What is the Effect of Resistors on Current Limit in a Power Supply Circuit?

[tomizzo]

The circuit schematic attached is one for a power supply. This power supply is restricted to 5 amps. However, I am having a difficult time understanding the calculations on how the resistor values are chosen to achieve this limit of 5 amps.

When using the simulator's oscilloscope, the limit is closer to 6 amps. Regardless, there has to be a way to determine this value of 5-6 amps.

The first transistor in the circuit is meant to do most of the current limiting. Can someone explain to me how the resistors affect the current allowed to flow to the load?

 

[mjhilger]

The key is to see that R2 & D2 set up the bias to the current limiting Q2. If you see that the 4 10K resistors R3 R4 R6 & R7 set up a voltage divider of approximately 1/2 rail voltage. The difference is that the divider with D2 sets up the + side of the OP-AMP. The voltage drop across D2 is approximately 0.6 volts. As long as the current through R2 is such that its voltage drop is SMALLER than the drop across D2 then the OP-AMP is peggeg to the negative rail (well not really, but it is pulling the base current through Q2 to allow it to conduct. When the voltage drop across R2 becomes greater than that across D2, then the OP-AMP output swings more to the positive rail or it basically allows a base current through Q2 that balances the voltage drop across R2 to equal the voltage drop across D2. So if you actually put a pot across D2 you could provide a current control with more fine control. But the key is to see that the circuit is basically monitoring the voltage drop across D2 compared to R2 and will not let the drop across R2 exceed that across D2. The remainder of the circuit is a voltage follower through the darlington to provide a stable voltage output.

Hope this helps.
Mitch

 

[berkeman]

The circuit schematic attached is one for a power supply. This power supply is restricted to 5 amps. However, I am having a difficult time understanding the calculations on how the resistor values are chosen to achieve this limit of 5 amps.

When using the simulator's oscilloscope, the limit is closer to 6 amps. Regardless, there has to be a way to determine this value of 5-6 amps.

The first transistor in the circuit is meant to do most of the current limiting. Can someone explain to me how the resistors affect the current allowed to flow to the load?

Good reply by Mitch.

Where did you get this circuit from? It's got some pretty serious problems, IMO. For one thing, the short-circuit recovery transient at the voltage output may be big enough to blow up whatever it is connected to...

 

[tomizzo]

The key is to see that R2 & D2 set up the bias to the current limiting Q2. If you see that the 4 10K resistors R3 R4 R6 & R7 set up a voltage divider of approximately 1/2 rail voltage. The difference is that the divider with D2 sets up the + side of the OP-AMP. The voltage drop across D2 is approximately 0.6 volts. As long as the current through R2 is such that its voltage drop is SMALLER than the drop across D2 then the OP-AMP is peggeg to the negative rail (well not really, but it is pulling the base current through Q2 to allow it to conduct. When the voltage drop across R2 becomes greater than that across D2, then the OP-AMP output swings more to the positive rail or it basically allows a base current through Q2 that balances the voltage drop across R2 to equal the voltage drop across D2. So if you actually put a pot across D2 you could provide a current control with more fine control. But the key is to see that the circuit is basically monitoring the voltage drop across D2 compared to R2 and will not let the drop across R2 exceed that across D2. The remainder of the circuit is a voltage follower through the darlington to provide a stable voltage output.

Hope this helps.
Mitch

You are saying Q2, I assume you meant Q1?

So say the voltage at the junction above R3 is 16.5 Volts. With a voltage drop of .6V across the diode D2, this would force the Op-Amps V+ terminal to be (16.5V-.6V)/2. This is the voltage across both resistor R3 and R4. The voltage across R4 with respect to ground is Op-Amp V+.


The rail voltage of the first op-amp will be (16.5V-(current through R1). Therefore, the voltage of the op-amps V- terminal should be half of that.

I understand that when the current through R1 increases, the voltage of V- will decrease and the op-amp's output will bias the transistor Q1 so that it limits the current. This in turn will increase V+ and will balance the op-amp.

However, I'm confused at how you're simply looking at the voltage across R2 and D2 while not even consider the voltage dividers that are connected to the Op-Amps input terminal. Would you care to elaborate more on the calculations which allow you to simply look at these two voltages?

 

[tomizzo]

Good reply by Mitch.

Where did you get this circuit from? It's got some pretty serious problems, IMO. For one thing, the short-circuit recovery transient at the voltage output may be big enough to blow up whatever it is connected to...

Thank you for your input. This is an educational diagram that I assume is only meant to introduce certain concepts while overlooking protection circuitry that should exist.

Also, could you by chance expand on mjhiliger's response regarding looking at the voltages across D2 and R2 as opposed to directly looking at the input voltages of the op-amp?

 

[mjhilger]

Tom, sorry about the delayed response, I've been traveling. And I apologize for the Q1 Q2 mix - it should be Q1
We will call the output of the rectifier circuit V+
So we have the following equations
V+ = R3 * I1 + 0.6 + R4 * I1; where the 0.6 is the approximate voltage drop across D2
V+ = R2 * Iout + R6 * I2 + R7 * I2
That identifies the main voltages & currents through the paths feeding the op-amp inputs controlling the base of Q1 and allows you to determine the I1 & I2 numerically. Also, the input to an op-amp is very very little current so for the analysis we ignore any current entering the + & - inputs.
And we will state that Iout must near its cutoff of about 6 amps before the circuit can cause a change, therefore the current I2 that is also flowing through R2 is much much less than Iout and basically can be ignored unless you are carrying voltage measurements out to 5 or more significant digits.
So the inputs to the op-amp are
+ input = (V+) - (R3 * I1) - 0.6
- input = (V+) - ( R2 * Iout) - (R6 * I2)
You can see from the equations that the voltage dividers are the same with the difference being the drop across R2 compared to D2.
Let me know if that does not clarify the picture.
Mitch

 

[Jony130]

U1A op amp will work as a comparator. The voltage at non-inverting input is equal to
Vref = (Vs - Vd2) * R4/(R3 + R4)
Where
Vs - the output voltage of the rectifier circuit (supply voltage).
Vd2 - D1 diode forward voltage drop (0.6V)

The voltage at inverting input is equal to:

Vn = (Vs - Ilod*R2) * R7/(R6 + R7)

And until Vn > Vref the op amp output voltage is at his negative saturation voltage (close to ground). So Q1 is turn on.
For Vn < Vref - op amp voltage reach positive saturation voltage.

And for Vn = Vref op amp will change his output from high to low or low to high.

So Iload_max ≈ (R7 *Vs - (R6 + R7)*Vref)/(R2*R7)

I Ignore R6 and R7 current.
Also you need current limiter resistor for Q1 base. Also this circuit is potentially unstable so it can oscillates badly.

 

[Averagesupernova]

Also this circuit is potentially unstable so it can oscillates badly.

Most voltage regulators have a potential to oscillate. At some frequency, intended negative feedback can turn into positive feedback causing oscillation. Generally I think op-amps in a voltage regulator is a bad idea. Regulators like the 1723 will utilize something that appears like an op-amp but the designers of this part researched many many things before releasing it for production. Following the designers recommendations on something like this is a much better choice than throwing some op-amps together with pass transistors. Discreet regulators with a transistor for gain in the negative feedback path can work ok but I have found they do not have the control of voltage transients like something wired up with a 1723 does.