# Constant Current Discharge circuit question

#### Lionel C

Joined Feb 1, 2017
1
I was wondering if someone could help me learn the theory behind this as I'm pretty new to electronics and op-amps aren't my strength. I'm building a constant current discharge device and I am very curious why this works. The picture is from a video I was watching [ see bottom]

Why is I = Vin/R? I cant figure that out. If I attach a lipo battery (4.2V) to the load, shouldn't the current going down through the 1 ohm/5W resistor be I = 4.2/1ohm = 4.2A? Changing the trim pot changes the current, I don't fully understand that either. I've listened to the video over and over but it's not making any sense. Thank you in advance!

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#### dl324

Joined Mar 30, 2015
16,162
Welcome to AAC!

One of the fundamental laws of ideal opamps is that the voltage difference between the inputs is zero.

If you apply a voltage to the noninverting input, the opamp will try to make the voltage at the inverting input the same.

With a non ideal opamp like LM358, the input and output voltage ranges don't include the positive supply.

#### OBW0549

Joined Mar 2, 2015
3,566
This is a classic "constant current sink" circuit. When an op amp is connected in a circuit giving it negative feedback (like here), the op amp always tries to adjust its output so as to keep the voltages on its (+) and (-) inputs equal. In this case, if the current through R1 is such that the voltage across R1 is less than the voltage coming from potentiometer R2, the op amp output goes positive, turning MOSFET Q1 on more and letting more current through. If the current through R1 is such that the voltage across R1 is less than that coming from the potentiometer, the op amp outputs a less positive voltage, causing Q1 to conduct less. In this way, the op amp regulates the current drawn from your battery to a constant value, as set by the potentiometer.

Note that with a MOSFET, the current will be the same regardless of the battery voltage so long as the voltage is sufficient (i.e., the voltage across R1 plus a few tenths of a volt). MOSFET drain current changes little with changes in drain-to-source voltage.

#### crutschow

Joined Mar 14, 2008
33,358
Basically the op amp will adjust its output voltage and thus Q1's current until the voltage across R1 equals Vin (due to the negative feedback loop).
That's because the high gain of the op amp causes it to try to keep the difference in voltage between the two inputs very close to 0V.

So since the voltage across R1 now essentially equals Vin, then the current through R1 equals Vin/R1.
This will change very little with any battery voltage change since Q1 is absorbing any voltage difference between Vbat and VR1.
Thus the battery current is essentially a constant-current.

Make sense?

#### WBahn

Joined Mar 31, 2012
29,514
I was wondering if someone could help me learn the theory behind this as I'm pretty new to electronics and op-amps aren't my strength. I'm building a constant current discharge device and I am very curious why this works. The picture is from a video I was watching [ see bottom]

Why is I = Vin/R? I cant figure that out. If I attach a lipo battery (4.2V) to the load, shouldn't the current going down through the 1 ohm/5W resistor be I = 4.2/1ohm = 4.2A?
The problem with saying that the current would be 4.2 V / 1 Ω is that Ohm's Law relates the resistance of a resistor to the current through that resistor and the voltage across THAT resistor. 4.2 V is not the voltage across the resistor. It is the sum of the voltage across the resistor and the drain/source terminals of the transistor. The opamp controls the effective resistance of the transistor channel so that the voltage across the resistor is Vref. Hence I = Vref / 1 Ω.

#### BobaMosfet

Joined Jul 1, 2009
2,102
Welcome to AAC!

One of the fundamental laws of ideal opamps is that the voltage difference between the inputs is zero.

If you apply a voltage to the noninverting input, the opamp will try to make the voltage at the inverting input the same.

With a non ideal opamp like LM358, the input and output voltage ranges don't include the positive supply.
Um... no. Your statement "If you apply a voltage to the noninverting input, the opamp will try to make the voltage at the inverting input the same." isn't fully qualified, and isn't correct.

An OpAmp without a negative feedback loop, with the inverting input grounded, and the non-inverting input taking a voltage (say anything higher than the reference signal) will output as close to the high-rail as it can go. It can't do anything else, as there is no opposite feedback loop attempting to drive it to the same level as the reference input.