Tim Green

Here is the moment you have all been waiting for – the answers to the six op amp puzzles I posted last week. You know, the ones that you have all been puzzling and puzzling until your puzzler is sore! Score your number of right answers and bestow upon yourself the appropriate ranking at the end of this blog!

Repeated for reference are some key op amp specifications in Figure 1. By the way, the specifications given in Figure 1 are, like college and the real world, not all needed to solve the puzzles!













unless otherwise noted





Operating Voltage Range







Common-Mode Voltage Range



(V-) -0.1


(V+) - 1.5


Short Circuit Current







Voltage Output Swing from Rail


IOUT=+5ma (source)







IOUT= +10ma (source)







IOUT= +20ma (source)







IOUT= -5ma (sink)







IOUT= -10ma (sink)







IOUT= -20ma (sink)





Input Offset Voltage







Input Bias Current







Input Offset Current







 Figure 1: Key Op Amp Specification for OPA735

Puzzle #1: What is Vout?

Answer: Vin=1V, Vout=2.97V

I would accept Vout = 3V for the ideal first look. This topology is a V-I (Voltage-to-Current) converter. It yields Iout for Vin and is known as the Improved Howland Current Pump. For R5=R2 and R4=R3, the single-ended to differential gain from Vin to voltage across R1 is V_R1 = Vin *(R5/R4). Then Iout = V_R1/R1. In our case, then Iout = Vin/R1 or Iout = Vin/100. And so Vout = Iout * RLoad.

Puzzle #2: What is Vout?

Answer: Vin =1V, Vout= -4V 

This is a BTL (Bridge-Tied-Load) circuit. Voltage gain across the load, Rload, is double that out of U1. This is because the output of U1 is inverted through U2.

For our given resistor values, gain of U1 is +2 and gain of U2 is -1, so we expect Vout_U1 = 2V and Vout_U2 = -1*Vout_U1 = -2V. Given the polarity of the meter, Vout, across Rload 


This is a great circuit to get positive and negative voltage across a floating load on a single supply. It will also give you double the slew rate and double the voltage across the load.

Puzzle #3: What is Vout?

Answer: 23.5mA X 150 ohms = 3.52V = Vout

Closed loop gain on this is x4. Vin of 1 means you should get 4V at Vout.  Iout = 4V/150 ohms =26.67mA. But from our specifications in Figure 1, there is a current limit of +/-23.5mA, so 23.5mA X 150 ohms = 3.52V = Vout.

We are still trying to design the virtual op amp that has unlimited output current!

Puzzle #4: What is Vout?

Answer: Anything less than Vout=4V–100mV

This is a unity gain buffer amplifier circuit, so closed loop gain is 1. So why isn’t Vout = 4V?

Look at Figure 1 and the common-mode voltage range, which is (V+) -1.5V. So for a Vcc = +5V my input-positive common-mode voltage range is only up to +3.5V. You should especially be aware of the input common-mode voltage range in unity gain buffer circuits

The real IC may not clamp at the input common mode voltage range, as the simulation macromodel did. The real part will not be as accurate in this common mode violation as it would be if you were within the common mode voltage range. So, I will take a right answer as anything less than 4V–100mV!!

Puzzle #5: What is Vout?

Answer: Indeterminate

Why isn’t Vout = 10mV *100 = 1V?  What in the heck is going on? The front-end instrumentation amplifier, U1 and U2, has a differential gain of 100. And the output difference amplifier, U3, rejects common mode and only amplifies the differential voltage out of U1 and U2 by 1. 

Well, my esteemed readers, the input thermocouple, modeled as a battery connected differentially across Vin, is floating with respect to the ground reference for the power supplies that power the input op amps, U1 and U2. This means there is no DC bias current path for the +inputs of U1 and U2. Therefore, these op amp inputs are floating.

If you build this and test it, you may find any number of Vout voltages. Most likely a saturated U3 output near Vcc or Vee. So I will take for the right answer something like – indeterminate or any answer other than 1V!

Puzzle #6: What is Vout?

Answer: Vout=3.85V

Closed loop gain on this is x4. Vin of 1 means one should get 4V at Vout. But from our specifications in Figure 1, there is a voltage output swing from rail specification of Vout=3.85V at Iout=20mA. 

In this circuit, current limit is not the limiting factor, but swing to rail versus output current is. So you need to check the “Output Voltage Swing to Rail vs Output Current” data sheet curve for your large signal swing limitations at higher currents. 

I added the specifications from the data sheet graph into Figure 1 so you wouldn’t need to look further to solve Puzzle #6.


Correct answers

Bestow-able title


Software engineer or PHD


Digital engineer or firmware engineer


Electrical engineering college intern or MSEE


BSEE recent graduate


Real-world analog board/system design engineer


Honorary Texas Instruments analog applications engineer

Thanks for taking a break from your “real work” to puzzle a bit on these op amp puzzles.  Perhaps next year you can participate and postulate again whilst I perpetuate the op amp puzzlement to a proper portrayal of preventative procedures when using op amps.

I’m Vout-ta here for now!

PS: If you missed last week’s blog and thought this was fun, you might want to check out these posts from long-time TIer and retiree Bruce Trump.


  • Hello Sreeraj,

    Thanks for reading this post!  I agree with your observations.  One should look at the "Output Voltage Swing to Rail vs. Output Current" plot in the data sheet.

  • I have to disagree with the answer for puzzle #6 too. At 20mA output current (sourcing), 3.85V is "typical". Practically it can be higher or lower than this. Definitely not an exact 3.85V!!

  • Hi,

    For puzzle #3, I believe the answer is not accurate since the short circuit current is used - which is the maximum current that flows when output voltage is forced to  zero volts. But, there is a higher voltage present at the output, and the current that is sourced in this case cannot be assumed to be the short circuit current. Without knowledge of the opamp internals, we can only guess, and conclude that the voltage will be lesser than 3.85V (specified at 20 mA) with a minimum of 3.52V.  

  • Ken:

    Not sure if we should award you points if you used SPICE!  Hope you had some entertainment and learned a few new things about op amps as I continue to do!

  • I got them all save for the thermocouple example. I allowed SPICE to trick me in knowing that it will force a reference node. Half credit!