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[SOLVED] Voltage follower with Current Sinking

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Mokthar

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Hi,

I designed a voltage follower with current sinking circuit. The schematic is attached here View attachment Current Sink.pdf

Here there are 2 sources. One is for opamp operation. And the other is 11V@150uA source (say a led dimmer). The objective is to sink the current in the 11v source and bring down the voltage, so that it equals the Opamp output.

The output is fine, i.e. when i give a voltage of 1 to 10 v at the non inverting pin of the opamp the same voltage is reflected at the transistor collector.

But when there is 0 v at the non inverting pin or if the non inverting pin is left open, the output at the transistor collector becomes min (1.6V). The voltage at the base of the transistor is 0v but still the transistor conducts and sinks the current.

Why this happens and how to overcome this.

Thanks,
Mokthar
 

You can't leave the input of an op amp open.
If you connect the NI op amp pin to ground, then the transistor should not conduct.
In that case collector voltage should equal the collector supply voltage

The circuit you have is typically called a constant-current source, with the collector current being proportional to the op amp input voltage with current equal to Vin / Remitter.

But you say "when i give a voltage of 1 to 10 v at the non inverting pin of the opamp the same voltage is reflected at the transistor collector."
That is not proper circuit operation.
The transistor should carry a current equal to the input voltage divided by 1k ohm, with the collector voltage being whatever that current will give.
For 150µA, the input voltage would only be 150µA * 1kΩ = 150mV.
So if your source is limited to 150µA, then anything above 150mV input, will saturate the transistor and give you the observed behavior.

What exactly does the collector load look like?
Saying it's 11V, 150µA source is not sufficient information.
 
The voltage at the base of the transistor is 0v but still the transistor conducts and sinks the current.
Not possible with healthy and correctly connected transistor. Probably something is different from the reported conditions.

OPA336 has by the way a rated power supply range up to 5.5V. It will be surely burned with 12V supply.
 
Hi,

The Opamp NI pin is not left floating. The complete application circuit is given here.

F to V with Current Sink.png

The objective is to DIM the Led Light Intensity by controlling the Led Driver D+ and D- pins.

Here a PWM signal is generated from the uC. That signal is converted into a voltage with reference to PWM frequency.

That converted voltage is given to the OPamp circuit to sink the current of the LED Driver for reducing the light intensity. (i.e. 1 to 10 V Led Dimming).

In this circuit the 1 to 10V for dimming is acheived and working well. But when there is no PWM signal from uC, the output of LM231 is 0v. At this point the opamp NI is 0v. Also opamp output too is 0v. So there should not be any drop across the transistor but the voltage drops to min also the light intensity.

Note: When i keep the transistor and the resistor alone in between the D+ and D- pin of the Led Driver removing the Opamp. It works fine and the brightness is full either the base of transistor is connected to the D- pin (i.e. ground) or left floated.

May be the mixing of supply is the issue. But i cant get a solution to solve this. This occurs only when 0v is given to the opamp NI pin.

Mokthar
 

It's not clear to me which voltage and current conditions you see exactly in the unwanted/unexpected state. Did you take oscilloscope measurements to assure that the output stage is not somehow oscillating as guessed in post #3?

It's a simple circuit and it's state can be determined by basic measurements.
 

The LM331 as configured is a frequency to voltage converter. It is not a PWM to voltage converter.

So what are you feeding it with? Variable frequency or variable duty cycle?
 
Input is Variable frequency.

it is solved. Isolating the ground with a star point solved the issue.
 

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