I'm curious to know the role of bleeder resistance of 1k ohm at non-inverting terminal used in this circuit attached below. Is there any calculations to set the right value for this bleeder resistance and will this reduce any offset in the circuit? Kindly help me to understand this.

 

That's not typically called a bleeder resistance, perhaps pull-down resistance would be closer to what it does.
It just a resistance to keep the plus input at ground potential when there's no other input, otherwise the op amp output would go to one of the rails.
Usually it's minimum value is selected to not significantly load the source driving that input.
1k seems lower than necessary, 10k-100k would be more typical.

 

FYI. A bleeder resistor is intended to discharge a capacitor when the power is removed. It is connected across the terminals of the capacitor and is sized to discharge that capacitor to a "safe" voltage in some small number of seconds. For example, if I have a 100 uF capacitor in a +50 VDC power supply . I might use a 600 Ω bleeder resistor across the terminals. The time constant is the product of R & C and is 60 msec. This tells me that the capacitor will discharge to 36.8% of its original value in that time. After 5 time constants it will be down to 0.67% of it's original value. So 0.67% of 50 is 337 mV. So our time constant RC = (600)(100E-6) = 60 milliseconds, and five of those is 300 milliseconds. So this capacitor will have much less than 1 volt after 300 milliseconds. Now while the power is on this resistor will draw 83.3... mA and have to dissipate 4.1667 watts. That might be too much to manage inside a product. Let us say the we want to shoot for 20 mA of operating current through the bleeder resistor which implies that we should choose 2.5K.

With a 2.5K bleeder resistor the time constant becomes 250 milliseconds. five of those puts us at 1.25 seconds. After 1.25 seconds we are down to our 337 mV left on the capacitor. Now the power dissipation in the bleeder while the power is on is only 1 watt = (.020)^2 * 2500. There are of course other compromise solution you could pick in between those two values.

Just to inflict maximum pain, lets say that 1.25 seconds is too much and we want five time constants to be exactly 1 second. So 0.2 seconds divided by 100 uF = 2K. Now with 2K across 50 volts our operating current is 25 mA. This resistor will dissipate 1.25 watts. The best way to implement this would be 2 x 4K, 1 watt resistors in parallel, so each resistor dissipates 625 mW.

Naturally the bigger the supply voltage and the bigger the capacitor(s), the more care must be taken with safety. Also pray that there is no technician named "Fast Eddie" around who is hell bent on probing your power supply with a screwdriver only to weld it to the capacitor terminals and spray noxious oil all over the place.

Picky little problems can be fun when you understand what is going on.

 

FYI. A bleeder resistor is intended to discharge a capacitor when the power is removed. It is connected across the terminals of the capacitor and is sized to discharge that capacitor to a "safe" voltage in some small number of seconds. For example, if I have a 100 uF capacitor in a +50 VDC power supply . I might use a 600 Ω bleeder resistor across the terminals. The time constant is the product of R & C and is 60 msec. This tells me that the capacitor will discharge to 36.8% of its original value in that time. After 5 time constants it will be down to 0.67% of it's original value. So 0.67% of 50 is 337 mV. So our time constant RC = (600)(100E-6) = 60 milliseconds, and five of those is 300 milliseconds. So this capacitor will have much less than 1 volt after 300 milliseconds. Now while the power is on this resistor will draw 83.3... mA and have to dissipate 4.1667 watts. That might be too much to manage inside a product. Let us say the we want to shoot for 20 mA of operating current through the bleeder resistor which implies that we should choose 2.5K.

With a 2.5K bleeder resistor the time constant becomes 250 milliseconds. five of those puts us at 1.25 seconds. After 1.25 seconds we are down to our 337 mV left on the capacitor. Now the power dissipation in the bleeder while the power is on is only 1 watt = (.020)^2 * 2500. There are of course other compromise solution you could pick in between those two values.

Just to inflict maximum pain, lets say that 1.25 seconds is too much and we want five time constants to be exactly 1 second. So 0.2 seconds divided by 100 uF = 2K. Now with 2K across 50 volts our operating current is 25 mA. This resistor will dissipate 1.25 watts. The best way to implement this would be 2 x 4K, 1 watt resistors in parallel, so each resistor dissipates 625 mW.

Naturally the bigger the supply voltage and the bigger the capacitor(s), the more care must be taken with safety. Also pray that there is no technician named "Fast Eddie" around who is hell bent on probing your power supply with a screwdriver only to weld it to the capacitor terminals and spray noxious oil all over the place.

Picky little problems can be fun when you understand what is going on.

Thank you for the in-depth explanation.

 

EDIT: Oops: I read 0.67% as 67%.

At 5TC your almost fully discharged. https://www.electronics-tutorials.ws/rc/rc_2.html

That circuit is a current to voltage converter. The non-inverting input would normally be connected to ground.

Ib and Vos are the most important parameters for an I-V converter. he resistor in the feedback loop is very high so a small amount of Vos will result in a large output signal. Vout = -I*Rf; but you also have I=Vos/1000 as an offset signal. Vos/(Zero) is a bad large number.

Non-idealalities will always get you. In general, you have to provide a place for the input bias current to drop across.

As resistors get larger, the noise voltages get larger.

Sometimes you have to add series resistors in the inverting and non-inverting inputs so, the OP amp can handle voltages greater than the power supply. Especially when the supply is zero or unpowered.