# voltage regulator at 4.3V and 0.7V from 5V input that can handle inputs up to +-12V

#### mpopeang

Aug 5, 2017
17
Hi

I am trying to create a replacement for an old device which is not produced anymore to be used into a digital circuit.

This device It was able to sustain 2 voltages at 4.3V and 0.7V connected to clipping diodes to protect input lines from between -12V : +12 V. I want to be able do this this in a circuit which gets power from +5V, It has also inputs for -5V and +12V.
This old device was also using the -5V as input to itself to produce both 4.3V and 0.7V. I will prefer to not depend on the other power lines so i can just dropped them from the design and simplify it if possible

I come across this voltage regulator that can give 4.3V. And also some other using an LM137.
http://www.electronics-tutorials.ws/diode/diode_7.html

The clipping diodes are connected to the output system via a 100 ohm resistor.

The problem that i am not able to solve, is that once the input voltage gets increased the voltage generated by that zener diode or the LM137 also gets increased (more or less depending on the resistors i use there). I am not able to get something stable regardless of this voltage variation.
Plus i also need as the ground the +0.7V.

If i get resistors to handle the +12V, when the input voltage is the one expect in good working conditions +5V, then the input line voltage drops to 3.3-3.5V which i am not sure its correct. It can still be seen as a logic high by the circuit connected to it.

Plus i noticed (via simulation where i can set the exact voltages) that if these +4.3V and +0.7V are not at these values clipping diodes are not working properly. Powering the line with -10V will not go to almost 0v, but actually -1 or -1.9V which will damage the digital circuit.

I am sure this must have been implemented already and some schematics might exist for it.
Ideally if there is a single new device which does this will be perfect. If not i can implement it via a schematic.

Any help is appreciated.
Thanks

#### davenn

Moderator
Sep 5, 2009
14,263
I am trying to create a replacement for an old device which is not produced anymore to be used into a digital circuit.

This device It was able to sustain 2 voltages at 4.3V and 0.7V connected to clipping diodes to protect input lines from between -12V : +12 V. I want to be able do this this in a circuit which gets power from +5V, It has also inputs for -5V and +12V.

what is the device ?
show a schematic or a much better explanation

#### mpopeang

Aug 5, 2017
17

The device name is called IC Protector - 9000A-8076 MFG part5 85992.
See the U10 in the schematic. The A1..A5 are the clamping diones circuit. These are described into the second image.

As you can see it drives the 4.3V to 2 of the A1, A4, A5 and its drives the 0.7 ground to all of them.

#### Harald Kapp

##### Moderator
Moderator
Nov 17, 2011
13,728
You can use any voltage regulator, e.g. LP38841 from this table.
The relevant parameters to look for are:
input voltage >= 5 V
output voltage 0.7 V or 4.3 V, respectively, or adjustable within this range.

#### kellys_eye

Jun 25, 2010
6,514
I don't see that device as a power supply - I see it as a power MONITORING device that alarms at the loss or exceeding of certain set parameters.

U12 is driving a line (U10 pin 13) to logic 1/0 (given it seems to be supplied from -5V and +5V, dropped by CR2 to give 4.3V) and U10 pin 12 is driven from the +5V rail via R7.

Accordingly you could replace the device with simple comparators and/or window comparators.

Either way, 4.3V and 0.7V can be derived from simple silicon diode volt drops across a 5V supply.

#### Harald Kapp

##### Moderator
Moderator
Nov 17, 2011
13,728
It seems the originator of that circuit wanted to clip the input voltage to exactly 0 V and +5 V which will happen with the supply voltages shown for the diodes. Modern devices have similar input protection diodes but connected to 0 V and +5 V (or whatever Vcc there is used). Thus the input voltage is clamped not to 0 V and 5 V but to -0.7 V and +5.7 V. The chips tolerate this.
If older chips don't t tolerate this overvoltage or undervoltage, connecting the protection diodes as shown in the op's diagram is a remedy.

#### Alec_t

Jul 7, 2015
3,591
As Kelly suggested :-

#### mpopeang

Aug 5, 2017
17
That's the problem with this implementation that i was describing. This is not correct actually.
And it makes sense since the designer of this circuit dint used this simple approach but had to provide 4.3V and 0.7 using that power/monitoring IC. Otherwise why to bother. ...

For normal voltages at output which is +5V, this circuit will provide only 2.8V at that input. While this is much low that 5V can still be detected as a digital high. But for voltages in 0V - +0.8V you will have +2.1 V, which will fall out of range. (you should get here a voltage between 0-08V for a digital low.
You want to have values to read close to real voltages at outputs, specially when the inputs are normal.

#### kellys_eye

Jun 25, 2010
6,514
The power monitoring device U10 is being used to check the logic level signal inputs and ensure that they are within the +/-10% tolerance allowed for logic 1 and 0 in a 5V logic system.

Pin 23 of A3, A4 and A5 (et al) is tied to ground via R6 (200Ω) - this gives the 'logic 0' level of 0.7V that is fed to one input of the multi-input power-fail detector (U10). The +5V line is monitored via R7 (200Ω) tied to +5V to give the 4.3V that is tied to another input of U10 to give the 'logic 1' level.

You don't need to 'generate' ANY voltages in this area of the schematic - if U10 isn't working i.e. "power fail" (U10, pin 11) is active then you're either monitoring an incorrect logic 0 or logic 1 OR the monitoring device (U10) is faulty.

If U10 is faulty it can be replaced with a set of simple comparators having set-inputs of 0.7V and 4.3V with their outputs wired-OR to provide the fault signal.

Worst case you can 'ignore' the 'power fail' signal and tie it high (or low) but lose the ability of the equipment to detect monitoring problematic logic signal levels.

Your understanding and/or explanation fo the circuit seems to be lacking clarity.

#### mpopeang

Aug 5, 2017
17
That U10 is both a power monitoring a power supply device. Lets clarify that. OK?
And forget about that monitoring thing as this can be replaced. Atmega can do it easily.

The problem is how you do it to replace the other logic. The provided schematic above is not working.
If you have a better schematic, lets see it.

#### kellys_eye

Jun 25, 2010
6,514
That U10 is both a power monitoring a power supply device. Lets clarify that. OK?
Please do. How is it a power supply?

As I see it the circuitry around U10 shows a voltage reference used to compare the relevant logic levels and U10 pin 13 is even connected to a logic gate output (U12) - when has a logic gate output ever been driven by a 'supply'?

U12 output can be 1 (5V - or 4.3V if you like) or 0 (0V or -0.7V if you like).

As already mentioned, pins 6 and 12 are derived from across R6 and R7.

So where is the 'output' of U10?

If you have a better schematic, lets see it.

If that comment is meant to be the 'insult' I see it as I reckon you consider yourself more than qualified to sort the problem out yourself.

Good luck.

Goodbye

#### mpopeang

Aug 5, 2017
17
First, it is a power supply based on description in the manual that i am looking at.
Secondly , if you set the voltages +4.3 V and 0.7V on that line it works perfectly (keeping the ground and +5V connected via the resistors as in the schematic of course). Even when external voltage is at +12V, the line reading will be close to 5V. While if you dont have that power at 4.3V this wont happen. As seen in the schematic present earlier for example, when 0.5 V at input it doesn't get you a logical low as it should.

Having said that lets get over this part. The requirement of the circuit is simple :
You have a range of {-12 , + 12V }, that are connected to a digital line for which you want protection.
The result from that protection circuit should be one that translate this to 0- 5V. range (+-05 V depending on maximum input values for that digial logic). While keeping the digital circuit working, where accepted inputs for it as low: {0 , +0.8V| and high {+2.4V , +5V}

For the schematic, of course i want something that will work. Talking it talking but practice kills it.
I want to see something working.What better way to express your idea that having it working.

#### davenn

Moderator
Sep 5, 2009
14,263
First, it is a power supply based on description in the manual that i am looking at.

show us the exact description as I don't see it being used as a power supply in the circuit you provided above

maybe you are misunderstanding something ?

Aug 5, 2017
17
Here is it.

#### davenn

Moderator
Sep 5, 2009
14,263

you are misreading the pin ID's

The word POWER goes with the word FAIL below it as an ID for pin 11 .... Power Fail

so pins ...
Pin 11 .... POWER FAIL
Pin 2 ... REF HI
Pin 10 ... REF LO
Pin 13 .... SINK1
Pin 6 .... SOURCE2
Pin 12 .... SINK2
Pin 18 .... SENSE3
Pin 7 .... SENSE2
Pin 9 .... SENSE1

you just need to learn to read things more clearly

unless you can show me something different in an actual datasheet, I will go with the obvious markings on the schematic

#### mpopeang

Aug 5, 2017
17
As i mentioned a few times already...give it a try and see how can you come up with 4.3V and 0.7V on those lines others wise. ?

#### (*steve*)

##### ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,510
The +0.7V is a source, the +4.3 is a sink.

I think they are just logic outputs from the chip. My rationale is that one of these lines is connected to the output of a logic gate.

#### kellys_eye

Jun 25, 2010
6,514
My rationale is that one of these lines is connected to the output of a logic gate.
U12 pin 8.....

#### mpopeang

Aug 5, 2017
17
Yes, just one of the lines, but not the other 2.

When i've tried to put a voltage regulator with a output of 4.3V to replace that line from U10 (or other variants with diodes, or voltage dividers) , the circuit is not able to keep the voltage at that level (4.3V), once i connect the external line to +12V. I was using the diodes with 100 ohm resistor as the protection on it (as seen in that small schematic). And the 200 ohm to +5V connected to this line as seen in this large schematic.

I was hoping someone could explain why this IC is able to do it and a simple circuit like those is not ?

#### kellys_eye

Jun 25, 2010
6,514
but not the other 2.
...they come from the HARDWIRED R6 and R7 connected to +5V and ground..... sheesh...

Try Googling 90004-8076 and see what others have to say about it..... same as us, basically.

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