ADC on high voltages

I'm thinking of using the delta-sigma method of doing ADC as demonstrated in the counter application note and else where for measuring a volage between 0-90v. Obviously I need to reduce this voltage otherwise everything will go pop.
Is my best bet a potential divider along with a zener diode so that the range is 0-3.3 and cannot go above?
Graham
Is my best bet a potential divider along with a zener diode so that the range is 0-3.3 and cannot go above?
Graham
Comments
thanks
Graham
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Chip Gracey
Parallax, Inc.
Graham
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Regards.
Alberto.
The main constitutive equation comes from Kirkoff's current law. The sum of currents at the summing junction is zero. The average current consists of i1 from the signal source, plus a contribution when the bpin is high, minus a contribution when bpin is low. The contributions from bpin are weighted by a fraction H high and h low. The total duty cycle is H + h = 1.
That leads to equation 1, and then with a little algebra to get rid of h, directly to equation 2. At that point there is no assumption about the souce of signal current. It could be a true current source such as a photodiode, or it could be as is often the case a voltage in series with a resistor, or it could be a high voltage, for example, -90 volts in series with a high resistance.
Note that the value of the capacitor does not enter into the equations. 1 nf is recommended, but it is not critical. The little box with a * is a direct connection from the summing junction to the apin. All connections in the feedback loop have to be short and direct.
Note that the balance point is H = 0.5, 50% duty cycle, when the input signal current is zero. If the input current increases in a positive direction, the duty cycle decreases, H decreases and h increases in proportion. The relation between H or h and the input current is linear. When H=0, the input current is i1 = 1.65/R2, and that is the saturation point. The ADC will not operate for higher currents. On the other end, the limit is i1 = -1.65/R2.
Equation 3 is written for a voltage V1 in series with a resistor, R1. This is the voltage difference between the input signal V1 and the voltage Vx=1.65 at the summing junction, divided by the input resistance. Equation 4 solves for V1.
Again, if V1=1.65 volts, there is no steady state input current and the duty cycle is 50%, H=h=0.5.
V1 can vary within non-clipping limits that are found by solving with H=0 and H=1. Or, to accomodate a particular maximum input voltage, solve for R1 and R2.
Suppose R1 = 150 kohms and R2 = 100 kohms as suggested in AN001.
Solving for the input voltage when H=0 and H=1, it is a total span from 4.125 volts down to -0.825 volts. That covers the full scale of 0 to 3.3 volts, with a margin above and below. It is probably not good to push the duty cycle too near 0% or 100%, but with a 32 bit counter and a high frequency you can probably push it pretty close. You would probably calibrate the system with a known input voltage, which covers the effects of resistor tolerance, threshold variation etc.
To get V1 = -100 volts input, with R2=100 kohms, plug in H=1 and solve for R1. I come up with 6.16 megaohms. At -90 volts, it will be within range, duty cycle H<1.
I too would highly recommend an input clamp in addition to the Prop's built in substrate diodes. When the ADC is operating properly, the feedback will hold the input at 1.65 volts, but if the -90 volts is present when the prop is turned off, the current of about 15 microamps will flow through the 6 meg resistor. That is not much, but can you be sure someone won't short circuit the resistor?
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Tracy Allen
www.emesystems.com
Post Edited (Tracy Allen) : 2/20/2007 9:39:56 PM GMT
Tracy,
Wow thanks for that. I also now understand delta-sigma from your description of the analysis. I just realized that saying 0-90v was not a good idea, I meant 0v to 90v. Its not negative.
Graham
It doesn't much matter if it's positive or negative 90 volts. The range is symmetrical around 1.65 volts. With +90 volts, it will be driven to a duty cycle with more percent of time low, whereas with -90 volts it would be larger percent high.
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Tracy Allen
www.emesystems.com
Graham
www.goldmine-elec-products.com/products.asp?dept=1127
The confusion about -90 volts was my bad memory of your original question. Still it does emphasize that the circuit is bipolar.
If the range only has to be unipolar, an op-amp circuit like Mike suggested could ground reference the signal and scale it into the full range. There are lots of op amps that will go rail to rail at the inputs or the outputs or both, and operate well at 3.3 volts. I use the LT1490 a lot, and the LTC2055 for highest precision. The venerable LM10 is good, too, and even has an internal reference amplifier that can provide an offset if for example you only need to cover a part of the range like 30 to 90 volts.
There could also be a resistor to ground in the original circuit, which would provide additional protection. For example, a voltage divider is formed with a 6.49 megaohm resistor to +90 volts and 150k ohms to Vss, that has a Thevenin equivalent of 2.2 volts with a source resistance of 146kohms and could be connected to the ADC input. Then if the substrate becomes disconnected, nothing really bad will happen. Also with that arrangement, the 90 volts can not provide parasite power via the substrate, as might happen if the Propeller goes into RCSLOW mode running at microamps. In that case its probably best to leave the ADC pins as low outputs when not in use, to avoid parasite power.
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Tracy Allen
www.emesystems.com
Quattro
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'Necessity is the mother of invention'