What is purpose of pulldown to -5V on op amp circuits?

Thread Starter

matthej

Joined Oct 10, 2020
38
Hi,
I am trying to analyze a circuit. It looks like it is trying to determine the voltage across a thermistor by adding it to the feedback path of the first op amp. After that it feeds a couple of more op amps.

I am trying to figure out what the reason is behind adding the two pulldown to -5V lines on the circuits?Slide1.JPG
 
Last edited:

MrChips

Joined Oct 2, 2009
23,524
The purpose of the resistors to -5V is to inject a negative voltage in order to shift the DC operating voltages.
What are the voltages on the power supply rails to the opamp?
 

AnalogKid

Joined Aug 1, 2013
9,254
To discuss individual circuit elements, *please* add reference designators to each component.

Also - where it says "5+8.25K", is the 5 supposed to be 5 ohms or 5 K ohms? Same for the "130+". It matters.

ak
 

BassNotes

Joined Jan 26, 2021
1
I don't see what purpose the "5+8.25K" resistors to -5 V serve. The first op amp is going to determine the voltage at the opposite end regardless of the resistors.
 

Thread Starter

matthej

Joined Oct 10, 2020
38
To discuss individual circuit elements, *please* add reference designators to each component.

Also - where it says "5+8.25K", is the 5 supposed to be 5 ohms or 5 K ohms? Same for the "130+". It matters.

ak
Updated with reference designators. 5 is 5 ohms and 130 is 130 ohms
 

AnalogKid

Joined Aug 1, 2013
9,254
Updated with reference designators. 5 is 5 ohms and 130 is 130 ohms
Don't think so.

For example, the 4177 output stage is rated for +/-10 mA in normal operation, and +/-25 mA short circuit. 10 mA through 5 ohms is only 50 mV, so the voltage at the U1A output is -4.95 V, no matter what the inputs and thermistor are meant to do. The output of U1B would raise this to a whopping 80 mV, or . What is the rest of the system supposed to do with that? Note that all of this is completely independent of whatever the thermistor is supposed to be doing. Without R9 and R10 in place, U1B would attempt to turn this into +8.0 V - if - the opamp power rails are greater than +/-9 V. R9 and R10 would increase this output voltage.

Waaaaait a minit - I'm assuming each of the "2 resistors" pairs are in parallel. Are they? If they are in series, then it is false precision. The difference between 8.25K and 8.255K is 0.06%. Since 8.25K is a standard 1% value, it has an error band of +/-82.5 ohms. This completely swamps the contribution of the 5 ohm resistor. I ignored R8 in my analysis above for the same reason in the parallel case.

If R7 and R8 are in series, then there is no output current issue. However, the resistors also become meaningless because the opamp output stage presents a zero-ohm voltage source to them, so they do not affect the U1A output voltage.

Do you have access to the person who designed this?

ak
 

Thread Starter

matthej

Joined Oct 10, 2020
38
Don't think so.

For example, the 4177 output stage is rated for +/-10 mA in normal operation, and +/-25 mA short circuit. 10 mA through 5 ohms is only 50 mV, so the voltage at the U1A output is -4.95 V, no matter what the inputs and thermistor are meant to do. The output of U1B would raise this to a whopping 80 mV, or . What is the rest of the system supposed to do with that? Note that all of this is completely independent of whatever the thermistor is supposed to be doing. Without R9 and R10 in place, U1B would attempt to turn this into +8.0 V - if - the opamp power rails are greater than +/-9 V. R9 and R10 would increase this output voltage.

Waaaaait a minit - I'm assuming each of the "2 resistors" pairs are in parallel. Are they? If they are in series, then it is false precision. The difference between 8.25K and 8.255K is 0.06%. Since 8.25K is a standard 1% value, it has an error band of +/-82.5 ohms. This completely swamps the contribution of the 5 ohm resistor. I ignored R8 in my analysis above for the same reason in the parallel case.

If R7 and R8 are in series, then there is no output current issue. However, the resistors also become meaningless because the opamp output stage presents a zero-ohm voltage source to them, so they do not affect the U1A output voltage.

Do you have access to the person who designed this?

ak
Ok I apologize. I had R6 in the wrong location which will make a difference (updated)

The two resistors for R7/R8 and R9/R10 are in series. The thermistor has a resistance of 10K at 25 degC. The op amps operate off +12V and -12V and I do not have access to the person who designed this....
 

MisterBill2

Joined Jan 23, 2018
8,760
In most applications the opamp inverting input is used as a summing junction, and so the resistor to the negative supply is a constant input. In this circuit there is another opamp not shown, I am guessing. So the purpose of those resistors to the negative supply is just adding in a constant term.
 

Thread Starter

matthej

Joined Oct 10, 2020
38
In most applications the opamp inverting input is used as a summing junction, and so the resistor to the negative supply is a constant input. In this circuit there is another opamp not shown, I am guessing. So the purpose of those resistors to the negative supply is just adding in a constant term.
But U1A is already supplying a constant output if its connected to -5V input. What purpose would the -5V pullup serve?
 

MisterBill2

Joined Jan 23, 2018
8,760
But U1A is already supplying a constant output if its connected to -5V input. What purpose would the -5V pullup serve?
No, U1A iis NOT supplying a constant output. The output changes are the value of the thermistor changes. In addition, the two leads exiting the bottom of the drawing go to the summing input of another device, which we are not shown. And in every one of the instances where there is a resistor to the minus 5 volt source, it is an input to a summing junction, NOT a "pullup resistor." This is an analog circuit, not a digital one.
 

Thread Starter

matthej

Joined Oct 10, 2020
38
No, U1A iis NOT supplying a constant output. The output changes are the value of the thermistor changes. In addition, the two leads exiting the bottom of the drawing go to the summing input of another device, which we are not shown. And in every one of the instances where there is a resistor to the minus 5 volt source, it is an input to a summing junction, NOT a "pullup resistor." This is an analog circuit, not a digital one.
Sorry, I meant to say that if U1A input is a DC level (-5V), wouldn't the output always be a DC level (changing based on thermistor value). If thermistor was 6.19K for example, gain would be around .6 so output would be around +3.3V. So how do I calculate the output value of U1B? Even if we can simplify it and say the other connection (going to R6) is not there?

Would it just be

Vout= -Rf/Rin (vin): -10K/5.3K*-5 +(-10K/6.19K)*Vout of U1A?
 

MisterBill2

Joined Jan 23, 2018
8,760
Yes, the output is a DC value based on the thermister resistance. That is how opamps work. The output depends on the currents into the summing junction. A thermister in the feedback loop is one way to use the changing resistance of a thermister to get a proportional voltage related to temperature.
 
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