# Tolerance value of SMD resistors

#### Amateur Developer

Joined May 7, 2018
11
We are using an encoder circuit, where a SMD resistor is used and i am unable to know about it's tolerance value. I am replacing the same based on the value of the healthy resister, i.e. 220Ω, so, replacement is done after 5% deviation from the same. The value written on the resistor is 2200. Please guide me about the tolerance value.

#### mcgyvr

Joined Oct 15, 2009
5,394
How can i confirm that the SMD resistor is a E96 series resistor
Does it have 2 numbers and a letter?
If not then its not EIA-96 as it only uses 2 numbers and a letter..

It has 4 numbers (2200) which means it falls into the 4 digit system and indicates its a precision surface mount so its 220 Ohm 1%

#### ericgibbs

Joined Jan 29, 2010
9,806
How can i confirm that the SMD resistor is a E96 series resistor
Did you not type the 2200 into the calculator link I posted.?
E

#### ebp

Joined Feb 8, 2018
2,332
What is marked on surface mount resistors of 1% or better tolerance varies from manufacturer to manufacturer. Some use four numerals. Some use the three-character code (which also includes the E24 values that are not also found in E96). Some do it the former way on "big" parts and the latter on "small" parts. In general, you can't go by appearance and marking to distinguish a 1% tolerance 100 ppm/°C part from a 0.05% 10 ppm/°C part, though you may be able to if you know the specific manufacturer.

220 is not an E96 value, but that does not mean it is not available in 1% or better tolerance. 4 digits implies 1% or better tolerance.

With industrial products that are not manufactured in high volume is is quite common to find 1% tolerance resistors in places where 5% tolerance would be adequate. The cost difference for "ordinary" 1% resistors is very small and it is often simplifies the bill of materials and/or reduces the cost of inventory and assembly to use 1% parts.

Joined Feb 20, 2016
3,427
In all my boards, only 1% resistors are used. As ebp says, the cost difference in not great. Particluly when getting them by the reel.

#### Amateur Developer

Joined May 7, 2018
11
Thank you for the knowledge. I will get back if i have any more queries.

Joined Feb 20, 2016
3,427
Really you need to have a go at drawing it yourself. The aim of this forum is to help, not do it for you.
That board is pretty simple so is a good one to start with.
Come back with your attempt at drawing it and we can have a look at it for you.
Also, pictures from straight above and below will help.

#### Amateur Developer

Joined May 7, 2018
11
ok i will try and get back

Joined Feb 20, 2016
3,427
Not too bad. My take on it is...

I think it is a diode on the +ve line for reverse polarity protection.
Then, each Hall sensor has a 220R decoupling resistor and a bypass cap.
The outputs are pulled up to the Hall supply with a 2K7 resistor.
So this is a quadrature rotary encoder that will give a directional 2 phase signal allowing position and direction of the shaft/motor to be determined. The other unused holes are for a different model I think.
Have a search for Quadrature Encoders to see how it all works.
The are a couple on the old type of ball mouse used with computers.

#### Amateur Developer

Joined May 7, 2018
11
You all are great, i think bull eye's being hit, will get back to you after working further

#### MrChips

Joined Oct 2, 2009
20,880
For the given application, resistor tolerance is not significant.
You can be off by 100% and the circuit will still work ok.

#### Amateur Developer

Joined May 7, 2018
11
For the given application, resistor tolerance is not significant.
You can be off by 100% and the circuit will still work ok.
Can the short capacitors or the diode's open circuit resistance fall (i.e. very less than ideal 250 Kohm) create problems

#### MrChips

Joined Oct 2, 2009
20,880
If a component is faulty that is a different matter.
Yes, a faulty component can make the circuit fail. It depends on the nature of the fault.

Joined Feb 20, 2016
3,427
Any cap that measures 0R, that is, short circuit, is a problem.
To test the diode. you need a diode test function on your multimeter. If good, one way will be open, that is the max reading displayed on the diode test, like the leads no connected, and the other way it will read something like 0.6volts. Measuring the resistance is not the way to go generally. And if a cap measures 0R, it may nor be the cap but the Hall sensor. When you measure in circuit there is always a question of what part you are actually reading.

#### Amateur Developer

Joined May 7, 2018
11
If a component is faulty that is a different matter.
Yes, a faulty component can make the circuit fail. It depends on the nature of the fault.
Ok i will come back after study. Thank you all for the knowledge

#### Amateur Developer

Joined May 7, 2018
11
Any cap that measures 0R, that is, short circuit, is a problem.
To test the diode. you need a diode test function on your multimeter. If good, one way will be open, that is the max reading displayed on the diode test, like the leads no connected, and the other way it will read something like 0.6volts. Measuring the resistance is not the way to go generally. And if a cap measures 0R, it may nor be the cap but the Hall sensor. When you measure in circuit there is always a question of what part you are actually reading.
The capacitors in some stock are showing 0R while in some stock, the diode is giving 2k resistance instead of 250k but showing 0.6V, but the maximum problem is the 220ohm resistors, giving Kohm in multimeter

#### Amateur Developer

Joined May 7, 2018
11
Either a pul
If a component is faulty that is a different matter.
Yes, a faulty component can make the circuit fail. It depends on the nature of the fault.
Either the direction is wrongly sensed or the pulse may be missed for a few seconds creating faults

#### MrChips

Joined Oct 2, 2009
20,880
I have no clue what you are saying.

You cannot measure the values of the components while in-circuit.

To observe the dynamic behaviour of the circuit you need to use an oscilloscope.