# Portable DC Bench Power Supply

Discussion in 'Power Electronics' started by Anila, Feb 4, 2016.

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Feb 4, 2016
2. ### Harald KappModeratorModerator

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Welcome to electronicspoint.

The output current is given by I=Vout/Rload. You can control this current by either changing Vout or by changing Rload. Nothing else.
Current limiting is used to protect the power supply (or the load) from overload (too high current). When the output current rises above the limit value, the output voltage is automatically reduced such that by I=V/R the current is below the limit value.

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Feb 4, 2016
4. ### Harald KappModeratorModerator

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Nov 17, 2011
Then what did you expect, what was the behavior simulated? How did you simulate? Where or what are the differences?

I did the simulation with load resistors 100Ω, 50Ω, 20Ω, 10Ω, 5Ω, 2Ω and 1Ω. This is my result: One can clearly see that the voltage is regulated to 8 V as long as the load current is less than ~3.8 A.
When the load current reaches 3.8 A, the output voltage is reduced such that by I=V/R the current is limited to 3.8 A.

The problem may not be with the circuit but the way you simulate it.

5. ### Anila

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Feb 4, 2016
The circuit output voltage is 8V and current limit is 3.8A, if i want to change the current limit to 2A with Vout=8V, According to the data sheet of LT3081 Ilim adjust resistor can be used to change the current limit , typical ratio of current limit to resistor value is 360mA/kΩ. Then to change the current limit to 2A,Ilim adjust resistor =5.5K, but if I put this value, I didn't get the Ilim as 2A.

6. ### Harald KappModeratorModerator

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Nov 17, 2011
The value for Rlim in the downloadable .asc file differs widely from the one in the article: 104kΩ vs. 5kΩ.

1. It seems that the value in the schematic is way too high. Therefore the internal current limit of the LT3081 sets in at ~2A per IC which adds up to ~4A (3.8A) with two ICs used.
2. The equation from the datasheet is not directly applicable as two ICs share a common limit resistor in this schematic. Using values from 100Ω...5kΩ for Rlim you get different current limits. Simulate a few and figure out how Rlim relates to Ilim using linear interpolation.
Regards,
Harald

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Feb 4, 2016
8. ### Harald KappModeratorModerator

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Nov 17, 2011
To limit the current the voltage needs to be reduced to fulfill Ohm's law: I=V/R or V=I*R (which is the same).

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Feb 4, 2016
10. ### Harald KappModeratorModerator

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Nov 17, 2011
You need to learn how to read a diagram. Voltage does by no means stay constant. Due to I=V/R it cannot!
Look at e.g. the red line for Rilim=1.5k.
Voltage is 1 V for current less than ~0.4A. At I=0.4A the voltage drops very steeply (almost vertically) to zero because any increase in current needs to be countered by a fall in voltage to keep the current at the limit.

11. ### Anila

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Feb 4, 2016
For RIlim=4.53K, the diagram shows that voltage is 1V for current less than 1.45A, ie regulate to 1V when load changes from 0A to 1.45A, at 1.45A the voltage drops steeply. And If RIlim=3.01K, the voltage is 1V for current less than 0.9A.

(Vout= Rset * 50uA -------------------(1)
Rset
is the resistor connected at the set pin of IC & RIlim is the resistor connected at the Ilim pin of the IC
Rset is used to change the output voltage and RIlim is used to change the output current limit)

If I reduce RIlim from 4.53K to 3.01K , Rset=20K (remain constant) then as per the eqn(1) Vout remain at 1V for current upto 0.9A, But the Vout also reduces. ie, for current less than 0.9A, vout is less than 1V,it doesn,t regulate to 1V, why?

12. ### Harald KappModeratorModerator

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Nov 17, 2011
Show me your simulation results. In my simulation the regulator works well. You have toskip the first 80µs of the simulation, during this time the regulator settles to the set output value. Afte ~80µs, the output voltage is stable - and 1V.

13. ### Harald KappModeratorModerator

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Nov 17, 2011  