How to Measure LDO's Output Current

Measure Output Current on SIMetrix

LDO model packet and test circuit for output voltage

 

Fig 1. LDO model packet and test circuit for output current

Objective

We want to know how much current our LDO can supply until OCP (Over Current Protection) circuit work and disable V_OUT. 

Condition

• V_IN = 5V because test circuit has high dropout.

• V_OUT(S) = 1.5V

  

Procedure 

To simulate output current we will need to sweep resistor R₁ from maximum to minimum value so that I_OUT will increase from minimum to maximum due to Ohm’s Law I = V / R

• R --> ∞ then I --> 0
• R --> 0  then I --> ∞  

But we can’t sweep R₁ in SIMetrix, so we are going to do it in another way. We connect V_OUT__Pto a voltage source V₂ in this case, V­_{OUT_P} = V₂– I­_OUT x R₁. The value of R₁ is set to be very small (R₁ = 1µΩ) so that V_{OUT_P} is equal to V₂(and to be able to probe the output current).

We know that V_­OUT will drop to zero when I_OUT reach its maximum value, so we will sweep V₂, which is equal to V_OUT, from 1.5V (V_OUT(S) = 1.5V) to zero and probe V_OUT and I_OUT we will get the curves of the two.

Fig 2. Choose analysis voltage

 

Result

Run SIMetrix circuit simulator

Then, we select the two curves and compare them on the graph to see how V_OUT acts when we sweep I_OUT.

As we can see, I_OUT is now the X axis which sweeps from 0 to 140mA and the red curve is V_OUT . The output current the current at which the output voltage becomes 95% of V_OUT(E)­ after gradually increasing the output current. V_OUT(E) = 1.497V, so 95% of V_OUT(E) is 1.41972598V. As we can on the graph, I_OUT  = 136mA at 95% of V_OUT(E). After 136mA, V_OUT is drawn to zero and I_OUT is drawn back to 15mA by OCP.
Hence, output current is I_OUT = 136mA.