0
\$\begingroup\$

I would you to help me to understand a circuit I have in my notes which is reported in the following picture.

enter image description here

First of all it's my first time in which I see the supply voltage in some ways connected to one of the opamp inputs, which looks a little bit weird to me. Secondly I would expect a 2.5 V as output voltage instead of the 5V reported here.

In fact, according to my knowledge, the opamp is in a kind of non inverting configuration, where the gain should be

\$ G = 1 + \frac{R_g}{R} \$

since \$R_g\rightarrow 0\$ and \$R\rightarrow \infty\$ then the gain is \$G = 1\$.

Furthermore the LM336 has a breakdown voltage of 2.5 V, honestly I am not sure the voltage drop is large enough to guarantee a breakdown configuration, but I don't see any sense in putting a Zener diode in confinguration which is not the breakdown one.

From these considerations, I argue that the opamp, with its gain \$G = 1\$ provides a stable tension of 2.5 V in its output, and that the current passing through R should be \$(12V-2.5V)/1k\Omega\$ which is something less than 1 mA.

Is there anything wrong with my considerations? Furthermore, why do I need this configuration to have a stable 2.5 (or 5 V in the case in which I am wrong) power supply voltage? Is there any difference with a power supply obtained with a much more usual non inverting op amp?

\$\endgroup\$
1
  • \$\begingroup\$ Your calculation for the current through the LM336 is incorrect. If it is a LM336-2.5 the current would be (12-2.5) = 9.5V. 9.5V across 1kOhm is 9.5mA. \$\endgroup\$ Feb 25 at 18:14
2
\$\begingroup\$

The LM336 comes in different voltages. Perhaps the part being used here is an LM336-5 which is a precision 5V Zener diode.

That would make the rest of your circuit agree.

Datasheet here: TI LM336-5 Datasheet

LM336-5

\$\endgroup\$
2
\$\begingroup\$

You missed a decimal point in your calculation. (12-2.5)/1K = 9.5mA , but it's actually (12-5)/1K = 7mA

It's a voltage follower and a 5V reference. The (unseen) op-amp power connections will be 0/12V.

The LM336 is old and has so-so performance, but is still available, and the LM759 is obsolete, but it's a power op-amp with unusual 325mA output current capability.

It's unclear why they would operate the LM336 at such a high current, given that it is specified at 1.0mA. The dynamic resistance is about 25% better but the 7:1 reduction in series resistance means that it will be 5x worse for line regulation. Self-heating drift will also be much worse.

A modern circuit would likely use different components, and use less power.

The circuit may be prone to oscillation with a substantial capacitive load on the 5V output, as the op-amp is not designed to drive a capacitive load. Oscillation would tend to destroy 5V devices attached to the output if you are trying to use it as a power supply.

You can get similar performance for pennies by using an LM7805 or 78M05 regulator and it is virtually impossible to coax it into oscillation.

\$\endgroup\$
1
\$\begingroup\$

There are both LM336-2.5 and LM336-5.0 reference diodes, so your circuit implies that the LM336 is the 5V variety. The op-amp is just a unity-gain buffer of the 5V reference voltage. It provides lower source impedance for the 5V signal. This can be a useful when the signal is being distributed to a few places in a circuit and is loaded down. If you change the 2.5V in your current calculation to 5V you will get the correct current.

\$\endgroup\$

This site is temporarily in read only mode and not accepting new answers.

Not the answer you're looking for? Browse other questions tagged .