBMI = 0.5V

This should satisfy the requirement of having a common mode voltage (Vcmbb) of 0.5V?

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It's a bit more complicated than that. Have you seen the example driver filter circuit?

It's designed to be driven by a differential signal, centered on 0.5V. The datasheet talks about a swing of 0V to 1V. That should be your swing. So a 40% amplitude wave should have a peak at (0.5*0.4) + 0.5 on the positive rail and -(0.5*0.4) + 0.5 on the negative rail.

Additionally you need to provide a suitably terminated driver. The "negative input impedance" is a bit weird, but what it means is that the inputs will run away and self-destruct unless their bias current goes somewhere. The datasheet says "it is reccomended to have termination to ground or 0.5V through a resistor less than 1k", and indeed their example driver filter terminates to ground through a 100 ohm resistor.

(Why is the filter there? It's there to round off the edges of your signal coming from the DAC in order that it modulates cleanly).

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  • \$\begingroup\$ Ok, that makes more sense. Yes, I did see the example driving circuit, however, it seems to be driven by current. I wanted to drive via a voltage based ADC \$\endgroup\$ – hassan789 Apr 5 '13 at 16:55
  • \$\begingroup\$ Which DAC did you have in mind? Picking one with differential outputs is pretty much required. \$\endgroup\$ – pjc50 Apr 5 '13 at 18:06
  • \$\begingroup\$ Understood. I guess I will use a Balun (en.wikipedia.org/wiki/Balun) \$\endgroup\$ – hassan789 Apr 8 '13 at 16:46
  • \$\begingroup\$ Where's the signal coming from, ultimately? Could you just choose a different DAC? \$\endgroup\$ – pjc50 Apr 8 '13 at 17:01
  • \$\begingroup\$ Yes, i can choose a different dac. Actually, the source would be a microcontroler with a 10-bit dac. But, my baseband with bee purly digital. It will be either +0.4V or -0.4V. So I am now thinking about doing it without a DAC, and through OP-amps with built in LPF \$\endgroup\$ – hassan789 Apr 8 '13 at 17:04
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No, you have to be balanced; a 0.5V offset means BBPI = 0.7V and BBMI = 0.3V. The average of the two is still 0.5V. Having said that don't overdrive them too much either. The spec say these inputs can be run at 0.86Vp-p (page 4) and that voltage is superimposed on the 0.5V BUT remember they shouldn't be driven single ended or damage may occur.

At the peak of the 0.86Vp-p, one input will have 0.93V and the other will have 0.07V. Both average at 0.5V.

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