I need to reduce 5v to 1v for use as a reference voltage in an op-amp.

What would be the best way to do this? The simplest solution to me is to use a voltage divider, although I understand others have reservations about this.

Alternatively, a voltage regulator of sorts, although a quick Google, Mouser, Digikey, Farnell search yielded very few results...

Ideally I'm looking for a basic, and easy to understand solution.

  • 2
    \$\begingroup\$ How accurate does your reference voltage need to be? A simple solution (in part count) would be a 1.024V reference, but you could do it far cheaper with two resistors and an opamp buffer depending on requirements. \$\endgroup\$ – David Jul 8 '18 at 21:13
  • \$\begingroup\$ It is to be used as a differential amplifier, with 1v on the inverting equal resulting in 0v, 2v in 10v, and 0v in -10v. So, I should imagine it needs to be quite accurate. \$\endgroup\$ – 19172281 Jul 8 '18 at 21:25
  • \$\begingroup\$ How accurate do you need? 20% of precision input +\- 0.x% of +\-1.00V +\- y% regardless of input used as a comparator? x or y?? \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jul 8 '18 at 22:40

A voltage divider can work well if:

  1. Absolute accuracy is not that critical
  2. The 5V source supply is stable and accurate
  3. The current load you put on the central point of the voltage divider is 100 times less than the current you put through the two resistors of the divider will allow for ~1% variation
  4. You use decent 0.1% or 1% resistors in the voltage divider
  5. And you place a capacitor across the lower resistor in the divider to filter some of the voltage changes that are caused by sudden changes in the load current.

A very simple way to get a very decent reference is to use what is called a "shunt regulator" or "voltage reference" IC. One of these would connect to GND and to a pullup resistor to the +5V supply. The top side of the reference would be your stable voltage. If you need 1V you could look at the Analog Devices ADR510ARTZ. You can shunt up to 10mA through this part.

This picture from the Analog Devices data sheet for the ADR510 shows the typical operational behavior of the shunt type regulator.

enter image description here

Keep in mind that the shunt regulator will adjust the amount of current that it sinks (IQ) in order to keep the voltage drop across itself equal to 1V. The size of the bias resistor (RBIAS) sets the nominal current through the resistor/shunt regulator assuming no load current (IL). Once that operating point is established if the load is such that it wants to sink current then that additional current will be supplied through the bias resistor. If the load wants to source current then that extra current gets sunk through the shunt regulator.

As with any linear regulator (shunt like this one or other series pass type regulators) power dissipation is always something to consider. So if the load current nominal value is say 100uA because it just feeds impedance loads it would not be necessary to have there being 10mA flowing through the shunt regulator all the time. On the other hand if the load current had a dynamic change of say +/-5mA during operation you would want carefully pick the bias resistor such that the current variation to the load keeps the shunt regulator within its operating range. For the ADR510 that operating range of shunt current is from 100uA up to 10mA.

  • \$\begingroup\$ The ADR510ARTZ looks like it might do the job. Can you explain the need for a pull-up resistor? Would the positive terminal be the 5v supply, and the negative be the 1v output? \$\endgroup\$ – 19172281 Jul 8 '18 at 21:42
  • \$\begingroup\$ The shunt regulator has to have current flowing through it to produce the reference voltage. You tie the negative terminal to GND. The positive terminal then goes to the pullup bias resistor to your 5V supply. So for a 1V reference device the pullup bias resistor will have a 4V drop across it. So if you wanted to bias the shunt regulator to 4mA then use 1K resistor as the pullup bias. If you are familiar with a zener diode and how they work then you can think of a shunt regulator as an almost perfect zener diode. \$\endgroup\$ – Michael Karas Jul 8 '18 at 22:01
  • \$\begingroup\$ But if the positive terminal is always tied to 5v, then surely there is always current flowing? \$\endgroup\$ – 19172281 Jul 8 '18 at 22:02
  • \$\begingroup\$ The shunt regulator will stay very accurate as long as the shunt current through the device is larger than the current you extract from the node between the bias resistor and the positive side of the shunt regulator. Since the shunt regulator regulates the voltage across itself the load current variation is well accommodated for and not sensitive like it is with a resistor voltage divider. \$\endgroup\$ – Michael Karas Jul 8 '18 at 22:05
  • \$\begingroup\$ The + side of the shunt regulator does not go directly to the supply. And yes, the bias pullup resistor and the shunt regulator will always have current flowing through them. Just make sure that this shunt current is greater than the current you try source or sink from the node between the resistor and + terminal. \$\endgroup\$ – Michael Karas Jul 8 '18 at 22:08

Use a specialised voltage reference IC, like this one.


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