OPAMP Voltage Follower/regulator

By using a emitter-follower or a Darlington pair,

A voltage-follower op amp configuration can source higher currents than the OP amp otherwise could.

With high-voltage regulators, powering the IC through the drive resistor for the pass transistors can become quite ineffic Ient.

This was avoided with the circuit in Figure 25.

The supply current to the IC is derived from Q1.

This allows R4 is increased by an order of magnitude without affecting the dropout voltage.

Selection of the output transistors would depend on voltage requirements.

For output voltages above 200V, it could be more economical to cascade lower-voltage transistors.

Current Buffer(Voltage follower)

This was really simple-use an N channel FET and has it as a source follower.

You can even use a BJT.

The one below have gain due to the 3k3 feedback and the 1k to ground from-vin.

If you don ' t want gain connect the output directly To-vin and omit the 1k.

Quite simply the feedback on the Op-amp tries to maintain the emitter at precisely the same voltage as Vin.

A unity gain buffer on the output of an op-amp was either an emitter follower or a source follower.

Simple as That-feedback from the Emitter/source back to inverting input of the op-amp.

Additionally, because the source/emitter voltage "follows" the Op-amps output signal,

The gate/base loading effects is minimal hence when using a mosfets you don ' t need to worry about gate capacitance.

Think about this sensibly-analog Devices or TI or MAXIM of Lt-their marketing team is not going to wake up O Ne morning

And say to their designers-why can ' t i design an op-amp, allows someone to add a gain stage on it and expect it to Be stable.

If They did, the designers would say that they ' d has to reduce the performance of the Op-amp for it to be stable

-Just how would this op-amp compete in the "the market" against all the Op-amps "that take the sensible road and keep building W Hat they is good at.

The output equals the input because the Op-amp has to have both inverting and non-inverting inputs at the same voltage.

OK Choose an op-amp, has a low vos and a low vos drift.

Choose an op-amp that have a very low noise figure if the sensor circuits is feeding is used for very small measureme Nts.

If your goal is simply to increase the current sourcing capabilityof an op-amp, the canonical-by-doing it is:

A transistor used in this configuration is known as an emitter follower.

The configuration is also known as "common Collector", since the collector is shared between the input and the output of T He transistor.

Your feedback elements go in the dotted box according to what are want the op-amp to does, just as if you hadn ' t added the T Ransistors.

By incorporating Q1 in the feedback loop, the Op-amp ' s gain would make significantly reduce any non-linearities in the Tran Sistor.

Further, if you already has an op-amp circuit that does what do you want, except you need it to source + current,

Adding the transistor this won ' t significantly alter the function of your circuit.

If you also need to sink:

This configuration, where one transistor sources and the other sinks current, is known as Push-pull.

Op AMP regulator with Series-pass transistor

What is the function of a voltage regulator circuit? It ' s basically this-

Maintain a precise voltage regardless of the current drawn by the load. Three basic components is needed to achieve good voltage regulation.

1. A Precision Reference (Zener diode) to set the output voltage.

2. A muscle Component (transistor) to deliver, the required current.

3. An automatic controller (OPAMP) to adjust the transistor drive.

The "Prime directive" of the OP amp is to adjust the base drive of Q1 delivering the required load current

While keeping the output voltage at a fixed value.

OUTPUT VOLTAGE

Resistors RF1 and RF2 feed a fraction of the regulator output Vo to the op amp ' s negative input v.

The op amp then adjusts the drive to Q1 such that v-is equal to the Zener voltage Vz.

When this occurs, the output voltage are related to the Zener voltage through the RF1, RF2 divider by

With the Zener voltage at about 3V and rf2=10k, rf1=5k,

The output voltage should be 3 x (1+10/5) = 9V.

If you are need to source current and not sink it as well,

You could get by with a garden-variety op amp and a pass transistor.

Here ' s the basic topology:

Here's another circuit, similar to Phil's, except it uses an NPN transistor

With the output taken from the emitter instead of a PNP with output taken from the collector.

This circuit uses one of the both DAC outputs of the The TLV5618-bit DAC.

The power supply have to at least 3 or 4 Volts above the desired output,

Because of overhead of the LM358 Op-amp plus the overhead of the transistor.

There is better choices for the output transistor and op-amp if you had constraints on the input power supply.

In terms of circuit configurations, in Phil's circuit with the PNP, the transistor common emitter gain is inside the Feedb ACK Loop,

Which makes it harder to stabilize and it'll require a relatively large output capacitor.

The NPN follower gain is X1, which makes it easier-stabilize with a small capacitor.,

and base current contributes to the output of current.

But the PNP circuit are the best if you need to operate very close to the power supply rail.

Another option, if you really don ' t need to get the down to zero volts,

is to use a standard voltage regulator controlled by a digital potentiometer.

Those usually has a minimum voltage of something like 1.25 volts.

Voltage Regulator with op amp (negative feedback) and MOSFET

The schematic above is a very good voltage regulator, excepting of the high power dissipation for Q1 (also I ' d use of an NPN ins Tead of a PNP).

Can I replace Q1 and Q2 with a high current n-mosfet and expect the same behavior?

A MOSFET has lower power dissipation than a BJT.

It ' s also not a very good voltage regulator unless Vin is already regulated:

The voltage across Zener D1 is somewhat dependant on the current, and which is (Vin-v (D1))/R.

That's dependence, or "slope resistance" of the Zener differs for different voltages of Zener diode.

I ' d also say it ' s not guaranteed to be a good regulator in general applications

Because there is a good chance it would oscillate unless A1 is a very poor op-amp.

You can ' t say it's a good regulator without knowing what the component values is too.

Ditto Brian's comment about the Zener and I ' d add that if there is significant impedance in the line feeding it's d be E Ven worse!!

The new circuit work the same as the existing one,

You ' d need to replace the Q1 with a p-channel MOSFETs, not a n-channel one.

However, you would not save any power dissipation by doing this.

Because wouldn ' t operate the MOSFET fully switched, you ' d be operating it as a variable resistor.

Can see that there's no-significantly improve the power dissipation of this circuit by the equation for device Power

P = I * V.

In this circuit, the current through Q1 are equal to the load current and so you can ' t change that.

and the voltage across Q1 (collector to emitter) are equal to Vin-vout, and you can ' t.

So no matter what do you replace Q1 with (however good a mosfets you find), the power dissipation would still be

Iload * (vin-vout).

If you want to reduce the power dissipation from your regulator,

You need-to-a totally different type of circuit, like a buck switching regulator.

Here is another op-amp based voltage Regulator/power supply:

The Zener diode (Z1) provides a stable reference at the positive input of the OP amp.

The OP amp adjusts its output until the sample voltage at the negative input formed by the R2-R3 voltage divider equals th e Zener voltage.

Output voltage control is then maintained through a wide range of load current and input voltage.

Unfortunately, this circuit was not very efficient.

Efficiency can significantly improved with switching regulators.

Fixed Output Voltage feedback networks

A linear regulator includes an error amplifier that compares the voltage applied to its feedback pin to an internal re Ference.

The regulator contains a controller with a feedback voltage amplifier this compares an external voltage,

Derived from the regulator's output voltage through a resistive divider network, to a fixed internal voltage reference.

The output of the error amplifier drives a series pass element, or power stage, which delivers current to the output.

By setting the external resistors, Rf1 and Rf2, you can set the regulator ' output voltage

to any voltage greater or equal to the internal reference.

In this case, the operational amplifier opens the transistor more if the voltage at its inverting input drops Significantl Y

Below the output of the voltage reference at the non-inverting input.

Using the voltage divider (R1, R2 and R3) allows choice of the arbitrary output voltage between Uz and Uin.

is opamps as a voltage regulator effcient?

If I setup an opamp in a non-inverting configuration as follows:

Vrail+ is 7.5V, Vrail-is GND (0 V)

Vin is 2.5V, Vout = 3.3V (in other words, the gain, aka, 1+r2/r1 is 1.32 [v/v.])

Iout = 100mA (connected to some load)

What is the source of my losses? How efficient is these types of configurations?

Does I only take the quiescent current as the loss?

Opamps can ' t usually supply mA.

But, they can still is used to control a power voltage if you add some current gain to their output.

If you really want a opamp to control a power voltage the can supply 100s of MA, here are a simple-to:

The Opamp still does the controlling and still provides voltage gain from Vin to Vout.

However, Q1 provides most of the Vout current.

R2 and R3 are the voltage feedback and the overall gain is the reciprocal of their attenuation ratio.

In this example R2 and R3 be equal, so the voltage gain from Vin to Vout is 2.

You might think all need are the transistor and R2 and R3, but this could be unstable.

To make the Opamp stable, you put a small cap directly between its output and inverting input.

But, you also need to let the opamp output isn't being loaded directly with the end load, especially if-load is capacitive .

R1 decouples the Opamp output from the load such, the Opamp should is able to reach whatever output voltage it wants t O.

Without that, the stability feedback provided to C1 won ' t work right and the thing can oscillate anyway.

The best value of C1 are hard-to-predict, so-start with something vaguely plausible, like I-show, and then experiment.

Making it lower would invite instability and result in oscillations.

Making it higher would dampen the response to transient loads.

Test by connecting the worst case non-resistive load you expect,

Find the value of C1 just where it starts to is unstable, and then double it in the real circuit.

However, stepping back a couple of layers in the case, you don ' t need a opamp voltage regulator

Since the voltage you want are both a common value and known up front.

There is plenty of 3-terminal linear regulators specifically designed for this task the can source.

If It's OK to waste the extra power as heat in the linear regulator, then this is the clearly the the-to go-your case.

Dropping 7.5v-3.3v = 4.2V, which times is 420 MW.

A linear regulator in a TO-220 or similar case would get warm, but won ' t require a heat sink.

If efficiency or getting rid of heat is a issue and then use a buck regulator to make the 3.3 V supply from the existing 7.5 V Supply.

You won ' t is able to supply 100mA by all standard Op-amps-

Maybe 20mA for most and up to 50mA for more specialist devices.

There is such things as power op-amps that can deliver amps but these is specialist devices.

Regarding Efficiency-the power delivered to the load might be 3.3V x 20mA = 66mW

And the power into the Opamp would be a 7.5V x (20mA + ~5ma for the Op-amp) = 187.5mW.

The quiescent current of the device is 3mA rising to 5mA to produce the output current.

There ' s a bit of hand-waving but even if you ignored any current taken by the Op-amp

You still has 7.5V powering it and 20mA being delivered to the load = 150mW.

If you ran the op-amp from a 5V supply it would was better but remember it was a linear device

That isn ' t like a switch mode power supply-if your output was 20mA this 20mA gets taken from the Op-amps power rails.

This circuit gives the output voltage as was set for (maximum output are limited by the rails)

But it'll is not being well regulated anymore at higher load.

This also depends on the strength of the op amp of interest.

There is lots of high current capability (amperes) OP amps available now out there.

This circuit isn ' t a voltage regulator by any chance.

It ' ll oscillate with high capacitive load, and can ' t sustain line or load transients.

Voltage Regulator Calculator

This calculator is created to give the designer of the optimum values for voltage regulator set resistors,

Given a value for the reference voltage. See the description below the calculator for more

Can I use a op-amp in voltage follower configuration
As variable voltage power supply?

I see the This works in theory, but I am gut feeling says there is a drawback or caveat of which I ' m not aware.

Basically what I want to achieve are described best by the circuit diagram below.

I have the A power supply (2 a @ 12VDC) that I want to the make a variable voltage power supply.

In the voltage follower configuration, whatever are on the non-inverting terminal on the Op-amp are present on the output te Rminal.

The extra current in the output terminal needed to keep the voltage on is provided by the Op-amp ' s supply.

Since I ' m using a linear pot as a voltage divider to set the input voltage to the Op-amp,

Would this work as a effective variable power supply from 0V to almost 12V?

If not, what can this configuration is not being used this?

The problem is, and the output current from most op amps is weak at best.

If you use the OP amp to control a p-type high-side transistor instead then you'll have yourself a LDO regulator.

It can be used op-amps the maximum current

You would is able to drive from the output would be limited to less than 50mA.

Op-amps is not miracle workers!

On the other hand you can attach an NPN and PNP transistor in a push-pull configuration and get more current:

OPAMP Voltage Follower/regulator