What is the best way to obtain a regulated low voltage (0.3Volts)? (There are no zener diodes with)

Thread Starter

Ultravioletta

Joined Feb 17, 2017
21
Such Values.

I Have an acquisition card that does not take up more than 0.4V. However, the input signal can sometimes due to disturbances exceed this value (it goes up to few volts). So I want to eliminate the excess I thought of using a zener Diode. Is there such small value for a Zener breakdown voltage? If not, how do I proceed?
 

BobTPH

Joined Jun 5, 2013
8,807
A Schottky diode will have a forward drop of about 0.3V. If the signal is always positive, one will do, if it goes negative, you will need two in anti-parallel.

Bob
 

MisterBill2

Joined Jan 23, 2018
18,167
Indeed shunt diodes can provide the protection that you want, but there are a few potential problems with that way of doing it. First is the fact that most diodes have a fair amount of input capacitance, and that capacitance varies with the applied voltage. So unless you are working with DC for the input, that may be a challenge. Also, diodes mostly have a voltage range where they star conducting some, a ways below the saturation voltage.
So what I recommend is a shunt diode with a saturated forward voltage drop of 0.7 volts, and using two of them in opposite directions, in parallel. That will protect your analog input while not interfering with signals of an amplitude that you are interested in. A 1N4007 type is a common one that would work.
 

Plamen

Joined Mar 29, 2015
101
Such Values.

I Have an acquisition card that does not take up more than 0.4V. However, the input signal can sometimes due to disturbances exceed this value (it goes up to few volts). So I want to eliminate the excess I thought of using a zener Diode. Is there such small value for a Zener breakdown voltage? If not, how do I proceed?
Petkan:
Classical 3 terminal voltage regulators output voltage is defined by its internal reference and a voltage divider (external or internal) Vo=Vref(1+R1/R2). They cannot produce voltage lower than the reference. However there is a class or regulators from Linear (now Analog) like LT3080. Instead of constant voltage source as reference, they have a constant current, scaled by a single resistor. These regulators can start from 0V and certainly could be scaled to 0.4V
 

Ya’akov

Joined Jan 27, 2019
9,069
Such Values.

I Have an acquisition card that does not take up more than 0.4V. However, the input signal can sometimes due to disturbances exceed this value (it goes up to few volts). So I want to eliminate the excess I thought of using a zener Diode. Is there such small value for a Zener breakdown voltage? If not, how do I proceed?
Is this an analog varying input? If so, at what frequency?
 

MisterBill2

Joined Jan 23, 2018
18,167
Petkan:
Classical 3 terminal voltage regulators output voltage is defined by its internal reference and a voltage divider (external or internal) Vo=Vref(1+R1/R2). They cannot produce voltage lower than the reference. However there is a class or regulators from Linear (now Analog) like LT3080. Instead of constant voltage source as reference, they have a constant current, scaled by a single resistor. These regulators can start from 0V and certainly could be scaled to 0.4V
Regardless of the the of this thread, the TS wants to limit the over-voltage that gets to the data acquisition input. So providing a regulated voltage source is not a solution.
 

Plamen

Joined Mar 29, 2015
101
Is this an analog varying input? If so, at what frequency?
Petkan:
Go to analog.com and download the datasheet. Also download the free LTSpice and simulate it there.
This is an analog voltage regulator, capable of sourcing up to 1.1 A. It could be used a fixed voltage regulator, or adjustable (tracking the reference input). It could also be used as current source (scaled by 2 resistors). Play in LTSpice!!
 

crutschow

Joined Mar 14, 2008
34,281
A forward biased diode to ground, such as a 1N4148 should work, depending upon the source and load impedances.

Note that the diode voltage is a logarithmic function of the forward current (below) so the clip voltage will depend upon the source resistance/current, and the diode will conduct some below the desired clip point, which can degrade (lower the voltage) of the signal

The simulation shows the input and output voltages for a 1kΩ series resistor and a 1N4148 diode to ground.
At a 300mV input there's only about a 1.8mV drop across the resistor (yellow trace), so only a small degradation of the signal level, but at a 3V input there's about a 2.4V drop across the resistor, with the diode limiting the output voltage to about 0.62V (blue trace).

Does that seem acceptable?

upload_2019-4-23_11-37-2.png
 

Plamen

Joined Mar 29, 2015
101
A forward biased diode to ground, such as a 1N4148 should work, depending upon the source and load impedances.

Note that the diode voltage is a logarithmic function of the forward current (below) so the clip voltage will depend upon the source resistance/current, and the diode will conduct some below the desired clip point, which can degrade (lower the voltage) of the signal

The simulation shows the input and output voltages for a 1kΩ series resistor and a 1N4148 diode to ground.
At a 300mV input there's only about a 1.8mV drop across the resistor (yellow trace), so only a small degradation of the signal level, but at a 3V input there's about a 2.4V drop across the resistor, with the diode limiting the output voltage to about 0.62V (blue trace).

Does that seem acceptable?

View attachment 175670
Petkan:
Why not follow the LT3080 idea? This diode changes by 2mV/degC, has part to part variation, load dependence, varying between 0.5V and 0.8V
 

WBahn

Joined Mar 31, 2012
29,976
Petkan:
Why not follow the LT3080 idea? This diode changes by 2mV/degC, has part to part variation, load dependence, varying between 0.5V and 0.8V
Who the heck is "Petkan"?

The TS is not looking for a regulated signal -- he's looking to limit a data signal so that it doesn't exceed the amount he wants to expose the input of his data acquisition card to. Read the first post and not just the thread title.

He wants to shunt current away from the input so as to limit the applied voltage. Using shunting diodes is a perfectly reasonable way to do this provided the shunted current isn't too disturbing at the top end of the measurement range and the clipped voltage isn't too much at the top end of the shunting range.
 

MisterBill2

Joined Jan 23, 2018
18,167
Who the heck is "Petkan"?

The TS is not looking for a regulated signal -- he's looking to limit a data signal so that it doesn't exceed the amount he wants to expose the input of his data acquisition card to. Read the first post and not just the thread title.

He wants to shunt current away from the input so as to limit the applied voltage. Using shunting diodes is a perfectly reasonable way to do this provided the shunted current isn't too disturbing at the top end of the measurement range and the clipped voltage isn't too much at the top end of the shunting range.
That is the same thing that I said back in post #6. evidently speed reading misses a whole lot of important details. Of course, I am also used to having my advice ignored by some folks. Possibly the problem is with the name of the thread.
So I am wondering what sort of problems a small overvoltage would cause with the data collection system. Perhaps the TS would explain why it is a big problem.
 

ebeowulf17

Joined Aug 12, 2014
3,307
If the cutoff needs to be extra clean, precise, and low, a dual op amp with a few passives can serve this function with basically any arbitrary input limit:
7924F591-F397-4E51-B71A-922A091917B8.png
This idea is a modified version of an idea I stole from @crutschow a while ago.
 

Thread Starter

Ultravioletta

Joined Feb 17, 2017
21
Is this an analog varying input? If so, at what frequency?
Yes. The signal is a 0-20Hz signal + the external disturbances of 50 Hz. My problem is that I want to measure the signal with an acquisition board that cannot take more than 0.3 V and the 50 Hz disturbances may go up to 2-3V. So I wanted to limit those.
 

Ya’akov

Joined Jan 27, 2019
9,069
Yes. The signal is a 0-20Hz signal + the external disturbances of 50 Hz. My problem is that I want to measure the signal with an acquisition board that cannot take more than 0.3 V and the 50 Hz disturbances may go up to 2-3V. So I wanted to limit those.
You should try the opamp circuit mentioned by @ebeowulf17 to avoid distorting the signal. Passive methods will change the signal in way that may not be linear. Ideally, you want the signal to go full input when it exceeds the threshold so you can tell that happened, ad probably max-N, where N is some distinguishable difference from full when it hasn’t clipped. Then your data will be cleanable.
 

DickCappels

Joined Aug 21, 2008
10,153
This circuit should work well. The signal to be clipped enters through the 10k resistor. The circuit below includes two clippers but you can just build the one you need.
upload_2019-4-24_16-16-17.png
 

MisterBill2

Joined Jan 23, 2018
18,167
A different version of the input clamping scheme could be to use germanium diodes, such as the 1N34, which have a much lower forward voltage drop.. And I am wondering if your data acquisition input is subject to damage from a small over-voltage, or just incorrect operation? Does it have any internal over voltage protection, or not? Most of the systems that I have used did have some protection, but they also needed current limiting on the inputs for over voltage protection. So I am wondering.
 

ebeowulf17

Joined Aug 12, 2014
3,307
Yes. The signal is a 0-20Hz signal + the external disturbances of 50 Hz. My problem is that I want to measure the signal with an acquisition board that cannot take more than 0.3 V and the 50 Hz disturbances may go up to 2-3V. So I wanted to limit those.
Does the signal go negative in normal operation? Is it AC, centered around 0V? Or are we talking about signals that only range from 0-0.3VDC in normal operation, with 2-3V excursions that must be removed?

If it's AC, including negative components, that might change my recommended circuit.
 
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