Hello all,
I play a little electric bass. Electric basses often have onboard preamps or simple buffers which have fascinated me for a while so I am starting to look a bit deeper into them.
My aim at the moment is to create a simple dual buffer, each buffer will "hide" one of the two passive pickups of the bass, before I combine the low impedance outputs to a specialized 25K MN taper pot. There's quite a few reasons I want to keep my passive pickups, but also don't want to directly connect them in parallel, because of all kinds of nasty phenomena I'm reading about and experiencing.
To this end, let's assume we have your average audio op-amp, not examining if it's really suitable for the specific characteristics of the signal coming from the passive pickup while the electric bass is being played. For example let's say we have a TL072 or NE5532 op-amp.
I would like to power the op-amp using two common alkaline (or even lithium) 9V cells. One battery provides the positive rail through its positive connector, the other provides the negative rail through its negative connector. The other connectors are connected, and in fact their junction is the actual ground of the system, our reference for audio. This keeps things nice and simple for starters (as opposed to not having to provide a bias voltage at the center of a single battery power supply). We also get double the battery capacity, double the headroom (although it's much more than is actually needed in this case) plus simplicity - reliability.
What I would like to ask about is a very specific scenario. This would involve one of the batteries not being present, or a battery lead being severed by accident. This would, at least initially, leave either the positive rail or the negative rail at zero volts, with the other one where it should be. Since we are speaking a bit generally, not referring to a very specific op amp, I would like to ask: If the two op-amps suddenly start receiving 0V positive and -9V negative, is it even remotely possible that they will get worn-damaged or completely die? If the possibility is even remotely possible, is there some power protection scheme that can be implemented using simple additional components (like diodes)? One such reaction of a safety circuit would be for example to somehow power the op-amps with +-4.5V from the battery that remains properly connected when the other one's lead is severed. This is just an example, there are probably a lot of best practices on protecting op amps at their power supply pins. Of course, all this may not be much cause for concern in the first place.
I am not thinking about input signal protection as each op-amp's input will be a passive pickup that can not really cause much damage with its output. I plan to place a 1M resistor from the op-amp's input to ground, to set the preferred input impedance and likely stabilize stuff in general.
My understanding of basic electronics, let alone op amps is pretty superficial but I think I can build and test something that will work. Any help with the input power protection aspect of this will be greatly appreciated.
I play a little electric bass. Electric basses often have onboard preamps or simple buffers which have fascinated me for a while so I am starting to look a bit deeper into them.
My aim at the moment is to create a simple dual buffer, each buffer will "hide" one of the two passive pickups of the bass, before I combine the low impedance outputs to a specialized 25K MN taper pot. There's quite a few reasons I want to keep my passive pickups, but also don't want to directly connect them in parallel, because of all kinds of nasty phenomena I'm reading about and experiencing.
To this end, let's assume we have your average audio op-amp, not examining if it's really suitable for the specific characteristics of the signal coming from the passive pickup while the electric bass is being played. For example let's say we have a TL072 or NE5532 op-amp.
I would like to power the op-amp using two common alkaline (or even lithium) 9V cells. One battery provides the positive rail through its positive connector, the other provides the negative rail through its negative connector. The other connectors are connected, and in fact their junction is the actual ground of the system, our reference for audio. This keeps things nice and simple for starters (as opposed to not having to provide a bias voltage at the center of a single battery power supply). We also get double the battery capacity, double the headroom (although it's much more than is actually needed in this case) plus simplicity - reliability.
What I would like to ask about is a very specific scenario. This would involve one of the batteries not being present, or a battery lead being severed by accident. This would, at least initially, leave either the positive rail or the negative rail at zero volts, with the other one where it should be. Since we are speaking a bit generally, not referring to a very specific op amp, I would like to ask: If the two op-amps suddenly start receiving 0V positive and -9V negative, is it even remotely possible that they will get worn-damaged or completely die? If the possibility is even remotely possible, is there some power protection scheme that can be implemented using simple additional components (like diodes)? One such reaction of a safety circuit would be for example to somehow power the op-amps with +-4.5V from the battery that remains properly connected when the other one's lead is severed. This is just an example, there are probably a lot of best practices on protecting op amps at their power supply pins. Of course, all this may not be much cause for concern in the first place.
I am not thinking about input signal protection as each op-amp's input will be a passive pickup that can not really cause much damage with its output. I plan to place a 1M resistor from the op-amp's input to ground, to set the preferred input impedance and likely stabilize stuff in general.
My understanding of basic electronics, let alone op amps is pretty superficial but I think I can build and test something that will work. Any help with the input power protection aspect of this will be greatly appreciated.
Split supplies are usually protected by two reverse-biased diodes, each connected between a supply rail and ground. The diodes prevent the supply rails from being pulled to the wrong polarity. They are conveniently placed in parallel with the decoupling capacitors.
Ed
Ed
Thanks for your response @EdGr.
In my case, power rails are provided by two 9V alkaline batteries. The battery connectors provide mechanical protection against wrong polarity, so I am not very worried about that scenario.
The fault case I would like to specifically investigate is powering the circuit up with only one battery, or equivalently, with one of the four battery leads severed.
In other words, I'm investigating if an op amp like the TL072 can get damaged if Vcc is floating due to a missing positive rail battery, while the negative rail battery continues to provide Vee = -9V. And vice versa. What do you guys/gals think?
In my case, power rails are provided by two 9V alkaline batteries. The battery connectors provide mechanical protection against wrong polarity, so I am not very worried about that scenario.
The fault case I would like to specifically investigate is powering the circuit up with only one battery, or equivalently, with one of the four battery leads severed.
In other words, I'm investigating if an op amp like the TL072 can get damaged if Vcc is floating due to a missing positive rail battery, while the negative rail battery continues to provide Vee = -9V. And vice versa. What do you guys/gals think?
@MusicLover, do you mean the coupling capacitor for the output signal? Do you know how that affects the op-amp's power supply?
Yes, the coupling capacitor for the output signal.
Depending on the design of an Op Amp and its surrounding circuitry, removing a battery can make its output go to + or - 9 volts.
However, with an output coupling capacitor this voltage can't flow as current into the next stage. The actual Op Amp won't care about +/- 9V.
Depending on the design of an Op Amp and its surrounding circuitry, removing a battery can make its output go to + or - 9 volts.
However, with an output coupling capacitor this voltage can't flow as current into the next stage. The actual Op Amp won't care about +/- 9V.
@MusicLover I'm glad you generalized a bit because it's not 100% certain I will use the TL072 as I would like something suitable for the application with even lower power consumption.
I will make sure to use output coupling capacitors. I was initially thinking of connecting the outputs directly to the 25k blend pot, but honestly, the idea of 9V going downstream towards a bass amp is scary.
It's reassuring to know that the op amp won't care (read: won't get damaged) if one of its power pins starts floating. I hope this applies to most op-amps designed for audio.
I will make sure to use output coupling capacitors. I was initially thinking of connecting the outputs directly to the 25k blend pot, but honestly, the idea of 9V going downstream towards a bass amp is scary.
It's reassuring to know that the op amp won't care (read: won't get damaged) if one of its power pins starts floating. I hope this applies to most op-amps designed for audio.
@WhiteDragon I'm specifically going after this to avoid directly combining two high impedance passive pickups with a blend pot (or even a three way switch). But let's not go to the reasons behind this here.
Thanks for your pointer to single supply. I am aware of it and plan to go for it in the future. A 9V or even 18V single supply would work fine and be fool proof with just one battery connector (9V) or two connectors in series (18V) - just a short comment related to the specific fault condition we're discussing in this thread.
I realize 18V of headroom (after taking into account the op amp's swing limits) is much more than I need for this application. The passive pickup signal amplitudes are much narrower. But there's the advantage of simplicity and also twice the capacity from two batteries powering the circuit, therefore double the battery life. The TL072 is low power but not ultra low power. The quest for an ultra low power op amp suitable for this application I leave for the future and perhaps a different thread.
Again thanks for your recommendations.
Thanks for your pointer to single supply. I am aware of it and plan to go for it in the future. A 9V or even 18V single supply would work fine and be fool proof with just one battery connector (9V) or two connectors in series (18V) - just a short comment related to the specific fault condition we're discussing in this thread.
I realize 18V of headroom (after taking into account the op amp's swing limits) is much more than I need for this application. The passive pickup signal amplitudes are much narrower. But there's the advantage of simplicity and also twice the capacity from two batteries powering the circuit, therefore double the battery life. The TL072 is low power but not ultra low power. The quest for an ultra low power op amp suitable for this application I leave for the future and perhaps a different thread.
Again thanks for your recommendations.
They have ultra low power rail to rail op amps stable at 3 to 5 volts.
Just shuffle through the Texas Instruments website.
The instrument jack would use a switching jack. So if you unplug the bass
the power is also released. Battery life can be months to years with low power opamps
even 10 years ago. So should be easy to find a opamp.
Since they are stable to 3 to 5 volts. Surprised if not already why not just normal
cell phone battery or RC servo battery. And just have a USB charger and never remove the battery again.
Just repeating the 3 to 5 volt range because those opamps are the ones with extremely low power
consumption. And often designed to be single supply because the cell phone market is what drove
such a market for more ideal low power opamps.
Just shuffle through the Texas Instruments website.
The instrument jack would use a switching jack. So if you unplug the bass
the power is also released. Battery life can be months to years with low power opamps
even 10 years ago. So should be easy to find a opamp.
Since they are stable to 3 to 5 volts. Surprised if not already why not just normal
cell phone battery or RC servo battery. And just have a USB charger and never remove the battery again.
Just repeating the 3 to 5 volt range because those opamps are the ones with extremely low power
consumption. And often designed to be single supply because the cell phone market is what drove
such a market for more ideal low power opamps.
I think you are missing the point. With, for example, the negative battery disconnected, the op-amp's supply current will pull the negative rail positive. (Besides, with an ordinary 9 V battery, it isn't difficult to temporarily connect it with reversed polarity, just push it to the contacts the wrong way around.)
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@MarcelvdG thanks for your input. Both valid points. Of course I was oblivious about the first one, which lead me to post this thread.
I presume pulling the negative rail positive is not exactly good for the op amp's health. If I may ask, how would you wire this thing up exactly? I am interested in a robust configuration (with two batteries), no missing parts, no redundant parts.
Each op amp's input signal source is a magnetic pickup coil, and destination a 25k blend pot where the two output signals meet.
I presume pulling the negative rail positive is not exactly good for the op amp's health. If I may ask, how would you wire this thing up exactly? I am interested in a robust configuration (with two batteries), no missing parts, no redundant parts.
Each op amp's input signal source is a magnetic pickup coil, and destination a 25k blend pot where the two output signals meet.
There are two things to watch out for:
1. If the positive supply goes negative with respect to the negative supply, parasitic diodes all over the chip will go in forward conduction and may very well overheat the chip. This would happen when someone tries to sabotage your amplifier by connecting both batteries the wrong way around, but it may also happen when one battery is accidentally connected the wrong way around and that battery has a higher voltage than its colleague.
2. When any signal input or output is forced above the positive or below the negative supply, this may trigger a parasitic thyristor that shorts the supplies and blows up the chip (latch-up). Usually ICs are designed not to latch up unless the current into or out of the input or output pin exceeds 100 mA.
Whether 2 is possible depends on the circuit. Ed's antiparallel diodes protect against all of it except the two wrongly connected batteries. Ideally, the diodes should be large Schottky diodes (such as SB1H90), as these have lower forward voltages than the junction diodes on the IC. Their disadvantage is their relatively high leakage current.
1. If the positive supply goes negative with respect to the negative supply, parasitic diodes all over the chip will go in forward conduction and may very well overheat the chip. This would happen when someone tries to sabotage your amplifier by connecting both batteries the wrong way around, but it may also happen when one battery is accidentally connected the wrong way around and that battery has a higher voltage than its colleague.
2. When any signal input or output is forced above the positive or below the negative supply, this may trigger a parasitic thyristor that shorts the supplies and blows up the chip (latch-up). Usually ICs are designed not to latch up unless the current into or out of the input or output pin exceeds 100 mA.
Whether 2 is possible depends on the circuit. Ed's antiparallel diodes protect against all of it except the two wrongly connected batteries. Ideally, the diodes should be large Schottky diodes (such as SB1H90), as these have lower forward voltages than the junction diodes on the IC. Their disadvantage is their relatively high leakage current.
By the way, when you supply a TL072 from two 9 V batteries in series, it will drain them at least as quickly as when you supply it from one 9 V battery, in the sense that the current drawn from each battery doesn't get smaller when you use two in series. Maybe you get a slightly longer battery lifetime with two batteries because you can discharge them down to a lower voltage per battery before running out of headroom.
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@MarcelvdG so if a TL072 is powered by two batteries, one positive rail, one negative rail, the current draw from each battery is stll 1.4mA (nominal) per channel and not half of it - 0.7mA? That's quite a surprise to someone thinking about it energy-wise (like myself).
I presume that to get almost double the lifetime you need a correctly implemented 9V single supply with two 9V batteries connected in parallel, sacrificing half the headroom.
I presume that to get almost double the lifetime you need a correctly implemented 9V single supply with two 9V batteries connected in parallel, sacrificing half the headroom.
You cant connect 2 batteries in parallel as one will be a higher voltage than the other and they will pass a lot of current.
@nigelwright7557 I have done it successfully in the past and didn't fry my electronics. If I am not horribly wrong, when connecting two identical fresh batteries in parallel you get same voltage, double capacity.
One battery will be at a higher voltage and will discharge into the lower voltage battery until both are equal.
A waste of battery power.
A waste of battery power.
@nigelwright7557 if I connect the positive leads together and the negative leads together, I don't see how one battery will be at a higher voltage than the other. Higher voltage for one battery would occur in a series connection.