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#### walters

Slew rate rise time

Is Slew rate a delay or offset from input to output?

Why don't transistors have slew rate?

If u input a Square waveform into a op-amp the slew rate makes the output "linear" to measure the upper and lower limits

If u input a exponential waveform or voltages does the slew rate change the exponential waveform into a "linear" voltages or waveform?

Does the Slew rate only affect the high frequencys?

Does the Slew rate change the input waveform into a Linear waveform?
Because the Slew rate produces a "Rise time" delay

Re: Slew rate rise time

Slew rate is the maximum speed at which the output can change, in V/us.
What that means is that any input waveform that varies faster than the slew rate of the opamp will be distorted and look like a ramp at the output. It does not depend on the waveform, only on how fast the voltage rises in time.

You have to understand it as a limit.
Imagine your car can do 100km/h in 10s. That means that as long as you accelerate more slowly than that, the car will follow your acceleration pattern.
But as soon as you try to accelerate faster than that, and no matter how you push on the gas, the maximum you will get is the above limit. You can step on the gas and you will not get better than that. You can start more slowly and then accelerate harder, you will still be kept below that limit.

Re: Slew rate rise time

Thanks alot VVV

If the slew rate is longer like 100us or 500us or longer what does that mean please? does that mean that lower frequency will have a ramp output?

Can baising transistors produce slew rates?

What causes slew rates inside the chip?

Slew rate rise time

Capacitance...charging and discharging leads to slew rate

Re: Slew rate rise time

Slew rate is measured in V/us.
That means that the output voltage can change at that maximum rate.
It also means that for sinusoidal outputs, the waveform gets distorted when the amplitude is higher, since the number of volts/ us increases (in the same amount of time you have to drive the output more volts).

Inside the chips, there are capacitors, used for frequency compensation, that need to be charged up and discharged. This charging/discharging is done by a current source, which represents an active load. However, the current is limited, so the capacitor takes time to charge up/ discharge.

Re: Slew rate rise time

Thanks for the help

But the charging and discharging slew rate is "linear" how can that be when capacitors charge and discharge exponential

The slew rate is linear not exponential

If i put 4 op-amps in series do i "add" the slew rates together?

If each slew rate is 100us X 4 op-amps= 400us

400us is a delay offset so slew rates cause offset time delays?

Re: Slew rate rise time

The charging/discharging of the capacitor comes from two parts - slew rate and the settling time. Slew rate is a non-linear effect due to the limitation of the current from the amplifier while the settling time which is a linear effect is due to the bandwidth of the amplifier.

The slew rate contributes to the initial linear charging and the settling time contribute to the exponential part.

Re: Slew rate rise time

Thanks for the information

What is settling time?

Which components produce the settling time please?

Re: Slew rate rise time

settle time is also caused by capacitance charge.
it should be a digital circuit related issue.

Re: Slew rate rise time

The charging of a capacitor is not "exponential" or "linear", it is given by what charges the cap. For an RC circuit, yes, the charging voltage is exponential, because the current varies exponentially, too.
But for a constant current, the voltage across the cap varies linearly.

That is because the voltage is u=1/C*∫i(t) dt

If the capacitor current is constant, the integral is a first-order function with respect to time, that is a ramp, so the voltage across the cap varies LINEARLY with time.

As I mentioned in the previous post, the internal capacitor is charged by a constant-current source, which acts as an active load. Since the current is constant, the capacitor charges linearly, so the slew rate is "linear." (I don't like this term; what I really mean is that the output voltage of an opamp experiencing slew rate limiting varies linearly with time, not being able to follow the input.)

Settling time is simply another parameter used to characterize opamps. It refers to the time needed for the output to reach the final value within a certain percentage-usually 0.1 or 0.01%. The response can be oscillatory (ringing), so this takes into effect all that and tells you when the output is within the specified percentage of the final value. It is useful for example in A/D signal conditioning, since it tells you when you can start the conversion and be certain that the result will be accurate. Look at the picture.

Re: Slew rate rise time

Thanks alof VVV for the pictures and the information

Whats the differences between a constant-current source VS a DC input to the capacitor?

I thought DC is a constant voltage and current source but it charges the capacitor exponential not linear

A constant current source chanrge the capacitor Linear but the voltage is Zero volts "Across" the capacitor?

Re: Slew rate rise time

A constant-current source is a circuit, built with active components, that maintains a current constant in time.
The name DC refers to the fact that the voltage or the current does not change its sign, that is opposed to AC.

Applying a constant current to the capacitor results in a ramp voltage across the capacitor, as shown in the previous post. The capacitor continues to charge and the voltage across it continues to increase until either the sonstant-current source can no longer follow that voltage and supply the current (has reached it voltage-compliance limit) or until the capacitor fails because the voltage is too high.

A DC voltage can be applied to the capacitor, usually through some current-limiting device, such as a resistor. At the end of the charging process (exponential if charging through a resistor), the capacitor has a DC voltage across it.

Re: Slew rate rise time

I think the Slew Rate reflects large small parameter of OPA when the OPA is overdrived by large step signal, so it is said that the Slew Rate has no relation with the band width of OPA. But the Settling Time reflects small signal parameter of OPA when the OPA is operating in the linear region, so the Settling Time has relation with the band width. Pls see the picture.

Pls correct me if i am wrong! Thks!

Re: Slew rate rise time

Hi VVV,

The settling time is a linear system effect but gives an exponential waveform. The slew rate is a non-linear effect which is due to a constant current (a limitation of current in the case of opamp) but gives a linear waveform.

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