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Voltage follower - voltage limiter for protection

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bmandl

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Hello!
I am designing voltage sensing circuit for 0-100V DC input with voltage divider at aprox. 5V output. But my requirements are, that anything higher than exactly 5V at output shouldn't go further to the ADC. I want to make some kind of a voltage follower and voltage limiter for anything higher than 5V. If there comes signal, higher than 5V, I want to drop that voltage across some resistor and LED for indication. Can I make circuit with decent accuracy with opamp, and how? I drew some kind of a block diagram.
**broken link removed**
 

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You can use a 5V zener diode between point A and ground. The zener will clamp the voltage to 5V.
A 5.1V zener diode is far more common than a 5.0V zener.

I doubt you can use a LED. The current is too small.

Hope this helps.
Tahmid.
 

A zenner diod is not certain. I need 5V, nothing more. And I am not building voltage regulator, but voltage limiter. I don't want fixed 5V at output, but 0-5V, but zenner will fix voltage at 5.1V, am I right? For LED, I was thinking of separate power supply with transistor (maybe FET) for driving it.
 

You could use a an op-amp set up as a comparator. A precise 5V on one input and the voltage at point A on the other. Its output would turn on a transistor, or transistor plus resistor in series, connected across the 0 - 5V, at point A, if that voltage tried to exceed 5.
 

Tahmid, with uncertain, I mean, that zenner voltage varies. Syncopator, can you please draw me your idea?
 

How do I calculate those three resistors?
 

Syncopator? Is this comparator with hysteresis of some kind of what? I don't have enough knowledge for understanding of this circuit.
 

No, there's no hysteresis.

It's similar to a standard voltage regulator circuit in which the output voltage is compared to a reference voltage, and any necesssary corrective action is taken if there is a difference.
In this circuit, the output voltage, which should be 5, is compared with a precision voltage reference.

If the output voltage rises above 5, the amplifier's output voltage will rise, turning on the transistor. This is effectively a another shunt resistor in parallel with the variable one you show in your circuit.

In this case the output voltage decreases until it is equal in value to the reference voltage.

If the output voltage sinks below 5, the additional circuit does nothing.

The two resistors around the amplifier set its gain. It's a circuit with which you should be familiar.
There is no need for high gain. It may be as low as one.

The resistor between the amplifier's output and the transistor's base is to limit the base current.

The amplifier needs both a positive and negative supplies, referenced to the system's zero volts point (which you have indicated to be connected to Earth).

If you are playing around with an a.d.c. I would have thought that this circuit was well within your ability to understand.
 

Thank you for rich explanation. But, can I use amplifier with single supply option? +5V for power supply and reference?
 

Unfortunately, no. The voltage reference i.c. will need a supply higher than 5V.

Depending on the particular i.c. you use for the ampliffier it could need at least a little bit of negative, otherwise its output may not go low enough to cut off the transistor when it isn't needed.
However, you could put a diode, perhaps two, or perhaps a low voltage zener in series with the amplifier's output and the transistor's base. That should overcome the inability of the amplifier's output to go very close to 0V.

I expect that you will experiment a little to see what works for you.

Analog Devices, amongst others, make a 5V reference i.c., type REF02. It has an output accuracy of ±3%, and a minimum supply voltage of 7.

So, with a bit of expriment you may dispence with a negative supply, but you will need something higher than 5V for the reference i.c.
 

This is my simulation. I also made DC sweep simulation and my circuit does not work as expected. I compared circuit to an ordinary voltage divider. Comparison should be equal, until input voltage exceedes 100V (when output is higher than 5V). Please comment this. Thank you.
 

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Unfortunately, no. The voltage reference i.c. will need a supply higher than 5V.

Depending on the particular i.c. you use for the ampliffier it could need at least a little bit of negative, otherwise its output may not go low enough to cut off the transistor when it isn't needed.
However, you could put a diode, perhaps two, or perhaps a low voltage zener in series with the amplifier's output and the transistor's base. That should overcome the inability of the amplifier's output to go very close to 0V.

I expect that you will experiment a little to see what works for you.

Analog Devices, amongst others, make a 5V reference i.c., type REF02. It has an output accuracy of ±3%, and a minimum supply voltage of 7.

So, with a bit of expriment you may dispence with a negative supply, but you will need something higher than 5V for the reference i.c.

I will use fixed 5V from other circuit for voltage reference, so this is not the problem. I already have fixed 5V. So can I use this reference voltage for VCC also? How can I do that?
 

I will use fixed 5V from other circuit for voltage reference, so this is not the problem. I already have fixed 5V. So can I use this reference voltage for VCC also? How can I do that?

You must use a supply voltage for the amplifier which is at least a little higher than the voltage at its inputs. If you can't give it a supply voltage greater than 5, you would need to lower the voltages at its inputs; using a divider across both the controlled voltage and the reference.
 

I finnaly came to an end. I will not use any op-amp, becouse of its unlinearity. I will just use resistors and two schottky diodes with verry low forward voltage, so I can come as close as 5.01V on the output. D2 is used for protection of negative voltage:
 

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