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Registered Member #190
Joined: Fri Feb 17 2006, 12:00AM
Location:
Posts: 1567
I have set up a voltage divider with some resistors going to an 18v regulator. I have 31kohms going from the rail to the input. I have a 0.33uf cap on the input and a 0.1uf cap on the output both going to ground. I have a 40k resistor from the input to ground. When I power up the rail to 200vdc the divider with the regulator connected is near 18-19v, and the regulator gives a steady output of 17.8v as I increase the rail. So far this is ok. I want the input to be near 18v when the rail is at 200vdc.
When I put a load on the regulator it does not go much above 5v. I can guess that the 31k of resistors is not letting enough current through to the regulator. If I decrease the resistance of the divider to allow more current, the input to the regulator hits 18v at too low a rail voltage. If I lower the values of the divider too much the unit will draw too much wattage.
Registered Member #1956
Joined: Wed Feb 04 2009, 01:22PM
Location: Jersey City
Posts: 172
IamSmooth
Do you have 200vdc just because you rectifies the mains? If it is so, I would try a capacitive voltage drop (a simple cap in series with your circuit, rectifying after the cap). It is better than a resistor divider and very simple to do... just calculate the capacitive reactance that you need and then the cap by Xc = 1/(2 * pi * f * C).
Use X2 class capacitors. I think Y2 would be fine too.
The only disadvantage that I see is that your circuit needs to have a constant current consumption. If it don't, you can try a capacitive voltage divider on AC mains too.
Registered Member #103
Joined: Thu Feb 09 2006, 08:16PM
Location: Derby, UK
Posts: 845
What are you actually trying to do, do you just need a regulated 18v supply from a ~200v rail? or does it need to vary with rail voltage? What are you powering from the 18v as well, or is it just a feedback signal?
Registered Member #15
Joined: Thu Feb 02 2006, 01:11PM
Location:
Posts: 3068
IamSmooth wrote ...
I have set up a voltage divider with some resistors going to a LM311 18v regulator. I have 31kohms going from the rail to the input. I have a 0.33uf cap on the input and a 0.1uf cap on the output both going to ground. I have a 40k resistor from the input to ground. When I power up the rail to 200vdc the divider with the regulator connected is near 18-19v, and the regulator gives a steady output of 17.8v as I increase the rail. So far this is ok. I want the input to be near 18v when the rail is at 200vdc.
When I put a load on the regulator it does not go much above 5v. I can guess that the 31k of resistors is not letting enough current through to the regulator. If I decrease the resistance of the divider to allow more current, the input to the regulator hits 18v at too low a rail voltage. If I lower the values of the divider too much the unit will draw too much wattage.
Do I have any reasonable options?
You have too much resistance.
I already stated how to design this.
You'll have a single resistor between your 200VDC and the input to your regulator. Knowing what your load current is very important here as this is how you size your series dropping resistor.
If you're output is 18V, then you need at least 20V at the input. If your load draws a maximum current of 100mA then you want to drop 200V-20V maximum across the series resistor at 100mA. This turns out to 1.8k resistor. Power dissipation of the resistor will be 18 Watts! Yes, it *is* a lot of power, but this is a linear regulator.
Your problem is your resistance is much too high, so with any load current, you drop too much voltage and don't have enough at the input to the regulator.
Either live with the power dissipation, or use a transformer or alternate way to get the 18V.
Registered Member #190
Joined: Fri Feb 17 2006, 12:00AM
Location:
Posts: 1567
Thanks Dr. G.
As I figured and you confirmed my resistance was too high. I set the input resistance to 5k/20w and I was able to get the voltage up to 240vcd to maintain a stable 18v to my comparator chip as it was drawing current.
When the chip stopped sinking current, the load dropped and the voltage going into the regulator shot up over 100v. There was no load on the regulator. The absolute maximum on the chip specs says it should not go above 36v. What is one to do if the load on the regulator drops? Is there an op amp or comparator chip that will maintain a constant load independent of its state?
Registered Member #1807
Joined: Tue Nov 11 2008, 07:36AM
Location: San Luis Obispo, California
Posts: 19
You could "make" your own regulator...and thereby design it to withstand/dissipate as much voltage/power as you want.
Here's a sim ramping the input voltage from 0 to 200V.
The output stays pretty close to 18V, regardless of load. It would probably take some experimentation to match up the zener voltage minus the MOSFET threshold voltage = 18V. Keep in mind that the MOSFET is acting as the dissipating element, burning up 17W at 200Vin, 200 Ohm load. Yowza! If anything, you could use something like this to preregulate the input prior to an actual voltage regulator (Vregs normally do not to so well against 100V!) 17W (or similar) could be eaten up by a cpu heatsink (free if from an old computer!)
Honestly, though, if this is going to dissipate more than a few watts steady state then your best bet is to find a transformer (old wall wart or similar).
Registered Member #190
Joined: Fri Feb 17 2006, 12:00AM
Location:
Posts: 1567
I've come to a partial solution. I have picked enough resistance (5k) going from the rail to the regulator input such that the voltage is between 18-33v when the load is present. I added somewhere between 700ohms and 1k going from the regulator input to ground. This way when the load is gone the resistive divider takes over and limits the regulator input voltage.
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voltage regulator problems
