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# Output Current Limit

## CIRCUIT

Schematic: Op_Amp_ilim.asc
Op Amp Symbol: Opamp_3.asy
Op Amp Shematic: Opamp_3.asc
Right Click on filename, select "Save link as...",

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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
• 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.
• 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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