7
\$\begingroup\$

While looking at the LM399 datasheet, I noticed on the portable calibrator application there is a 5k resistor between the output of the opamp and the noninverting input. It looks like it forms a voltage divider with the 200k resistor, but I cannot figure out why they have that in there?

Portable Calibrator Circuit LM399 - Datasheet

\$\endgroup\$
  • 2
    \$\begingroup\$ Some black magick trick. \$\endgroup\$ – Marko Buršič Sep 16 '17 at 16:05
7
\$\begingroup\$

The LM399 requires that its operating current is between 0.5 mA and 10 mA but the 200 kohm cannot supply enough current for proper operation but, it can be used to "start up" the device.

The voltage may not be very accurate during start up but it will be a few volts and the op-amp will produce some amplification and then deliver enough current through the 5 kohm to properly bias the LM399.

Given that the output will have stabilized at around 10 volts, the bias current into the LM399 will be approximately 10 volts minus 6.95 volts divided by 5 kohm = 0.61 mA i.e. in the range for proper operation.

Additionally, it will keep that current fairly constant because the 10 volt output will be constant. This means little self heating (current is 20% above minimum required) and this current will not change even if the supply input rises from 12 volts to 18 volts.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ I was going to ask why they don't have it on the other circuits link the one at the very top "10V Buffered Reference" but your last sentence answered that for me thanks! Can you explain "start-up the device" a little bit more? \$\endgroup\$ – Mark Omo Sep 16 '17 at 16:17
  • \$\begingroup\$ It's basically a zener diode and with not enough current it won't regulate very well. The 200k gets a couple of volts across and it is that I refer to as start up. A bit like when you don't supply enough voltage to a motor - if you give the rotor a bit of a spin you can get the motor to begin rotations i.e. you overcome stiction. \$\endgroup\$ – Andy aka Sep 16 '17 at 19:20
  • \$\begingroup\$ @Andy aka: why don't we reduce 200k resistor to provide the necessary bias current instead? \$\endgroup\$ – anhnha Sep 17 '17 at 3:55
  • 1
    \$\begingroup\$ @anhnha you could do that but then when the supply voltage changes you get a slight change in output voltage. It's all about making a design have better performance. \$\endgroup\$ – Andy aka Sep 17 '17 at 8:10
3
\$\begingroup\$

This application states the input can be 12~18V and the Zener needs a 200k pullup to get started and then the 10V+5k maintains the constant bias current.

Since this is positive feedback but with a gain in the OA of ~1.3, the feedback loop from 5K to the ESR of the Zener of <30 Ohms is more than enough attenuation to prevent oscillation for loop gain << 1 and constant Iz.

Iz = 3V/5k= 0.6mA which reduces self-heating helps the internal heater to stability temp in a few seconds.

The 6.95V is essentially a standard 6.3V Zener in series with Vbe of 0.65V at 0.6mA thus minimizing variation due to ESR from epi process controls.

| improve this answer | |
\$\endgroup\$
3
\$\begingroup\$

It's a bootstrapped bias circuit. The LM399 mostly determines its own bias current from the 10.00V output voltage.

The 200K is used to for start-up- it has to ensure that the input voltage to the non-inverting input exceeds any negative offset voltage of the op-amp, otherwise the circuit may not start up properly.

Because of the bootstrap, the zener impedance is of much less importance to line regulation than it would be if the 200K was reduced to provide all the bias. Worst-case the 200K provides 55uA (at 18V in) whereas the 5K resistor provides 610uA so the line regulation is improved by more than 10:1 simply by using one more inexpensive resistor.

| improve this answer | |
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.