Slew rate calculations - how much do I need?

[djk]

Considering the opamp delivers less than a volt to the voltage gain stages, I would say the 7V/µS opamp is fine.

 

[MagicBox]

The way I see it: You care enough to wanting 10Khz+ fundamental 2nd, 3rd and 4th ssuppressed well enough or not? If so, you want a higher slewrate than is required to produce a basic 20KHz sine. Negative feedback can 'erase' these harmonics well enough only if it has the needed slew for the loop regulation to be effective (GBWP)

 

[djk]

.5V/µS per peak volt of output is enough slew rated for a 250Khz sine wave.

Do you have a clue as to what the gain/bandwidth product is on the MJ15015 outputs?

 

[dirkwright]

Thanks. The second D-75A arrived today and when I popped the top the oem op amp was in a socket, so I thought "what the heck" I already have the LME chips on hand. Nothing blew up so far.

 

[dirkwright]

.5V/µS per peak volt of output is enough slew rated for a 250Khz sine wave.

Do you have a clue as to what the gain/bandwidth product is on the MJ15015 outputs?

There you go:
Current-Gain - Bandwidth-Product @ IC = 1.0 Adc
fT = 0.8 MHz (Min) - NPN
fT = 2.2 MHz (Min) - PNP
Safe Operating Area - Rated to 60 V and 120 V, Respectively

The Bipolar Power Transistor is designed for high power audio, stepping motor and other linear applications. It can also be used in power switching circuits such as relay or solenoid drivers, dc-to-dc converters, inverters or for inductive loads requiring higher safe operating area than the 2N3055.

 

[dirkwright]

Here's the data sheet.

 

[stratus46]

.5V/µS per peak volt of output is enough slew rated for a 250Khz sine wave.

Do you have a clue as to what the gain/bandwidth product is on the MJ15015 outputs?

When the amp is in slew limiting, it's effectively running open loop with no feedback. You want a lot more capability than the minimum.

G²

 

[djk]

That's why we pick 0.5V/µS per peak volt of output, not 0.05V

It's got about 100dB of feedback at 20Khz, you think it needs more?

 

[dirkwright]

That's why we pick 0.5V/µS per peak volt of output, not 0.05V

It's got about 100dB of feedback at 20Khz, you think it needs more?

Well, 40 watts into 8 ohms is about 18 volts rms (for a power factor of 1), the peak value of 18V is 25.5, requiring a slew rate of over 12V/us (25.5x0.5) according to your standard, or 16 for a bandwidth of 100KHz according to another calculator. The amp is rated for 6V/us however. That seems too low to me.

You seem to be awfully defensive of this particular amp. Do you work for Crown or what's your relationship with them?

 

[djk]

"You seem to be awfully defensive of this particular amp. Do you work for Crown or what's your relationship with them? "

None, you are imagining a relationship that does't exist.

My point was that there are other things holding this amp back, and the opamp is of secondary importance.

 

[dirkwright]

"You seem to be awfully defensive of this particular amp. Do you work for Crown or what's your relationship with them? "

None, you are imagining a relationship that does't exist.

My point was that there are other things holding this amp back, and the opamp is of secondary importance.

Ok.... so what in your opinion is "holding this amp back"???

 

[dirkwright]

Considering the opamp delivers less than a volt to the voltage gain stages, I would say the 7V/µS opamp is fine.

Doesn't the fact that this op amp is rated for a slew rate of 7V mean also that the entire amplifier is limited to that as well? Crown says the amp is rated for only 6V. I'm pretty sure it's not the output or the driver stages that limit the slew rate, since they appear to be wide bandwidth devices.

There are other reasons to use the LME46740 in place of the oem chip. It has half the noise and vanishing low distortion, for example. I believe the CMRR and PSRR are higher as well.

 

[MagicBox]

No, the entire amp is not limited by the slewrate of the opamp, that is, assuming the output swing is gain proportionally larger than that of the opamp output.