DC - High Voltage DC converter that acts like a sink
A good solution will be 'not too hard" to provide once the true requirement is fully described :-).

At present your description covers a range of possible requirements. Can you provide more details and/or a circuit diagram with suitable notes?

At present the following may describe what is wanted.

Current from a negative voltage potential of 0 to more than &minus;200V is required to be clamped at a user selectable potential of from 0 to &minus;200V.
max required current to be clamped = xxx mA
max power to be clamped is xxx mW



A solution to the above is a simple clamp regulator which can be set to a preset voltage.
Probably a P-channel MOSFET, opamp to drive it, "high side" (high = negative in this case) driver, voltage reference and voltage divider to allow reference and clamp voltage to be compared.



If the clamp point is to be held constant for driving sources above and below the set point (eg at &minus;150V when driving voltage goes from eg &minus;50V to &minus;200V) then the device may need to actively drive the output.

Depending on power level the driver could be a driver with drain or collector load, a "totem pole" pair,  or an active switching regulator drive.

Assume that this is low power and that a single ended driver with resistor will suffice.
This is identical to the clamp regulator above EXCEPT that a resistor is taken from the P-channel MOSFET drain to a negative source of at least &minus;200 V potential.

Example

A current of up to &minus;10 mA is sourced from a potential of 0V to &minus;300 V. (This may be supplied via a resistor "R1" but this need not matter as long as the maximum current is limited).

The current is to be clamped to a user preset voltage of 0 to &minus;200 V.

Solution

Provide an e.g. &minus;300 V, 10 mA or more supply. This is more negative than &minus;200 V to allow some "headroom" when running at &minus;200V sink. 300 VDC is a convenient value for derivation from rectified 230 V AC mains or from voltage doubled 110 VAC mains.

At &minus;200V and 10 mA, $R_{sink,max} = (300\mbox{ V}-200\mbox{ V})/10\mbox{ mA} = 10\mbox{ k}\Omega$

At $V_{clamp} = 0\mbox{ V}$, power in resistor is $V^2/R = (300\mbox{ V})^2/10\mbox{ k}\Omega = 9\mbox{ W}$.

Provide a P-Channel MOSFET > 300V rating. Best >= 400V.
Source to ground.
Drain to Rink_10k
Rsink other end to &minus;300V.
Monitor Rsin_MOSFET drain junction and use control circuit to drive MOSFET drain to required voltage.
Connect circuit to be clamped to MOSFET drain.

Control circuit is an opamp, probably operating between ground and say &minus;15V.
Vin is divided by say 50:1 making the 0 to &minus;200 V swing a 0 to &minus;4V.
Compare this value with a reference voltage.
Vref = Vclamp/50.
Use opamp to drive MOSFET gate.

QED.

$$I_{max}(R_{sink}) = 300 \mbox{ V}/10 \mbox{ k}\Omega = 30\mbox{ mA}$$

Unloaded:
$$P_{MOSFET,max} = \frac{V_{max}}{2}I = \frac{V_{max}}{2} \times \frac{V_{max}/2}{R_{sink}} 
= \frac{V_{max}{}^2}{4R_{sink}} = \frac{(V_{max}/2)^2}{R_{sink}} $$
$$ = \frac{(150 \mbox{ V})^2}{4 \times 10\mbox{ k}\Omega} = 9/16 \mbox{ W}$$

E&amp;OE.

Loaded is slightly more complex as clamped current gets in on act.
Easily enough calculated but lets see user true requirement first,\ 

If this does not meet the requirement then a better description of the requirement would help.