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AD5535B using all channels together

Hi, I need an ADC for 200V but only one output.

1. Adding output currents

On the AD5535B each channel can drive: 550 µA, I wonder if is possible to use all together to supply more current? 0.55uA*32 = 17.6mA. I know that 550uA is the maximum, not recommended for continuous use. Actually, my power supply has 8mA max.

The output buffer amplifier is specified to drive a load of 1 MΩ and 200 pF

So 200V/1M = 200uA is the expected usage.  

Is it possible to shortcircuit outputs to get more output current? 32*.2mA = 6.4mA?

Fig 8 shows that even with a load between outputs the short circuit protection turns off the output. But maybe because 200v/10k = 20mA and outputs can behave like sinks of 2.1mA which is more than the 550uA source.

Would it be possible to override the protection using a fast diode on each output? Forcing the outputs to behave like sources and preventing them to sink?

2. Max output current at 200V

From the specs table, I understand that with VPP = 215V and setting an output of 200V each output can source 550uA is it correct? Fig 10 doesn't mean that only for 70V we can get 550uA right? With an output of 200V, the maximum current is 550uA, ok?

3. Setting all the outputs at the same time

Setting all the outputs at the same time is not possible right? 

A4 to A0 Bits The A4 to A0 bits can address any one of the 32 channels. A4 is the MSB of the address, while A0 is the LSB.

There's no register for "all"

SCLK works up to 30Mhz, each set requires 19b but really becomes 3x8 packets, we could write each channel on approx 1us, ok? So for 32 channels = 32us

The user must allow 200 ns (minimum) between successive writes.

If we cannot set all at the same time since there's no latch option, when the first channel set from 0 to 200v will try to source all the current to the load and It'll protect himself going to 0v right? Will it recover once the other channels start to provide current to the load?

4. Price

Why is the evaluation Board cheaper than the chip alone? 


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