What could be the purpose of these diodes I actually saw in a bioamplifier circuit. The Vin+, Vin- at left are differential inputs, but why it is connected to the diodes? And why is the Vs+, Vs- (power supply) connected to the diodes at the right? AMP01 is the main amplifier with the V+, V- connected as shown.

 

they are used to clamp excessive input signals to +/- Vs. it is a basic form of protection

 

here is an example of such diodes used internally in ESD sensitive device (OPA186)

 

they are used to clamp excessive input signals to +/- Vs. it is a basic form of protection

How could the excessive input signal be suppressed using diodes connected to the power supply? I know in surge protection, it is clamped to ground, but not to the power supply itself. Is the circuit about removing slow excessive voltages or sudden surges? Pls. explain in words.

Also in my circuit. There are no output diodes or ESD Current-Steering diodes. Only the input diodes are there. Is the output diodes function separate from the input diodes? Thanks.

 

The effective bridge rectifier does indeed clamp to about the power supply voltage, which in this case appears to be 7.2 volts. For a number of IC devices the requirement for input voltage limits is the power supply voltage. If that is the case here, that is all the protection needed. Not my favorite scheme, but certainly valid. My preference is to have the clamp voltages much lower.

 

here is the circuit detail with components marked



If Vin+ is large enough and positive, R1 and D2 will conduct. therefore voltage on input (right side of R1) will never be able to go above about +8V which is Vs+ value plus 0.7V for the forward drop across D2. the excess voltage will be absorbed by R1.

if the Vin+ happens to be large but negative, R1 and D1 will conduct. so the input voltage (right side of R1) will never be able to go below about -8V which is Vs- value and the voltage drop across D1.
in other words input voltage is clamped in this case to +/-8V even if the Vin+ is excessive.

Similarly, R2,D3,D4 work for clamping input signal to +/-8V

it has to be noted that current through diodes will try increasing Vs values so if the supply cannot absorb the change, Vs will grow to potentially dangerous limits.

to prevent that, TVS diodes are added as shown in my other post

 

here is the circuit detail with components marked
View attachment 316120


If Vin+ is large enough and positive, R1 and D2 will conduct. therefore voltage on input (right side of R1) will never be able to go above about +8V which is Vs+ value plus 0.7V for the forward drop across D2. the excess voltage will be absorbed by R1.

if the Vin+ happens to be large but negative, R1 and D1 will conduct. so the input voltage (right side of R1) will never be able to go below about -8V which is Vs- value and the voltage drop across D1.
in other words input voltage is clamped in this case to +/-8V even if the Vin+ is excessive.

Similarly, R2,D3,D4 work for clamping input signal to +/-8V

it has to be noted that current through diodes will try increasing Vs values so if the supply cannot absorb the change, Vs will grow to potentially dangerous limits.

to prevent that, TVS diodes are added as shown in my other post

Thanks. When you said "(right side of R1)", did you mean the Vs on far right side (or power supply voltage Vs), or did you mean the voltage drop across R1? The diodes are marked H30 0A. What does 0A mean? see below. Also let's use actual example. For example the input voltage is 50V excessive, the R1 is 4.99k Ohm.. How is the 50V excessive input absorbed with interaction with the 7.2V Vs power supply?

 

this is R1



Vs does not connect to R1.
when i said right side of R1, i meant the terminal on the right... terminals of R1...that is the signal going up to LF.... that signal is clamped to +/-8V.

 

For example the input voltage is 50V excessive, the R1 is 4.99k Ohm.. How is the 50V excessive input absorbed with interaction with the 7.2V Vs power supply?

IF the Vs supply is steady enough and able to maintain 7.2V then 50V at input will cause current through R1 and D2 to Vs.
Since forward voltage of the diode is about 0.7V and Vs is 7.2V voltage drop across R1 is

Vr1 = 50V - (0.7V + 7.2V) = 42.1V

the current through R1 is then
Ir1 = Vr1/R1 = 42.1V/4990 Ohm = 0.0084368737475 A
or
Ir1=8.44mA

So resistor R1 will dissipate as hear power that is:
Pr1= Vr1*Ir1=0.355W

it will either do so or it will die trying...
if the resistor is only rated 0.25W for example, it will burn. not instantly, this will take few seconds - if the input stays at mentioned 50V.
but ESD and transients are not long lasting events, so such resistor would be fine (not even have time to warm up)

 

this is R1

View attachment 316122

Vs does not connect to R1.
when i said right side of R1, i meant the terminal on the right... terminals of R1...that is the signal going up to LF.... that signal is clamped to +/-8V.

Please tell me by what Ohm law principle why the resistor can never become 50V too when the input is 50V? And why the 50V can't get to the input of the IC at bottom.

 

resistor is linear device that follows Ohms law.
you put higher voltage across resistor and more current flows through it.

diode is a non-linear device.
you bring small voltage (less than 0.7V), and almost no current flows.
you bring a bit more and the current flows.
you bring a lot more, and the current flows (and it is MUCH larger) but the diode voltage barely changes - it only increases slightly (maybe to 0.8V).

 

Oh I just read your message the R1 becomes 42V. So this won't get to the IC input?

Isn't it the purpose of the R1 resistor is to limit the current to I=V/R = 7.2V/4.99kOhm = 3mA so if Vs accidentally contacts the input, the electrodes connected to skin won't get high amperage. But supposed the power supply got reversed voltages with Vs- becomes 7.2V and Vs becomes -7.2V and the diodes conduct. Won't the electrodes get current more than 3mA and electrocute (or stop the heart) of the person? See original diagram. Thanks.

 

your understanding is very flimsy...
Ohms Law applies to each device but... values are for the SAME device.
you cannot apply Ohms Law by picking value of R1 and voltage of something else like Vs. R1 does not even connect to Vs.
to apply Ohms Law to R1 you need I,V,R values that are relevant to R1. when there is 50V at the input, that is not 50V across R1. you need to work your way through to get them as shown above.

and your calculation is wrong. 7.2V/4.99k is not 3mA.

 

your understanding is very flimsy...
Ohms Law applies to each device but... values are for the SAME device.
you cannot apply Ohms Law by picking value of R1 and voltage of something else like Vs. R1 does not even connect to Vs.
to apply Ohms Law to R1 you need I,V,R values that are relevant to R1. when there is 50V at the input, that is not 50V across R1. you need to work your way through to get them as shown above.

and your calculation is wrong. 7.2V/4.99k is not 3mA.

I know. But I was asking what voltage will get to the input of the IC at bottom of my last figure. Is it 42V or 8V?

I was using 14.4/4990 Ohm before to come up with 3mA. So it's 7.2/4990 = 1.44mA

 

Won't the electrodes get current more than 3mA and electrocute (or stop the heart) of the person? See original diagram. Thanks.

when you wear synthetic clothing and brush your hand across it, large static charge can be obtained. 1000V is not much, could be much higher. in winter more so since the air is dry. shifting wait while sitting in a chair with foam cushion can produce 10-15 times more. so when you get up from that chair and touch something like light switch or door knob, you can experience zap. and that is all it is - a short zap. it is not harmful to humans. but for extremely delicate structure of precision amplifiers this could be devastating. to prevent circuit damage, protection is built. one form of protection is what you have seen discussed here.

so ESD voltage is real and high but does not pack a lot of energy. diodes are used to clamp the excess voltage at the input to safe limits. in your example, Vs+ is 7.2V so input is clamped to +8V (or 7.9V if you take the diode forward voltage literally as 0.7V).

the excess voltage is spread across input resistor (such as R1). the internal circuit (like OpAmp input) never sees more than 8V because of R1/R2 and the diodes.

 

I was using 14.4/4990 Ohm before to come up with 3mA. So it's 7.2/4990 = 1.44mA

no it isn't.. Vs is NOT the voltage across R1.
voltage across R1 is most of the time near 0V since input of OpAmp is pretty high impedance and current is VERY low.,
the only time any appreciable current flows through R1 is when voltage exceeds +/-8V. THEN diodes start conducting and a bit of current flows through that resistor. And even then, this is just a moment to equalize charge. after that, it is gone...
This is just a pulse to distribute charge and then it is over....

 

btw, if you think you understand Ohm and Kirchoffs Laws, try to chose resistor for following circuit. This can teach you a lot.

 

I know. But I was asking what voltage will get to the input of the IC at bottom of my last figure. Is it 42V or 8V?

I was using 14.4/4990 Ohm before to come up with 3mA. So it's 7.2/4990 = 1.44mA

There are 3 different scenerios:

1. What I meant was if there was accidental contact between Vs and input..The resistor was there to limit the current. Gets?

2. In the original diagram. What if the power supply got reverse polarity. D3 and D4 can conduct. Wont the current get to the input?
Why not? the resistor limits the current to 1.4mA?

3. In the voltage divider concept of R1/R2. what if the IC has internal resistance of 4.99kOhm too. Then it can get half the voltage or 25V of the 50v input. why not? Thanks.

 

Let me detail the above so my context makes sense.

1. Remember even 9V can electrocute a person. Imagine the heart is connected to the 9V battery with only 100Ohm resistance. I=V/R = 9V/100 Ohm = 90mA enough to stop the heart. Remember the story of the sailorman with faulty multimeter who died from 9V battery.

2. If power supply got reversed. Input may get voltage from the reversed power supply but the 4.99kOhm just limit the current to 7.2/4990 = 1.4mA so it is safe. Right?

3. This is what I want to focus on. I know your point 8V will be what the input to the bottom IC will always see. But what if there is no diode. In the bottom . Won't 7.2V still be seen by the input of the IC? Or 50V + 7.2V? 57.2? What voltage would it exactly see? Thanks.

 

1. What I meant was if there was accidental contact between Vs and input..The resistor was there to limit the current.

even then there is no current since both sides of diode is at the same potential.
to get the current through, potential difference need to exceed Vf of the diode.

2. In the original diagram. What if the power supply got reverse polarity. D3 and D4 can conduct.

that will give you bigger problem - the circuit would be destroyed unless there is a parasitic diode inside some of the ICs to work as a clamp and save the day

3. In the voltage divider concept of R1/R2. what if the IC has internal resistance of 4.99kOhm too. Then it can get half the voltage or 25V of the 50v input. why not? Thanks.

those resistors will certainly work as part of voltage divider but this is only when the input voltage is not limited by diodes (connected voltage is within +/-8V). but once the input voltage reaches or exceeds those +/-8V, diodes with R1/R2 will clamp the voltage on the right side of those resistors to said +/-8V. the left side of R1/R2 will be at whatever potential you bring to the circuit.