Hi,
This is LTS with the LED's.
E
View attachment 281856

May I see the similar graph where "vtmp" is connected to input terminal(+) and "Vref" is connected to input terminal(-) of the AD620 IC chip?

The graph came from transient analysis is it?

 

Hi,
This is LTS with the LED's.
E
View attachment 281856

May I know what's the explanation behind the "D1 LED" and "D2 LED" lighting up differently when the LM35 is below 0.5V and above 0.5V?

My understanding:

1. When the LM35 is below 0.5V, the Vo is in positive value thus overcoming the forward voltage of the "LED D2" and lighting it up.

2. When the LM35 is above 0.5V, the Vo is in negative value, thus not able to overcome the forward voltage of the "LED D2" and not lighting it up. During this period, the voltage of 5V connected in series to "LED D1" is lighting it up.

3. Or should I observe the current value passing through "LED D1" & "LED D2"?

4. For the AD620 IC at the above circuit, I saw that the Rg pins are connected to 10ohm resistor. Is there a reason? the difference between shorting it and connecting a 10 ohm resistor?


Thanks in advance,
Isaac

 

hi Isaac,
The Gain cannot be infinite, look at page #7 of this PDF, to show the gain limits.

The AD620 or any INA is not suitable for use as a Comparator.

E

If i observed the graph, I can observe the gain is upto 1000 only. If i were to put 10 ohm resistor at the Rg pin 1 & pin 8. Theoretically, through calculation the G = (49.4k/10) + 1 = 4941.

Can I assume the AD620 chip limit the gain to 1000 only?

 

hi Isaac,
Check out Post #20 for a LTSpice simulation result with a AD620 and LM35

May I know what's the explanation behind the "D1 LED" and "D2 LED" lighting up differently when the LM35 is below 0.5V and above 0.5V?

The AD620 is acting as a Comparator and the Vout switches over at the Vref versus Vtemperature voltage point.

4. For the AD620 IC at the above circuit, I saw that the Rg pins are connected to 10ohm resistor. Is there a reason? the difference between shorting it and connecting a 10 ohm resistor?

I considered it the practical solution.

2. When the LM35 is above 0.5V, the Vo is in negative value, thus not able to overcome the forward voltage of the "LED D2" and not lighting it up. During this period, the voltage of 5V connected in series to "LED D1" is lighting it up.

Yes.
Note: the 0.5V for Vref represents a temperature of +50Cdeg, ie: 10mV/Degree Centigrade at the LM35

E

 

hi Isaac,
Check out Post #20 for a LTSpice simulation result with a AD620 and LM35


The AD620 is acting as a Comparator and the Vout switches over at the Vref versus Vtemperature voltage point.


I considered it the practical solution.


Yes.
Note: the 0.5V for Vref represents a temperature of +50Cdeg, ie: 10mV/Degree Centigrade at the LM35

E

May I know if there is a calculation/theory on the voltage value captured by the multimeter at both LED1 & LED2 for the simulation below?

_______________________________________________________________________________________________________________________________________

When I detached the connection of AD620 output from the LED, the output voltage value is different:

Vtemp < Vref: Vdc = -14.0


Vtemp < Vref: Vdc = 13.8V

If I were to understand it, because the AD620 is used as a comparator. The output voltage value would be saturated towards the supply voltage +15 & -15.

What happens to the voltage value of the output of A620 when it is connected to the two LEDs?

 

hi Isaac,
Check out Post #20 for a LTSpice simulation result with a AD620 and LM35


The AD620 is acting as a Comparator and the Vout switches over at the Vref versus Vtemperature voltage point.


I considered it the practical solution.


Yes.
Note: the 0.5V for Vref represents a temperature of +50Cdeg, ie: 10mV/Degree Centigrade at the LM35

E

May I know how connecting a 10 ohm resistor to Rg pin 1 & pin 8 would be a practical solution?

 

May I know how connecting a 10 ohm resistor to Rg pin 1 & pin 8 would be a practical solution?

Hi Isaac,
What do think the value should and why.?
E

 

Hi Isaac,
What do think the value should and why.?
E

If i refer to the gain equation and the maximum gain limit given in the AD620 datasheet:






The Rg should be less than 49.45 ohm to achieve maximum gain. The reason I said this is because if I would like AD620 to act as a comparator, the gain would be amplified to ensure the output voltage reaches saturation at +15V or -15V if refer to my constructed circuit.


Btw, I saw the gain limit is different for these 2 datasheets of AD620:
1. https://forum.allaboutcircuits.com/attachments/ad620a-pdf.281829/



2. https://www.analog.com/media/en/technical-documentation/data-sheets/ad620.pdf



I apologize if my fundamentals knowledge on comparator is lacking.

Thanks in advance,
Isaac

 

hi Isaac,
With a Rg of 10 ohms and the Gain Equation, that is a Gain = 49400R/10R = 4940.
Which is roughly half the maximum gain stated in your second data clip of 10,000 and 5 times greater than the first data clip of 1000 maximum.
If you are confident that the second link is correct, you could use a 5R for RG, which gives a gain of ~10,000, attached simulation.
I have increased the LED resistors for the LED's to 1k5, this will give an LED current of approx 10mA which is OK.

E

 

hi Isaac,
With a Rg of 10 ohms and the Gain Equation, that is a Gain = 49400R/10R = 4940.
Which is roughly half the maximum gain stated in your second data clip of 10,000 and 5 times greater than the first data clip of 1000 maximum.
If you are confident that the second link is correct, you could use a 5R for RG, which gives a gain of ~10,000, attached simulation.
I have increased the LED resistors for the LED's to 1k5, this will give an LED current of approx 10mA which is OK.

E

View attachment 281917

My laboratory is only able to provide the smallest resistor value of 10ohm, unfortunately. So I'll stick with 10 ohm resistor as your previous LTspice simulation circuit.

May I know how did you manage to produce the current value of 18mA & 10mA from the two graphs of both your LTspice simulation?

How can I control the current value passing through the two LEDs? Is it through connecting a different value resistor? Is there a way to calculate? Ohms Law?

Thanks in advance,
Isaac

 

How can I control the current value passing through the two LEDs? Is it through connecting a different value resistor? Is there a way to calculate? Ohms Law?

hi,
May I ask what technical course you are enrolled on and at what year level you are studying.?
I ask this in order to know at what level I have to explain my help.

Calculating the current through a Series resistor and LED are very basic equations.
E

 

hi,
May I ask what technical course you are enrolled on and at what year level you are studying.?
I ask this in order to know at what level I have to explain my help.

Calculating the current through a Series resistor and LED are very basic equations.
E
View attachment 281926

Hi, I'm currently enrolled in BSc in Electrical & Electronics Engineering, 2nd year of studies.

If I were to calculate the current passing through the resistor, R1 would be:
V = IR

- V = ? [AD620 output after differing the voltage value between (+) & (-) terminal] *I dont get this part*
- I = ?[I am finding this value]
- R = 1k ohm

I(R1) = V/1k
= ?
______________________________________________________________________________________________________________________________________
If I were to calculate the current passing through the resistor, R3 would be:
V = IR

- V = 5V
- I = ?[I am finding this value]
- R = 1k ohm

I(R3) = V/R
= 5/1k
= 5mA *This is not tie-in to the value from your graph*

 

If I were to calculate the current passing through the resistor, R1 would be:
V = IR

Hi Isaac,
When driving an LED you must consider the forward voltage drop across the LED.
eg: Assume the LED is a Red type with a forward voltage drop of 2V and the supply is 5V, and we require 10mA through the LED.

Rser= (5v-2v)/0.01A , > 3V/0.01A = 300R.
Note: different coloured LED's have their own forward voltage drop, always refer to the datasheet.

- V = ? [AD620 output after differing the voltage value between (+) & (-) terminal] *I dont get this part*
- I = ?[I am finding this value]
- R = 1k ohm

The Vout terminal of the AD620 can either Source or Sink current at the junction of the LED's/Resistors network.
The AD620 Source/Sink current has to be considered in your calculations as well as the current from the +5V voltage source.


Added: Sim showing the AD620 Vout Source and Sink currents, I(V5) plot and the LED currents.

E

 

Hey Eric, can you comment on my calculations?

Based on this simulation I am currently doing, during the sink current at the junction of the LED's/Resistors network:


Calculations for I(R4) [Note: LED(Green) forward voltage is 2.13V from multisim]:


Based on this simulation I am currently doing, during the source current at the junction of the LED's/Resistors network:



Calculations for I(R3) [Note: LED(Red) forward voltage is 1.83V from multisim]:


Is this the correct method in calculating the current flowing through R4 & R3?

 

The Vout terminal of the AD620 can either Source or Sink current at the junction of the LED's/Resistors network.
The AD620 Source/Sink current has to be considered in your calculations as well as the current from the +5V voltage source.

Hi Eric, I am not clear regarding this part. I am trying to understand how the AD620's Vo of "10.2V" & "-5.22V" from my simulation circuit came to be.

My current understanding is that because the Rg inserted to AD620 is 10ohm. The Vo of AD620 is saturated towards the supply voltage(+15V & -15V). From my circuit, I'm getting +13.5V & -13.7V which I can consider theoretically correct?

 

My current understanding is that because the Rg inserted to AD620 is 10ohm. The Vo of AD620 is saturated towards the supply voltage(+15V & -15V). From my circuit, I'm getting +13.5V & -13.7V which I can consider theoretically correct?

Hi Isaac,
The AD620 is not a Rail to Rail output, it will saturate at approx Vout = +/-Vsupply -1.5V.

E
Clip from datasheet.

 

Hi Isaac,
The AD620 is not a Rail to Rail output, it will saturate at approx Vout = +/-Vsupply -1.5V.

E
Clip from datasheet.

View attachment 282029

Hi Eric,

Thanks for the clearance on the AD620 saturation output voltage part.

Is this the correct method in calculating the current flowing through R4 & R3?

May I ask if my calculation for the I(R4) and I(R3) is correct?

Thanks in advance,
Isaac

 

Hi Isaac,
This LTSpice sim plot shows the Voltage on Vout for both saturation levels, also the currents flowing in the Resistor/LEDs.
I have chosen 1k5 resistors in order to reduce the current loading on the AD620 output.
If you consider in the positive Saturation State that you have +13.5v across R1/D2 and in the negative saturation you have +5V at the high end of R3 and –13.5v at the lower end, a total of 18.5V across R3/D1.
Compare your calculations with this plot

E

 

hi Isaac,
For comparison using 470R resistors, give approx 15.6mA, so your calculations are close enough.
E

 

The Vout terminal of the AD620 can either Source or Sink current at the junction of the LED's/Resistors network.
The AD620 Source/Sink current has to be considered in your calculations as well as the current from the +5V voltage source.

Hi Eric,

While referring to my calculations made for I(R4) & I(R3), I am wondering about the voltage value of Vout terminal of the AD620.

1. When the Vout terminal is not connected to the LED part of the circuit, the Vout would be +13.5V & -13.7V. [I understood this one from your earlier reply, thanks]

2. When the Vout terminal is connected to the LED part of the circuit, the Vout would be +9.62V & -5.22V. [I am not clear on this part]





Is it because of "Current source" & "Current sink"? If I were to understand these 2 terms would be this way:



Or is it because of resistors and LEDs present after being connected to the AD620 output?


Thanks in advance,
Isaac