Output Current Limit

CIRCUIT

Schematic: Op_Amp_ilim.asc
Op Amp Symbol: Opamp_3.asy
Op Amp Shematic: Opamp_3.asc
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SPEC IT

Intro	The output stage of an op amp is designed to limit the output current to a maximum level. Why? If the output is accidentally short circuited, the current limit prevents the op amp from going up in smoke.
Definition	Ilim - defines the maximum output current
Design Goal	Deliver a 10V signal into the expected load
Design Spec	Vo ≥ 10V into a 200 Ω load

DESIGN IT

• Circuit Design
  • Unity Gain: Kcl = 1V/V
  • R2=10k, R1=open
  • Rload = 200 Ω
• Op Model Param
  • Aol=1e6 fu=1e6 slew=5 vlim=1.5 ilim=0.04 ro=1
  • Max output current is 40mA.
• Circuit Test
  • Drive the input to 10V
     .step param Vstep list 1V 5V 10V
      Repeats simulation for 3 levels of Vstep.
  • Expected Vo ≥ 10V

TEST IT

• Run a Transient simulation. (.TRAN)
• Plot v(vs), v(vo) and measure the highest output voltage level.
• Add another plot and add trace i(rload). Measure the highest output current level.
• Did it meet spec?
  • If Vo ≥ 10V, then PASS, else FAIL
•  DESIGN ISSUE: The output current fails to meet spec! 

SOLVE IT

• Incrementally increase Ilim.
• Rerun the SPICE sim.
• What Ilim is needed to meet spec?

THEORY REFRESH

• How does the output stage regulate a max current?
  Let's take a look at a typical positive output structure.

• What components make up the final output stage?
  • Q7 - the output transistor that delivers positive output current.
  • R5 - the resistor used to sense the output current.
  • Q9 - the transistor that regulates max current.
• Why is a current limit needed?
  • If a positive vout is shorted to GND (0V), then almost all of the VCC rail voltage falls across Q7!
  • Consequently the power in Q7 (at max current) rises to approximately P_Q7 = Io_max x Vcc.
  • Q7 must be a physically large enough device to keep its junction temperature low.
  •  Same holds true for the negative output transistor and supply.
• So how do they work together?
  • Q7 delivers Iout into the load.
  • Q7 requires a base current Ib ≈ Iout / Β where Β is it's current gain.
  • The previous stage (unseen to the left) raises the diode string voltage to allow enough of the 150uA to flow into the Q7 base for the required Iout.
  • During normal operation Q9 is OFF.
  • However, if the current through R5 develops a voltage large enough so that R5*Iout = Vbe_Q9 > 0.65V, then Q9's Vbe turns on and steals away Q7 base current to regulate Iout to a level: 
    Iout_max = Vbe / R5.
• The specific Ilim for an op amp device will vary depending on its output structure and actual devices.

• What happened to our original design?
  • The original op model had a current limit of Ilim = 40mA
  • The expected max output current required is
      Iout_max = 10V / 200 = 50mA

• Let's solve it
  • Set Ilim for the Iout_max greater than max expected max.
  • Retest the circuit with the new Vlim. Did your circuit meet spec?

   Select an op amp that meets or exceeds your required Ilim.

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