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Negative Voltage Sensing

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BKI

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I have a negative output voltage (-60V).
What i plan to do to regulate this voltage is to use a resistor divider at the output to receive a -2,4V signal instead of -60V.
Then, i want to compare this signal with a bandgap reference voltage source.

The problem is now, can i use a comparator to compare? If i use a standard bandgap, it only gives me a positive 2,4V right?

So, how can i sense a negative voltage?
 

BKI said:
I have a negative output voltage (-60V).
What i plan to do to regulate this voltage is to use a resistor divider at the output to receive a -2,4V signal instead of -60V.
Then, i want to compare this signal with a bandgap reference voltage source.

The problem is now, can i use a comparator to compare? If i use a standard bandgap, it only gives me a positive 2,4V right?

So, how can i sense a negative voltage?
Click to expand...

You can follow your idea, simply using circuits working "under zero". If you can afford to load your -60V source by one milliamp, you can use an optocoupler. Or use an opamp to transfer the sensed voltage to a positive level.
Comparators with internal reference can be used simply "under zero", with their Vcc at GND and Vdd at say -5...-15 V. Their output, negative TTL or CMOS, can be used thru the optocoupler as above..
 

jiripolivka said:
You can follow your idea, simply using circuits working "under zero". If you can afford to load your -60V source by one milliamp, you can use an optocoupler. Or use an opamp to transfer the sensed voltage to a positive level.
Comparators with internal reference can be used simply "under zero", with their Vcc at GND and Vdd at say -5...-15 V. Their output, negative TTL or CMOS, can be used thru the optocoupler as above..
Click to expand...

I wand to integrate it. So, i would prefer to avoid optocoupler. What do you think about this implementation with opamp?

Negative Output Voltage Sensing.png

If i choose R1=1k and R3=16,67k, then i should receive a positive output voltage of around 2,4V right?
 

You need to understand that the two resistors have the same current in them and the two opamp inputs are both at 0V.
Then use Ohm's Law to calculate the resistor values like this:
 

Attachments

  • negative to positive converter.png
    negative to positive converter.png
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You can use a positive bandgap reference of any voltage and your negative supply scaled to be equal and opposite. Then your inverting input is now 0V ,( with +Vin=0V) your (-) must now use the same source impedance with a common feedback gain.

You may add series RC values for derivative and proportional gain control and common RC feedback for integral control and now have a PID error control to optimize stability, overshoot and bandwidth for ideal step load response as you wish.

The feedback cap ought to have a series R for improved phase margin improvement but overall good noise reduction. You can also inject a soft start feedback.
 

The opamp way is good but you must use the two-polarity power supply for it, +/- 5 or +/- 15 V DC. Then your DC input can be 0 to -5 (-15) V to operate correctly. You can use the opamp as a comparator, too.
 

jiripolivka said:
The opamp way is good but you must use the two-polarity power supply for it, +/- 5 or +/- 15 V DC. Then your DC input can be 0 to -5 (-15) V to operate correctly. You can use the opamp as a comparator, too.
Click to expand...
No.
There are many opamps that have inputs that work perfectly down to 0V when it has only a positive supply. Even a lousy old LM358 dual or LM324 quad opamp will work fine (but slowly) in the circuit I posted when it has only a positive supply.

My opamp circuit is inverting so the inputs are always at 0V. Then the voltage to the input resistor can be negative many thousands of volts if the ratio of the resistor values is correct.
 

In case this might save some design effort...

This simple level-shifter will lift -60V input up into the 0 to +5V range. However notice peak input is not translated to corresponding peak output. If you want -60V to show as +2.4V, you'll need to add additional circuitry.

6130390500_1415327810.png


The 1M resistor is unnecessary. It is only there to create a reference to 0V ground.

- - - Updated - - -

Also, your -60V input needs to be able to sink a few mA of current flowing into it.
 

SunnySkyguy said:
You can use a positive bandgap reference of any voltage and your negative supply scaled to be equal and opposite. Then your inverting input is now 0V ,( with +Vin=0V) your (-) must now use the same source impedance with a common feedback gain.

You may add series RC values for derivative and proportional gain control and common RC feedback for integral control and now have a PID error control to optimize stability, overshoot and bandwidth for ideal step load response as you wish.

The feedback cap ought to have a series R for improved phase margin improvement but overall good noise reduction. You can also inject a soft start feedback.
Click to expand...

I believe you want V- sensing for regulating the voltage and want to use a single supply with a bandgap reference.

BEFORE you start any good design as always, you must define a list of specs; such as the range of inputs and output response required for DC and AC, supply voltage. regulation, noise and noise suppression.

Therefore you must decide the choose the overall gain and frequency response including the attenuation of the -60V INPUT, depending on the sensitivity of the output which drives another regulator.

reg OA.jpg

This is how I would start for your integrator with proportional gain (PI Controller). Then you will need to modify it to anticipate noise to make a PID controller with phase margin compensation as I described in previous text for stability.. After you define specs, I can modify the result.

I arbitrarily chose high gain of 1000 x 1/25 with a 2.5V reference. You can change Vref and adjust ratios to balanced for DC gain with null out or any desired offset.
 

I think the first thing to do is to convert the -60V into a positive change. So find a likely positive supply, say +5V. Now connect a 60 K resistor to the -60V line and a 5K resistor to the +5V line. the junction of them will be 0V when the -60v is present and +60/65 X 5V when its absent. So alter the resistors so that when the -60V is present the junction is +1V (61K + 4K). Now you have a change with in the voltage range of your comparators. It could be an idea to incorporate a clamping diode from the junction to earth (cathode to junction and anode to earth), just in case the -60 goes to -80V.
Frank
 

chuckey said:
I think the first thing to do is to convert the -60V into a positive change. So find a likely positive supply, say +5V. Now connect a 60 K resistor to the -60V line and a 5K resistor to the +5V line. the junction of them will be 0V when the -60v is present and +60/65 X 5V when its absent. So alter the resistors so that when the -60V is present the junction is +1V (61K + 4K). Now you have a change with in the voltage range of your comparators. It could be an idea to incorporate a clamping diode from the junction to earth (cathode to junction and anode to earth), just in case the -60 goes to -80V.
Frank[/QUOTE

Any OpAmp idea is pretty much obsolete, because my technology gives me only an opamp that works in a 0-5V range.

I thought about another simple idea (pic attached) where voltages are added at a node. When the voltage goes below -60V, the node is negative. The inverter gives a 5V output signal as a result. When the voltage lies above -60V, the inverter output is 0.

That is the theory. If i simulate it in Cadence with DC sweep, the node that should be -5V is +5V. What is the reason?

Negative Voltage Regulation.png
Click to expand...
 

bki said:
chuckey said:
........................
I thought about another simple idea (pic attached) where voltages are added at a node. When the voltage goes below -60v, the node is negative. The inverter gives a 5v output signal as a result. When the voltage lies above -60v, the inverter output is 0.

That is the theory. If i simulate it in cadence with dc sweep, the node that should be -5v is +5v. What is the reason?
Click to expand...
It's likely the equivalent impedance of the three resistors going to the -60V (whose values are not shown)is too high. Eliminate these three resistors. Change the bottom 1kΩ input resistor to 13.7kΩ and the top 1kΩ input resistor to 165kΩ. Connect the top resistor directly to the -60V. That will give you very close to 0V at the resistor junction.
Click to expand...
 

How are you supplying and regulating the -60V LDO or switched mode?
and the + supply?

What current rating?

What tolerance and ripple expected?

What load? linear or not?

Designing anything is getting the cart before the horse, until this is specified.
 

the problem is, i can not reduce the 5MOhm total output resistance (3 resistors in series), because it is the equivalent load.

The -60V are generated by a charge pump, that is supplied by a 5V battery.
ripply should be max. 10%
equivalent load is 5MOhm resistor and 10nF capacitor.
 

Hi,

the simple circuit in post3 is correct.
But the resistor values are not.

If you need a 5Mohms input resistor, then (if you want 2.4V output @ -60V input) use a 200k feedback resistor.
Choose an OPAMP that is able to work with common mode input voltage down to 0V (better -0.3V or so).

Klaus
 

KlausST said:
Hi,

the simple circuit in post3 is correct.
But the resistor values are not.

If you need a 5Mohms input resistor, then (if you want 2.4V output @ -60V input) use a 200k feedback resistor.
Choose an OPAMP that is able to work with common mode input voltage down to 0V (better -0.3V or so).

Klaus
Click to expand...

I can not choose any OpAmp. It is a integrated circuit and i only have 1 OpAmp in my technology available.
That is the reason why the OpAmp idea is out.
I am now looking for a solution without OpAmp.
 

Hi,

Ok, i see. If you like to use the internal Opamp... what is it's common mode input range?

Do you have access to your DAC's Ref voltage?
Using avoltage divider to VCC is possible, but VCC may change.
If it moves 50mV you will see an about 12 fold = 600mV error in your -60V reading.
Therefore it's better to connect the voltage divider to VRef.

For sure if VRef moves this gives error also, But the error is not 12 fold.

******
With knowing the lowes common mode input voltage, you can give this voltage to the +in of the Opamp to work correctely.
All you need is to do a simple subtraction in SW.

Klaus
 

KlausST said:
Hi,

Ok, i see. If you like to use the internal Opamp... what is it's common mode input range?

Do you have access to your DAC's Ref voltage?
Using avoltage divider to VCC is possible, but VCC may change.
If it moves 50mV you will see an about 12 fold = 600mV error in your -60V reading.
Therefore it's better to connect the voltage divider to VRef.

For sure if VRef moves this gives error also, But the error is not 12 fold.

******
With knowing the lowes common mode input voltage, you can give this voltage to the +in of the Opamp to work correctely.
All you need is to do a simple subtraction in SW.

Klaus
Click to expand...


VICR is : 0 to 3,4V
I am not using a DAC
 

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