I am looking for help finding a specific document that I found then lost!
Burr Brown published a description of paralleling the outputs of a dual op-amp, likely over a decade ago.
I had planned to follow the info to build a 10Watt amp - increasing the op-amp current output might be useful, I thought.
I'd sure be grateful if someone knows of this old article and could post its URL for me to download.
Cheers,
Burr Brown published a description of paralleling the outputs of a dual op-amp, likely over a decade ago.
I had planned to follow the info to build a 10Watt amp - increasing the op-amp current output might be useful, I thought.
I'd sure be grateful if someone knows of this old article and could post its URL for me to download.
Cheers,
Have you looked at Doug Selfs NE5532 power amp?
https://www.diyaudio.com/archive/blogs/alexcp/1016-ne5532-power-amplifier.html
https://www.diyaudio.com/archive/blogs/alexcp/1016-ne5532-power-amplifier.html
The "how" means electrically, not mechanically.
Anyhoo, many op amps are smt these days.
Have to use the sharing resistors.
It seems to work in the video, but what COULD happen? Are we talking smoke and fire or just poor sound or?
Perhaps using something like NJM2114 (quad of 5534 inside?) is the way to go?
Excessive current flows between the outputs of the OP-AMPs in parallel and the overcurrent detection (if equipped) is activated, preventing normal operation.
If not, it may lead to destruction. NJM2114 cannot be used in parallel too.
If you really want to operate in parallel, the 5534, which is not dual but has internal nodes drawn out, can be operated in parallel by connecting all terminals in parallel.
If not, it may lead to destruction. NJM2114 cannot be used in parallel too.
If you really want to operate in parallel, the 5534, which is not dual but has internal nodes drawn out, can be operated in parallel by connecting all terminals in parallel.
Poor sound, large supply current and reduced reliability due to the high junction temperatures if the op-amps have good built-in current limiters, as they usually have. Smoke when they haven't.
I'll give you an example:
Imagine you make an amplifier with a gain of ten using two paralleled op-amps of the same type, but due to the inevitable exemplaric spread, they behave slightly differently. The load resistance of the amplifier is 100 ohm including its feedback network.
Op-amp 1: open-loop gain 100 000, offset 0 V, current limit at +/- 21 mA. To keep things simple, I will assume an open-loop output impedance of zero.
Op-amp 2: open-loop gain 100 000, offset 100 uV, current limit at +/- 20 mA.To keep things simple, I will assume an open-loop output impedance of zero.
When the amplifier has zero input voltage, op-amp 1 will try to make the output voltage of the whole amplifier 0, while op-amp 2 would like it to be +1 mV (ten times its 100 uV offset; I've neglected the effect of the finite open-loop gain, otherwise it would be +0.9999000099990000... mV). Op-amp 1 will win because it has the highest current limit. Op-amp 2 will source as much as it can source, the full 20 mA, but op-amp 1 will sink that current and keep the output at 0 V. The current through the load will be 0, so that has no impact.
When the amplifier has an input voltage of +20 mV, op-amp 1 will try to make the output voltage of the whole amplifier 200 mV, while op-amp 2 would like it to be 201 mV. It will end up in a state where the output voltage is 200 mV, the current through the load will be 2 mA, op-amp 2 sources its full 20 mA and op-amp 1 sinks the difference, that is, 18 mA.
So far everything behaves as a single-ended class A amplifier with op-amp 2 as the current source, but that's no longer the case with negative input signals.
When the amplifier has an input voltage of -20 mV, op-amp 1 will try to make the output voltage of the whole amplifier -200 mV, while op-amp 2 would like it to be -199 mV. It will end up in a state where the output voltage is -199 mV, the current through the load will be -1.99 mA, that is, 1.99 mA into the output of the amplifier, op-amp 1 sinks its full 21 mA and op-amp 2 sources the difference, that is, 19.01 mA.
So, besides the inefficiency and the reduced reliability due to the resulting high junction temperatures, you actually get a jump in output voltage at the point where the load current is such that the winning op-amp changes.
I'll give you an example:
Imagine you make an amplifier with a gain of ten using two paralleled op-amps of the same type, but due to the inevitable exemplaric spread, they behave slightly differently. The load resistance of the amplifier is 100 ohm including its feedback network.
Op-amp 1: open-loop gain 100 000, offset 0 V, current limit at +/- 21 mA. To keep things simple, I will assume an open-loop output impedance of zero.
Op-amp 2: open-loop gain 100 000, offset 100 uV, current limit at +/- 20 mA.To keep things simple, I will assume an open-loop output impedance of zero.
When the amplifier has zero input voltage, op-amp 1 will try to make the output voltage of the whole amplifier 0, while op-amp 2 would like it to be +1 mV (ten times its 100 uV offset; I've neglected the effect of the finite open-loop gain, otherwise it would be +0.9999000099990000... mV). Op-amp 1 will win because it has the highest current limit. Op-amp 2 will source as much as it can source, the full 20 mA, but op-amp 1 will sink that current and keep the output at 0 V. The current through the load will be 0, so that has no impact.
When the amplifier has an input voltage of +20 mV, op-amp 1 will try to make the output voltage of the whole amplifier 200 mV, while op-amp 2 would like it to be 201 mV. It will end up in a state where the output voltage is 200 mV, the current through the load will be 2 mA, op-amp 2 sources its full 20 mA and op-amp 1 sinks the difference, that is, 18 mA.
So far everything behaves as a single-ended class A amplifier with op-amp 2 as the current source, but that's no longer the case with negative input signals.
When the amplifier has an input voltage of -20 mV, op-amp 1 will try to make the output voltage of the whole amplifier -200 mV, while op-amp 2 would like it to be -199 mV. It will end up in a state where the output voltage is -199 mV, the current through the load will be -1.99 mA, that is, 1.99 mA into the output of the amplifier, op-amp 1 sinks its full 21 mA and op-amp 2 sources the difference, that is, 19.01 mA.
So, besides the inefficiency and the reduced reliability due to the resulting high junction temperatures, you actually get a jump in output voltage at the point where the load current is such that the winning op-amp changes.
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I don't mean use NJM2114 in parallel, but since I understand it is essentially four NE5534 inside one NJM2114, why not just use the NJM2114?
Why did you think there were four 5534s in the package?
The NJM2114 is a dual OP-ANP and consumes 9mA (total, not each) compared to the 5534's 4mA.
I think it is reasonable to assume that it is a modification based on the 5532.
Voltage noise is slightly improved over the 5532 and is equivalent to that of the 5534. The maximum output current (the current at which the limit works) is almost unchanged.
The NJM2114 is a dual OP-ANP and consumes 9mA (total, not each) compared to the 5534's 4mA.
I think it is reasonable to assume that it is a modification based on the 5532.
Voltage noise is slightly improved over the 5532 and is equivalent to that of the 5534. The maximum output current (the current at which the limit works) is almost unchanged.