Hello there new to the forums
I'm designing a precision threshold detector for a high voltage application. To be more precise the voltage I want to detect is 65 Volts.
I will use the LMP7300 with independent hysteresis control. I want negative hysteresis so I can give the voltage it’s time to drop but I as the voltage reaches 65 I want to detect it so no positive hysteresis.
The inv input will go to voltage ref, the high voltage will be fed to the non inv input via a resistor divider. Top resistor 309K and bottom 10K so at 65,331Volt the divider will give 2,048 Voltage.
The divider for the negative hysteresis will be 15K top and 1M bottom so I have 30mV hysteresis. Multiply that with 10/319 which is the ratio of the high voltage divider and I have 0,94 V of hysteresis on the high voltage signal
Are the above correct, what do you think?
Yes. Your calculations are correct.
However, there is an issue....
The LMP7300 was originally designed for battery monitoring - where the monitored voltage never dropped below 1V.
The Input voltage cannot go below 1V. This is mentioned in the text of the pin descriptions on page 5 and inferred in the CMRR line in the table. The legal input range is 1V up to V+.
So input voltages from ground to just over 1V are "illegal" - so you will get a false output in this range...
This can be remedied by modifying the input divider to not have the input go below 1V. This could be done by using the reference voltage to lift the end of the input divider above 1V ("level shift").
But since high voltages are involved in the divider, it is best to have a physical resistor to ground on the divider and not rely on the 7300's output to "tie down" the end of the divider (what happens when the 7300's power is off?). To avoid loading the internal micropower reference, the divider resistor base values would need to be raised 2-3X.
Instead, it is better to use a third resistor to "pull-up" on the divider node (+IN), turning it into a summing network, and modifying the resistor values. The "pull-up" voltage source can be either Vref, or a local supply voltage if you believe it is clean and stable enough.
Is the supply current of the detector circuit an issue? Adding the pullup will add a few dozen uA's to the supply current.
Is the high voltage divider resistor value fixed? Or can it be changed.
This info would help in selecting the base values.
SVA Precision Applications
Integrated Signal Chain Applications, SVA
Paul thanks for the swift reply
To tell you the truth i haven't noticed that, i just run through the datasheet in a harry.
What you mean by saying network is something like the following?
From a dc sweep I made, I saw that the voltage divider isn’t affected by the pull up R3, with the above values I get almost 1.4 values from 0 volts on high voltage divider and the 2,048 volts goes at 65,3 volt.
The supply current is not an issue since this is not a portable application, also the voltage divider can be changed, what I want is to trigger it at 65 Volts. Any observations, tips are greatly appreciated.
Any other alternatives you could think of for high voltage triggering? A tried the classic LM193 at 3.3 V single supply but the threshold didn’t work correct, when I increased the supply voltage to 5V the device was working as it should
Paul there is something else you could help me. I want to make an upper and lower threshold detector (is this called a windows detector?)
There is another high voltage application I want to design. This time I want to center it at 150V +- 5 Volts. So when I have 155 the upper comparator will trigger, when I have 145 the lower one. The plan is to use another voltage divider, say 722K top and 10K bottom in order to produce 2.048 at 150Volts. This gives me a 5/366 ratio divider so for 5volt hysteresis I need a voltage divider that gives 1.98Volt at hysteresis pins (actually with the maximum 130mV hysteresis i can get 1,77 volts in the original voltage, any way to get at 5Volt hysteresis on original voltage? ( i know there is a 130mv hysteresis limit)
.I’m confused where I should connect the +in and –in pins. I have took a look at figure 7 of the datasheet but I’m confused more.
Yes. You have the idea. You need to fiddle with the resistor values to get correct ratios.
While the "classic" LM193 is operable down to 2V, but the common mode range (again!) comes crashing down fast. The common mode limit is 1.5V below V+, so at 3.3V, the "legal" input range range is only 0 to 1.8V (more like 1.7V to be safe). This is why it did not work correctly. So you need to keep your reference and divider voltages below 1.5V or raise the supply voltage to 2V more than your reference voltage - or use a R-R input comparator like the LMC6772 or LMC7221.
The LMP7300 datasheet Figure 7 design is a window comparator. The trip points are set by a tapped reference resistor divider (R2, R3, R4), sourced from the top comparators voltage reference.
Check out application note AN-74. It is a nice summary of comparator circuits and goes into the theory of hysteresis. The "Limit Comparator" circuit (Fig 9) is also known as a "Window Comparator".
In your bottom circuit, the reference inputs would be connected to the resistor taps on the reference divider string (see AN-74 fig 9). The Hysteresis dividers would then be connected to the comparator inputs, and not to the "reference" output. Your circuit would generally look like the LMP7300 figure 7 circuit, but "RT" would be your HV divider resistor string tied to the HV line.
You are correct about LM193 i haven't any idea about common mode range. There is an apothegm in Greece that says as long as you live you learn, thanks for the ap notes
Now about LMP7300 your opinion is that with a 4.7K pullup i can make it stay above 1 Volts, like in the LTspice simulation?
About the window detector i think i got it,
1st i made a simulation to find voltages from the voltage divider for 140 (1.83494) and 160 (1.89866) volts
i tried to follow the example on the datasheet, i assumed that i had a 81.4K Rtotal. For R3 i did 1,83495=(R3/Rtotal)*2,048Volts i solved for R3 and found 72.9K
For R1 1,89869=(R1/Rtotal)*2,048Volts, i solved for R1 and found 75,46K. I substracted it from Rtotal and found R1. For R2=(R1+R3)-Rtotal. R2=2,56K
For Hysterisis 1 Volt i need 100mv. HV divider ration 1/101*100mv hysteresis=0,99V before the divider.
Any suggestion? Any mistakes you have seen?
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