[SOLVED] overvoltage protection circuit - help needed!!!

[p72]

Hi,

I have 4-20mA Input circuit with the 200 ohm resistor. The output of this goes into the OpAmp (opa333) which operated on +5V supply.

I would like to design circuit/component which protect my device (opa333) from overvoltage (>5Volt.) (in case if someone applies more than 5volts.).

So that, as I said once 4-20mA current passes through 200ohm resistor, I get the 0.8 to 4.0 volts (my input range).
(I am keeping 1volt margin, so that I am looking to protect OpAmp from overvoltage condition.)

One more thing: - What can I do if someone applies reverse connection?

Any quick solution?

I will appreciate your response.

Thanks in advance.
P72

 

[KonstanU]

1. Connect resistor between sensor resistor (200 Om) and OpAmp input .
2. Input Op Amp must be pulled to Vcc and GND through Fast/Schottky diodes (in reverse bias). It restricts input voltage level on the range -0.7V and VCC+0.7V

 

[p72]

1. Connect resistor between sensor resistor (200 Om) and OpAmp input .
2. Input Op Amp must be pulled to Vcc and GND through Fast/Schottky diodes (in reverse bias). It restricts input voltage level on the range -0.7V and VCC+0.7V

Thanks KonStanU for the suggestion.

I have a two questions to you- In your first step, you have mentioned connect resistor between 200ohm and opamp input. what value? and what is the purpose of this resistor?
Second question- what kind of Schottky diode? (can you give me some specific part number?)

Thanks again!!!!!!
p72

 

[KonstanU]

Putting resistor, you limit current in the worst case. If some one will apple +15V to your input then resistor limits current on the value about 10V/R.

I do not know what exactly amplifier do you use. That is why I could just suggest to use 4.7k. (You must account also power dissipation on this resistance and current that will flow through it.).

Use some of the NXP Shottky diodes.

And the last note, resistor and diode form delay circuit for input signal. It could effect on the value resistor

 

[p72]

Hi KonstanU,
I have made circuit as per your suggestion. Please let me know if I did anything wrong.

In my circuit, I have 0.8V to 4.0V(max) as an input supply. I would like to protect circuit anything over 5V supply. Does this circuit serve my need?

 

[Georgy]

Input terminals are internally diode-clamped to the power-supply rails. Input signals that can swing more than 0.3V beyond the supply rails should be current limited to 10mA or less.
Only R2 = 5 kOhm is enouth.
See Figure 18. "Input Current Protection" of datasheet.

 

[p72]

Input terminals are internally diode-clamped to the power-supply rails. Input signals that can swing more than 0.3V beyond the supply rails should be current limited to 10mA or less.
Only R2 = 5 kOhm is enouth.
See Figure 18. "Input Current Protection" of datasheet.

Hi Georgy,
Thanks for the input.

I did not get your second sentence.
I think 4.7kOhm is close to 5kOhm then what is the reason behind your (5kOhm) suggetion?

I need protection before OpAmp itself and that is why I put protection circuit before OpAmp. Is it ok to you? Do you see anything wrong?

I wanted to confirm that, the circuit I designed will fullfill my requirements or not.
Thanks,
p72

 

[KonstanU]

Sorry, that I haven't answered early.
You have done one mistake. It is connection of diodes. It must be connected in reverse bias.
So your protection circuit should look like

- - - Updated - - -

Input terminals are internally diode-clamped to the power-supply rails. Input signals that can swing more than 0.3V beyond the supply rails should be current limited to 10mA or less.
Only R2 = 5 kOhm is enouth.
See Figure 18. "Input Current Protection" of datasheet.

I agree with you that this amplifier has internal diode protection. But it is good practice to make external protection. It allows change this amplifier on another one without any problems.

 

[p72]

Sorry, that I haven't answered early.
You have done one mistake. It is connection of diodes. It must be connected in reverse bias.
So your protection circuit should look like View attachment 85224

- - - Updated - - -



I agree with you that this amplifier has internal diode protection. But it is good practice to make external protection. It allows change this amplifier on another one without any problems.

Hello KonstanU,
Thank you for the help!!!

If you don't mind could you please explain operation of the protection circuit in words?
Thanks,
p72

 

[KonstanU]

Hello KonstanU,
Thank you for the help!!!

If you don't mind could you please explain operation of the protection circuit in words?
Thanks,
p72

Firstly, I need to notice that protection inside OpAmp is almost the same as circuit which we discussed .

If input voltage is more than 5V then upper diode is in forward bias and voltage of signal line is restricted on the level 5V+Vdiode.
If input voltage is less than 0V then low diode is in forward bias and voltage of signal line is restricted on the level 0V-Vdiode.
if input voltage is in the range between 0V and 5V then both diodes are in reverse bias and do not conduct current.

 

[p72]

Firstly, I need to notice that protection inside OpAmp is almost the same as circuit which we discussed .

If input voltage is more than 5V then upper diode is in forward bias and voltage of signal line is restricted on the level 5V+Vdiode.
If input voltage is less than 0V then low diode is in forward bias and voltage of signal line is restricted on the level 0V-Vdiode.
if input voltage is in the range between 0V and 5V then both diodes are in reverse bias and do not conduct current.

Last thing: - I have selected BAT85 as a schottky diode. Honestly speaking, I have taken reference from internet about this diode but I don't know the reason why I have selected this. Could you explain what are the parameters of the diode need to check for this circuit?

 

[KonstanU]

Last thing: - I have selected BAT85 as a schottky diode. Honestly speaking, I have taken reference from internet about this diode but I don't know the reason why I have selected this. Could you explain what are the parameters of the diode need to check for this circuit?

So here is list of parameters which you must account for Shottky diode.
1. Reverse voltage. At least 2 times more then supply voltage (5V*2 = 10V). You must have voltage margin and account also spike voltage in transient process.
2. Current must be more then max. predicted current (if you have input signal 15V and R = 4.7k than Imax =2.2mA ). So take diode with Irat =5 or 10 mA
3. Calculate power dissipation and temperature rise. For BAT85 Vdrop = 0.1V (Tamb = 85C) and Ploss = Vdrop * Imax = 0.1 * 0.0022 = 0.22mW. Tj = Tamb + Ploss *Rth = 85 + 0.00022*320 = 85.7C. It is very good.
4. Delay line is t = 2.2*R*(2*Cd) = 2.2 * 4.7k*(2*8pF) = 160 nS.

I would suggest to use BAS40-04. It is dual diode.

And I forgot to ask you, what is general consumption of your scheme? Is it more then 2.2mA or not?

 

[p72]

KonstanU,

This is 4-20mA input circuit so mostly, consumption should be around this range.
Thanks a lot for the suggestion.

 

[p72]

So here is list of parameters which you must account for Shottky diode.
1. Reverse voltage. At least 2 times more then supply voltage (5V*2 = 10V). You must have voltage margin and account also spike voltage in transient process.
2. Current must be more then max. predicted current (if you have input signal 15V and R = 4.7k than Imax =2.2mA ). So take diode with Irat =5 or 10 mA
3. Calculate power dissipation and temperature rise. For BAT85 Vdrop = 0.1V (Tamb = 85C) and Ploss = Vdrop * Imax = 0.1 * 0.0022 = 0.22mW. Tj = Tamb + Ploss *Rth = 85 + 0.00022*320 = 85.7C. It is very good.
4. Delay line is t = 2.2*R*(2*Cd) = 2.2 * 4.7k*(2*8pF) = 160 nS.

I would suggest to use BAS40-04. It is dual diode.

And I forgot to ask you, what is general consumption of your scheme? Is it more then 2.2mA or not?

Hi KonstanU,
just wondering one thing about the circuit- Can this circuit(BAS40-04) also protect from "reverse polarity" connection on input? (Incase, if someone connected incorrect wires that is reverse wires.)

Thank you,
p72

 

[KonstanU]

Yes it will protect.
As I wrote before, input signal less then -0.3V will be restricted.
I suggest you simulate circuit in LTSpice.
I believe, I have given to you comprehensive answer with respect to your topic.

With best regards,
Kostiantyn Ulitskyi

 

[Mustafatarhan49]

addition to what konsanU said, connect a resistor to the non inverting part of OPAMP and put one capacitor paralel to that resistor. try to calculate the over shoot time and use Tau=RC equation

 

[p72]

Yes it will protect.
As I wrote before, input signal less then -0.3V will be restricted.
I suggest you simulate circuit in LTSpice.
I believe, I have given to you comprehensive answer with respect to your topic.

With best regards,
Kostiantyn Ulitskyi

Yes, You have given me all answers which I wanted. I am really glad that we met here and you helped me.
Keep in touch for any analog discussion.

Thanks again!!!!!!
p72