They look like ne5532/tl072 , those use about 5-7ma per rail each (my
amp servo's use either) =
20 ma/rail - all 4 IC's here are used as filters/inverters .
Just use a 1.5K-2.2K as a dropping resistor for the zeners. Some of my 12 zener
regulated amps draw 18+ma per rail between the op-amps and input pairs ,
I had to tell builders to drop to 3.3-3.9K at the zeners for 45V main rails.
Even at 3.9K - a rock solid 12V.
OS
amp servo's use either) =
20 ma/rail - all 4 IC's here are used as filters/inverters .
Just use a 1.5K-2.2K as a dropping resistor for the zeners. Some of my 12 zener
regulated amps draw 18+ma per rail between the op-amps and input pairs ,
I had to tell builders to drop to 3.3-3.9K at the zeners for 45V main rails.
Even at 3.9K - a rock solid 12V.
OS
i always changed that dropping resistor and zener w/ a regulator 🙂
better performance and less noisy 😀
better performance and less noisy 😀
"Yes, but that is the current over the resistors, You don't know how much goes over the opamps, and how much over the zener..."
It doesn't mater, it's a shunt regulator. If you use opamps that draw less the zener will just draw more.
4x 8mA +10mA = 42mA, about what is going through the dropping resistor.
If you reduce the current through the opamps (change to another PN) the zener will just pull more current.
The answer is to reduce the current through the opamps (different PN) and then change the dropping resistor and maintain enough current through the zener for it to regulate well.
"i always changed that dropping resistor and zener w/ a regulator "
Sure, what PN do you use to handle in excess of 40V, don't forget we need a negative regulator too. The regulator will now get as hot as the resistors did, and they are not rated for as high of a temperature either (did you add a heatsink?).
It doesn't mater, it's a shunt regulator. If you use opamps that draw less the zener will just draw more.
4x 8mA +10mA = 42mA, about what is going through the dropping resistor.
If you reduce the current through the opamps (change to another PN) the zener will just pull more current.
The answer is to reduce the current through the opamps (different PN) and then change the dropping resistor and maintain enough current through the zener for it to regulate well.
"i always changed that dropping resistor and zener w/ a regulator "
Sure, what PN do you use to handle in excess of 40V, don't forget we need a negative regulator too. The regulator will now get as hot as the resistors did, and they are not rated for as high of a temperature either (did you add a heatsink?).
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Thats true. But why push more than 5mA over the zener? So the current running over the opamps is important to find the optimal value of the dropping resistors.
How did You calculate this value? There is about 29mA as it was measured.
Maybe it's a little bit crazy idea to redesign the whole preamp stage to save 2 resistors...
since all of my car amps didnt reached 30V, i never have problem here.but i found a few circuits that limits the input voltage on google 🙂
"Maybe it's a little bit crazy idea to redesign the whole preamp stage to save 2 resistors... "
Changing an opamp and resistor is a redesign?
Get real.
The diode is probably a 1N4744, nominally 15V. The Izt on this part is 17mA.
1N4743 pdf, 1N4743 description, 1N4743 datasheets, 1N4743 view ::: ALLDATASHEET :::
How well will it regulate with far less than its design test current (the suggested 5mA)?
I like to run things the way the manufacturer recommends.
4x 5532 at 8mA = 32mA, + 17mA Izt = 49mA.
Post #6 says 43V, post #7 says 14V, so there is 29V across the resistor.
29V/620Ω=47mA, very close to the 49mA estimate.
Typical supply current on the TL072 is 2.8mA, max is 5mA, $0.70 at Digi-Key.
http://www.ti.com/lit/ds/symlink/tl071a.pdf
Do what you like.
Changing an opamp and resistor is a redesign?
Get real.
The diode is probably a 1N4744, nominally 15V. The Izt on this part is 17mA.
1N4743 pdf, 1N4743 description, 1N4743 datasheets, 1N4743 view ::: ALLDATASHEET :::
How well will it regulate with far less than its design test current (the suggested 5mA)?
I like to run things the way the manufacturer recommends.
4x 5532 at 8mA = 32mA, + 17mA Izt = 49mA.
Post #6 says 43V, post #7 says 14V, so there is 29V across the resistor.
29V/620Ω=47mA, very close to the 49mA estimate.
Typical supply current on the TL072 is 2.8mA, max is 5mA, $0.70 at Digi-Key.
http://www.ti.com/lit/ds/symlink/tl071a.pdf
Do what you like.
Maybe yes. And especially replacing means a lot of work comparing to the replacement of the resistor.
The regulation is not critical in case of opamp psu, as it can be anything between 12-18V, without any problem. Anyway, the zener will work properly with 5mA. I never found any problem running the zener with lower current, than Izt. And if the zener is just 400-500mW version, the Izt is 5mA (check BZX55CV15).
Sorry it was my mistake to remember wrong. The current is about 47mA across the resistor. This means 1.37W dissipation. But You can use 750ohms as well, with about 5-6mA on the zener. This will reduce the dissipation to 1.21W. It should not be a problem with 5W resistor.
Why You go just to the half of the route? Use TL062! This means 1.6mA for the all opamps! If You like the 17mA over the zener, You can apply 1.5kohm resistor, which means 0,56W!!!!! Almost cold...
Sajti
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Alrighty!
So, ah, lets assume I have a budget of $0 to get this working in a reliable manner...
The owner wants it back soon, and I don't want to spend anymore money on it 😛
So, ah, lets assume I have a budget of $0 to get this working in a reliable manner...
The owner wants it back soon, and I don't want to spend anymore money on it 😛
Just give it back and don't work on other peoples equipment anymore 😉
Everything costs money and time.
Everything costs money and time.
Folks, If the original measurements were correct, i.e. 28V drop across 620 ohm, then one has a dissipation of 1,27W - never mind why/how that is used or whether one can decrease it. A 5W resistor should be in order.
Twice that is not an unusual amount of heat under a chassis. Having mentioned the Quad II, its cathode bias resistor dissipates 3,75W in a smaller enclosed space than appears to be the case here. (And it also had an electrolytic capacitor in close vicinity. One of the rare less bright moments Peter Walker had.) Many other amplifiers are in the same position. Power resistors can easily handle over 100° C as stated. If this defines the complete position, mount those resistors as advised with plenty of solder (I hope the solder pad is big enough, otherwise twist the leads a few small turns to get length) - and forget about it. I presume the unit worked before for some time, so there. You are back to better than a factory position.
Twice that is not an unusual amount of heat under a chassis. Having mentioned the Quad II, its cathode bias resistor dissipates 3,75W in a smaller enclosed space than appears to be the case here. (And it also had an electrolytic capacitor in close vicinity. One of the rare less bright moments Peter Walker had.) Many other amplifiers are in the same position. Power resistors can easily handle over 100° C as stated. If this defines the complete position, mount those resistors as advised with plenty of solder (I hope the solder pad is big enough, otherwise twist the leads a few small turns to get length) - and forget about it. I presume the unit worked before for some time, so there. You are back to better than a factory position.
Why not just bodge in a pair of 10W wirewound resistors and call it a day? Either that, or parallel some appropriate 5W parts. That way you can dissipate the required power with lower temperature rise. Resistor ratings these days are a total joke, unless you don't care about the PCB and adjacent components getting roasted.
Wow that's hot and you're in South Africa not the arctic so yikes. It will create reliability problems for nearby components (those poor electrolytics and the PCB).
I'd twist some wires (a pair for each resistor) and run them from the resistor locations to a pair of 620 ohm/5 to 10W resistors mounted to the case.
That's what I'd do.
I'd twist some wires (a pair for each resistor) and run them from the resistor locations to a pair of 620 ohm/5 to 10W resistors mounted to the case.
That's what I'd do.
Thanks for all the info.
I ended up ordering some cheap cheap 25watt, aluminium housed resistors that I'll mount on the heatsink internally... hopefully... lol
FIXED!
I ended up ordering some cheap cheap 25watt, aluminium housed resistors that I'll mount on the heatsink internally... hopefully... lol
FIXED!
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