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DC balanced?

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bananasiong

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okay... this is the defination of the input offset voltage of a LM393
Input Offset Voltage. The amount of input voltage change that is needed to make the input transistors perfectly DC balanced. Typically 1mV, 5mV max.
I want to know what is DC balance...
 

Nigel Goodwin

Super Moderator
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Essentially balancing the opamp/comparator so there's zero output when there's zero input - think of it has a pair of scales, the inputs are the two sides, the output is the pointer that shows it's balanced!.

Put equal weights on either side, and if it's NOT balanced the pointer won't be in the middle - you adjust it (balance it) until it IS in the middle, same principle!.
 

bananasiong

New Member
so..
the input bias current of the LM393 is 25nA, 250nA max. with the 100k resistor, it gives only 2.5mV. it should be more than 5mV according to the input offset voltage. right or wrong??
 

dknguyen

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It's so you balancing the DC-bias at both inputs (by adding caps and resistors to make both inputs see the same impedence to ground) so that they are common-mode rejected by the op-amp isnt it?
 

bananasiong

New Member
don't understand..
the voltage provided by the external resistor (by multiplying the resistance with the input bias current) should be more than the input offset voltage listed in the datasheet in order to make it balance???
 

Nigel Goodwin

Super Moderator
Most Helpful Member
bananasiong said:
don't understand..
the voltage provided by the external resistor (by multiplying the resistance with the input bias current) should be more than the input offset voltage listed in the datasheet in order to make it balance???

Perhaps you might like to explain why you're asking this?, is it still to do with your magnetic field sensor?.
 

bananasiong

New Member
Nigel Goodwin said:
Perhaps you might like to explain why you're asking this?, is it still to do with your magnetic field sensor?.
Nono... nothing to do with the magnetic field..
i just want to know the operation of the LM393. The "input offset voltage" part, i don't really understand.
 

dknguyen

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Most Helpful Member
INput offset and bias currents are related but not the same thing. If the DC bias of the + and- terminals are Ip and In, respectively, then:
Input Bias Current = 0.5(Ip+In)
Input Offset Current = |Ip-In|

My text says offset voltage is due to mismatches in components while the input offset/bias current is due to non-idealities that cause some current to enter through the op-amp inputs.

Input bias currents and offset voltage are two (let's say unrelated) things. You correct for input bias currents by adding resistors and caps to make both inverter inputs see the same impedence to ground.

To correct for offset voltage, you add a voltage source of equal magnitude and opposite polarity to the offset voltage in series with the non-inverting input of the op-amp. (A pain in the ass to do, so it's easier to get an op-amp that has two extra terminals that you use to trim the offset voltage with a potentiometer, but that doesn't consider that the offset voltage drifts with temperature).
 

bananasiong

New Member
To correct for offset voltage, you add a voltage source of equal magnitude and opposite polarity to the offset voltage in series with the non-inverting input of the op-amp.
is it caused by the resistor from the non-inverting to GND? How if the magnitude is higher?
 

dknguyen

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Offset voltage is ALWAYS there, and is not produced by external components (although I assume it could be affected by them). It comes from the halves of the differential inputs inside the op-amp not being perfectly matched.

However, a SECOND offset voltage can be made because the bias current of the two op-amp inputs is travelling through different resistances. You can correct for this by adding resistors/capacitors to the non-inverting input to make the bias currnts travel through the same resistance (and therefore get common-mode rejected by the op-amp differential inputs). That is probably why you connected the non-inverting input to ground through a resistor rather than just connecting it directly to ground. The resistor is there to make both opamp inputs see the same impedence. This offset voltage, however, is not in the datasheets because it depends on the components around the op-amp.
 
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audioguru

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You connected a 100k resistor to ground at one input to force that input to typically +2.5mV due to the typical 25nA of bias current through the resistor.

You are correct in thinking that it should be more than the max input offset voltage of 5mV, because at only 2.5mV the output could be the reverse voltage from what you want if the input bias voltage is max and the opposite polarity to that created by the 100k resistor.
 

bananasiong

New Member
dknguyen said:
Offset voltage is ALWAYS there, and is not produced by external components (although I assume it could be affected by them). It comes from the halves of the differential inputs inside the op-amp not being perfectly matched.

However, a SECOND offset voltage can be made because the bias current of the two op-amp inputs is travelling through different resistances. You can correct for this by adding resistors/capacitors to the non-inverting input to make the bias currnts travel through the same resistance (and therefore get common-mode rejected by the op-amp differential inputs). That is probably why you connected the non-inverting input to ground through a resistor rather than just connecting it directly to ground. The resistor is there to make both opamp inputs see the same impedence. This offset voltage, however, is not in the datasheets because it depends on the components around the op-amp.

audioguru said:
You connected a 100k resistor to ground at one input to force that input to typically +2.5mV due to the typical 25nA of bias current through the resistor.

You are correct in thinking that it should be more than the max input offset voltage of 5mV, because at only 2.5mV the output could be the reverse voltage from what you want if the input bias voltage is max and the opposite polarity to that created by the 100k resistor.


You mean the resistances for both the inputs of the comparator should be the same? But in my circuit, non inverting is conected to the 100k and the 0.1uF to the GND. While the inverting input is conected to a tank circuit. How can their resistance to be the same? The circuit works well if i replace the 100k with a 270k.
 

dknguyen

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bananasiong said:
You mean the resistances for both the inputs of the comparator should be the same? But in my circuit, non inverting is conected to the 100k and the 0.1uF to the GND. While the inverting input is conected to a tank circuit. How can their resistance to be the same? The circuit works well if i replace the 100k with a 270k.

Can I see the circuit? If you did not design the circuit (I am assuming you didn't since you don't seem to be sure about certain things), then the designer might have already compensated the impedences at the inputs for you. That is probably why the non-inverting input is connected to ground through a capacitor and resistor. From what it sounds like so far (as far as I can say without a schematic), it seems the circuit would work just fine if you connected the non-inverting input straight to ground. Then the reason the capacitor and resistor is there is to compensate for the impedences at the inverting input. It may have already be done for you.
 
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bananasiong

New Member
dknguyen said:
Can I see the circuit? If you did not design the circuit (I am assuming you didn't since you don't seem to be sure about certain things), then the designer might have already compensated the impedences at the inputs for you. That is probably why the non-inverting input is connected to ground through a capacitor and resistor. From what it sounds like so far (as far as I can say without a schematic), it seems the circuit would work just fine if you connected the non-inverting input straight to ground. Then the reason the capacitor and resistor is there is to compensate for the impedences at the inverting input. It may have already be done for you.
this is the circuit.
 

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audioguru

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The PNP transistors at the inputs of an LM393 have a typical input bias current of 25nA. You want the output to be low without a signal so the inverting input must be biased at least +5mV to cancel any input bias voltage and a little more to cancel any input offset current. The 270k resistor raises the voltage at the inverting input to typically 6.75mV.

Your circuit doesn't want the resistance at both inputs to be the same.
 
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dknguyen

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Yeah, it looks like your circuit wants a non-zero bias, so the impedences on + and - can't match. If you wanted to get as close to zero bias as possible by compensating for bias currents, that's when you make the + and - input impedences the same. But that's not your case.
 
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