# OPA2134 quiescenr current

#### greierasul

##### Member
Did you know how big is the quiescent current of the output stage of the OPA2134 ? How it work in class A , output stage ? How mA, or uA ?

#### kzeprf22

##### Member
In the datasheet states 4mA per channel. This means that for 2500 to 3000 ohms load the opamp will work in class a for 10 volt peak to peak (10/2500 = 0.004). You understand off coarse that this is rough estimations just adjust your values to be on the safe side.
regards

#### Samuel Groner

##### Member
In the datasheet states 4 mA per channel.

4 mA for the entire amplifier, not the output stage alone!

There's no way to tell an accurate figure for the output stage quiescent current without detailed schematic. I'd say 1 mA as a first guess.

Samuel

#### kzeprf22

##### Member
Usually the output stage of an amplifier is ten times bigger from all the other currents. In the worst case which is equal with the vas it would be 2mA

#### AndrewT

##### R.I.P.
4 mA for the entire amplifier, not the output stage alone!

There's no way to tell an accurate figure for the output stage quiescent current without detailed schematic. I'd say 1 mA as a first guess.

Usually the output stage of an amplifier is ten times bigger from all the other currents. In the worst case which is equal with the vas it would be 2mA
Sam is much more believable.
Ib of output stage could be as low a 50uA, leaving 3.95mA for the rest of the opamp.

#### Joachim Gerhard

##### Member
You can load the output stage with 500 Ohm and then put a 1kHz sine into the input. Measure the PSU current with 100 Ohm resistors inline the plus and minus supply. You should measure the voltage over one of them Then raise the level of the input signal until you can measure a rise in supply current. That is the transition into class B.
I did that with several Opamps some time ago and found that the ones with less idle current current had also less class A. You can force class A of cause with a resistor to one of the supplies or with more sofisticated methods. This had been discussed here many time so i will not go into details.

#### Samuel Groner

##### Member
Usually the output stage of an amplifier is ten times bigger from all the other currents.

You're thinking power amplifier not IC opamp. As a rough estimate we can say that output stage quiescent current scales with maximum output current and the quiescent current of the other stages with power supply voltage. The maximum output current of a power amp is easily 100x that of an IC opamp; the supply voltage however is only 2-3x higher. Hence IC opamps can proportionally waste much less current in the output stage.

Samuel

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#### greierasul

##### Member
Joachim your practical metod it is a good rough aproximation. I'll try that in the next days.Why I was asking about this problem ? Because I have in the input stage of the amplifier a parasitic capacitance and that absorb a lot of current in the high frequency region. If the opamp will function in the B-class region at the high frequency , then I 'll have a crossover distorsion and switch-off distorsion in that region of frequency, where the negative feedback it is minimal and can not correct this. How I can't modify (in this moment) the input stage of the amplifier for the lesser parasitic capacitance , I'll try to biasing the opamp in the A-class region, or XD-class.

#### kzeprf22

##### Member
Anyway I am not sure about the numbers but if you compare the quicient current of op2134 wit others, which I done some years ago, If I remember correctly (maybe I am wrong) the opa had grate difference in current. Thus I have the opinion that the extra current is used in the output. Either way is great opamp. I also used two jfets to biased it more but I didn't noticed any audible difference. Maybe someone can!
Regards

#### Joachim Gerhard

##### Member
I think Samuel Groner measured the OPA2134 or similiar and as far as i know it does not like to loaded with much less then 2kOhm. Yes, my method is a rough estimate and when you really want to know what is going on you have to measure distortion in the real circuit you use it.