How do rail-to-rail op amps get so close to the rail voltages?

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How do rail-to-rail op amps get so close to the rail voltages?
I'm just curious, I have looked for application notes and such but don't see a simple answer to this question. Given the band gap of semiconductors that are very obvious in devices like diodes and transistors and such, how do designers achieve 'rail-to-rail' performance in an op amp, and get outputs to within 50mV of the supply voltages?
 

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DaveE
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In short, MOSFETs. Look at figure 2 here, for example.
 
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Baluncore
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Given the band gap of semiconductors that are very obvious in devices like diodes and transistors and such, how do designers achieve 'rail-to-rail' performance in an op amp, and get outputs to within 50mV of the supply voltages?
You must differentiate between R-R inputs and R-R output.

Because a BJT can be saturated, Vce can fall below Vbe. Saturation voltages of 20 mV are quite possible at the output, while the base is still at 600 mV. That makes R-R with a BJT output stage almost possible.

Low quiescent current, and very low power, do not favour BJTs because a base current must flow, so MOSFETs have taken over from BJTs for R-R output stages.

MOSFETS can also reach further beyond the supply rails because the gate-source voltage can be designed to be much greater, but then offset voltage is more of a problem than with BJTs.

The advent of auto-zero switched capacitor circuits that work to cancel input offset voltage have significantly improved MOSFET op-amps.
 
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OK, thanks. I had assumed there was still a band gap drop once switched. So, in fact, I guess that means if you want to minimise a voltage drop in a rectifier, then don't use a diode but use a switching transistor instead!?
 
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DaveE
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OK, thanks. I had assumed there was still a band gap drop once switched. So, in fact, I guess that means if you want to minimise a voltage drop in a rectifier, then don't use a diode but use a switching transistor instead!?
Yes, exactly. You can look up "synchronous rectification" in switching power supplies.

FETs don't really have a saturation voltage like BJTs, the simplest model has the Drain-Source acting like a resistor in saturation.
 
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cmb
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OK, thanks for that, and before I try to invent another circuit already invented (!), and the reason I am thinking about this - can you tell me if there is an existing solution to this?-

Given the very high input impedance of opamps, is it possible somehow to read a high voltage (i.e. one where you normally have to have a potential divider) without drawing any current?

I am thinking the only way is to generate a known reference voltage that you can draw a current from, connect the two potentials via a floating comparator, and then adjust the reference voltage until there is zero current between them?

Is there any solution a wee but simpler than that?

thanks.
 
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Baluncore
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Given the very high input impedance of opamps, is it possible somehow to read a high voltage (i.e. one where you normally have to have a potential divider) without drawing any current?
What range of voltage do you actually require to measure?
 
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DaveE
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If you literally mean zero current, then that will be really, really hard. However, it can be done with nearly zero current. For example, you can buy this instrument, which can measure up to 1KV with only 0.01 fA. I don't know how they do it for this one, but there is a lot of information on the web if you search for high voltage electrometer.
 
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Tom.G
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Not terribly hard, there are commercial units that use optical sensing of high voltages. They generally use the Pockels effect, but both the Kerr effect and the Faraday effect could be used.

The general approach is taking advantage of light transmission through an optical fiber that is affected by an electrical field, for instance a fiber that rotates the polarization of light when exposed to an electric field.

Here is a 50 year old patent for the approach:
https://patents.google.com/patent/US3466541
 
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Baluncore
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Given the very high input impedance of opamps, is it possible somehow to read a high voltage (i.e. one where you normally have to have a potential divider) without drawing any current?
You can measure a high DC voltage by using a "flying capacitor" and an electrostatic commutator made from a movable or vibrating capacitive reed.

Start by generating a similar voltage, with a potential divider and a D-A converter in the voltage control loop, giving HV steps of about 1 volt. The top terminal of the flying capacitor is connected to the reference voltage, while the bottom terminal is connected to the temporarily grounded input of an electrometer buffer amplifier.

Next the grounding at the bottom is removed while the top terminal is moved from contact with the reference to be closer to the input voltage. The voltage gradient between the reference and the unknown input then appears on the electrometer amplifier. The amplifier output will change in proportion to the sign and magnitude of the error voltage. A binary search for the unknown input voltage can be made quickly.

Once the approximate input voltage is known, the commutator can contact the HT input, so it can measure and track the HV by doing an A-D conversion of the electrometer amplifier output. There needs to be a short recharge of the flying capacitor every so often, between readings, to cancel drift due to capacitive leakage. An alternating system that employs two flying capacitors can provide a continuous output.
 

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