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I am looking at the voltage reference circuit used in a source measurement unit, which is meant to be stable and precise:

Original reference circuit

The circuit above is said to be a 10V reference as described in the block diagram. (You can find it here on page 132.)

I have two questions about this.

  1. I understand that this will amplify the 6.4V Zener voltage to 10V, but the resulting voltage is not exactly 10V as I have calculated. It is about 10.3V instead. Even though the additional R76 is added, it will be even greater than 10.3V. Why is this referred to as 10V?
  2. The top of the Zener diode is not connected to 15V but the op-amp output instead. I can also see a low-pass filter along with it to filter out any high-frequency noises for stability. Why isn't it connected to +15V? Is there any difference?

I have done the simulation in terms of noise. The result shows no difference between both, see the image below. The noise signal is a 15V with noise, which is defined as V=(rand(time*1e18)-0.5)*1.0+15).

Noise addon comparison

EDIT 1:

I just noticed that the 15V might use a different ground reference point compared to the Zener diode's. Not sure whether this is the underlying reason of this.

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  • \$\begingroup\$ What are the accuracy requirements for the circuit that will use this reference? \$\endgroup\$
    – devnull
    Jul 19, 2022 at 10:25
  • \$\begingroup\$ There is no description of the spec of this 10V reference in the entire document so I have no idea about it. \$\endgroup\$
    – ONLYA
    Jul 19, 2022 at 10:28
  • \$\begingroup\$ I am not too deep into the accuracy spec. There are different accuracy specs with different output voltage ranges, but they all have the same percentage in the front, which is like +/-(0.033%+xxmV) or +/-(0.033%+xxuV). \$\endgroup\$
    – ONLYA
    Jul 19, 2022 at 10:45

2 Answers 2

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  1. devnull has a great answer so I have nothing to add.

  2. Think of this like self biasing. It's basically a stability improvement: We don't know how the 15V line is stable i.e. it may increase or decrease so if the R96 was tied to this 15V source then the zener voltage would move with supply since the zener was not biased through a stable current source. But we know that the 10V output will be stable regardless of the supply voltage (within the acceptable range, of course). So basically the zener will be driven through a current source formed by the stable 10V source and R96 (possibly 1% tolerance at max). Therefore, the 10V ref output will be rocksteady against the supply voltage changes.

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  • \$\begingroup\$ Why my simulation result shows similar noise variation in the outputs when there is noise added in the supply line? \$\endgroup\$
    – ONLYA
    Jul 19, 2022 at 11:32
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    \$\begingroup\$ @ONLYA I built up the circuit in LTSpice and simulated it (I downloaded the same zener model). I put 100 mVpp 100 kHz sine on top of 15V. There was no noise across the zener because 100n ate all the noise, and I observed some noise on the output but the amplitude was 0.3 mVpp. \$\endgroup\$
    – Rohat Kılıç
    Jul 19, 2022 at 12:08
  • \$\begingroup\$ I believe this will depend on different diode models while this principle should be universally applied to the circuits with all Zeners. I also used the same setup with the originally supplied 6.4V Zener diode. It comes with 0.08mV out of 2.48mV difference between the two circuits. \$\endgroup\$
    – ONLYA
    Jul 19, 2022 at 20:53
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As it can be seen in this datasheet, the 1N4579 (which, according to the document you linked, is the actual one being used) is a "6.4 Volt Temperature Compensated Zener". It is not supposed to be accurate around the 6.4V (available between 1% and 5%), and this is why the R76 resistor is added (for trimming).

This Zener is stable regarding temperature drift (but the circuit must be trimmed for 10V):

enter image description here

[...] it will be even greater than 10.3V. Why does this refer as 10V?

The Zener used in the simulated circuit doesn't present the same reverse polarization voltage at the given current.

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    \$\begingroup\$ So even if the accuracy variation makes it "greater" than 6.4V, the operating current will drag it down so this circuit will always be able to generate 10V. Thanks! This first question has confused me for a while. I don't know how to give you a half-answered status as this only answers my first question but this definitely should be an answer. \$\endgroup\$
    – ONLYA
    Jul 19, 2022 at 11:20

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