Hey, I'm building a small 9VAC to +/-15V supply circuit, and I'm using 7815 and 7915 voltage regulators. I was told to just "lay them down" on the board, but should they have heatsinks? Here's the schematic:
And if so, should I put them upright and screw heatsinks to them, or should I lay them down, put a hole beneath them in the PCB, and then screw them to brass standoffs attached to the chassis? I can't imagine that they'd be generating that much heat with this circuit.
An externally hosted image should be here but it was not working when we last tested it.
And if so, should I put them upright and screw heatsinks to them, or should I lay them down, put a hole beneath them in the PCB, and then screw them to brass standoffs attached to the chassis? I can't imagine that they'd be generating that much heat with this circuit.
Hi,
you need to calculate the power lost thro' each reg ( current times volts drop). A plain To220 will dissipate a maximum of about 2w without a heatsink thats about 400mA @ 5 volts dropped.
Is that a voltage doubler on the front end? I don't know the circuit but this layout is not familiar.
you need to calculate the power lost thro' each reg ( current times volts drop). A plain To220 will dissipate a maximum of about 2w without a heatsink thats about 400mA @ 5 volts dropped.
Is that a voltage doubler on the front end? I don't know the circuit but this layout is not familiar.
Yes, it's a voltage doubler. I'm not very knowledgable when it comes to the science and math, so please bear with me. I'm willing to learn.
I know the circuit works, as it is based on a Rane design and modified per the suggestions of a very knowledgable pro. I'm trying not to ask him questions every 5 minutes, so I'm going to some forums. The circuit is good- it's just my understanding that has huge holes in it.
Anyway, it takes 9V AC and turns it into +/-15VAC at the end. I assume the 9V is actually doubled to 18V and then the voltage regulators knock it down to 15W with a swing of 30V. If I'm using the wrong terminology, please correct me.
The circuit will only draw about 200ma, as the only things being powered are a few AD826 opamps (2 min, 6 max).
So if my fuzzy numbers are correct, 200ma @ 3 volts dropped would be how many watts? .6W, correct? And if I put 6 AD826s in, and the current goes to 300ma, then I'm still only looking at .9W, and still well within non-heatsink spec, correct?
I need to look at the AD826 datasheets again. Thanks for the formula!
I know the circuit works, as it is based on a Rane design and modified per the suggestions of a very knowledgable pro. I'm trying not to ask him questions every 5 minutes, so I'm going to some forums. The circuit is good- it's just my understanding that has huge holes in it.
Anyway, it takes 9V AC and turns it into +/-15VAC at the end. I assume the 9V is actually doubled to 18V and then the voltage regulators knock it down to 15W with a swing of 30V. If I'm using the wrong terminology, please correct me.
The circuit will only draw about 200ma, as the only things being powered are a few AD826 opamps (2 min, 6 max).
So if my fuzzy numbers are correct, 200ma @ 3 volts dropped would be how many watts? .6W, correct? And if I put 6 AD826s in, and the current goes to 300ma, then I'm still only looking at .9W, and still well within non-heatsink spec, correct?
I need to look at the AD826 datasheets again. Thanks for the formula!
1 W power loss in the regulator with TO220 case is hot so check how much power you must burn away in the regulators.
0.6-0.8 W is OK without heatsink.
Please note also that a voltage doubler is a hard load for a transformer but also for the caps. You may need to insert some resistance (1-10 ohms) in order to limit the charging currents.
0.6-0.8 W is OK without heatsink.
Please note also that a voltage doubler is a hard load for a transformer but also for the caps. You may need to insert some resistance (1-10 ohms) in order to limit the charging currents.
Well, I double checked the AD826 specs, and each opamp draws 7.5ma max input. So, assuming I use 6, that's 45ma. So, .045x3 gives me .135W. I should be fine there.
I'll double check the voltage doubler portion, but as it's a Rane design in a well known product, I'm not too concerned. But I will look at it again. Thanks.
I'll double check the voltage doubler portion, but as it's a Rane design in a well known product, I'm not too concerned. But I will look at it again. Thanks.
Lyle, those datasheet figures do not count any load current in the circuit, just the power drawn when nothing is happening.
Unless you are particularly tight on space, I would allow room for heatsinking the regs, then you can always install them later if needed when the circuit is up and running.
Unless you are particularly tight on space, I would allow room for heatsinking the regs, then you can always install them later if needed when the circuit is up and running.
I recommend to place one diode in parallel with each capacitor to prevent reverse polarisation during startup
pinkmouse said:
I shall, then. The datasheets specifically say "max current" draw is 7.5ma, but I'll err on the side of caution.
Hi,
a few problems to address
18v cannot supply plus 15vdc after passing through the reg.
Secondly the doubler is not very efficient and as current rises the ripple increases and the basic dc falls badly. The dc level becomes the feed for the reg and then you subtract the volts drop, but the ripple also causes a volts drop and added power dissipation.
Next you want 2 by 15v supplies, this needs about 40v dc before the regs and therefore you need a 28 to 30vac transformer not 9vac nor 18vac. Oh and you want the regs to work at minimum supply voltage and not overheat at maximum supply voltage!!!!
Finally I think the 7mA you quote for the opamp is the quiescent current with no load current flowing. The load current is determined by the feedback loop and the output load in parallel and ouput voltage to determine the output current. this needs to be added to the quiescent current to give PSU loading.
However a small help is that the output current comes on alternate half cycles from first the plus then the neg rails and average power from each of the rails is approx half total power. Gives time for the reg to cool slightly.
PS I will not dispute the other contributors advice if they say you should limit power to 600mw or 1000mw.
a few problems to address
18v cannot supply plus 15vdc after passing through the reg.
Secondly the doubler is not very efficient and as current rises the ripple increases and the basic dc falls badly. The dc level becomes the feed for the reg and then you subtract the volts drop, but the ripple also causes a volts drop and added power dissipation.
Next you want 2 by 15v supplies, this needs about 40v dc before the regs and therefore you need a 28 to 30vac transformer not 9vac nor 18vac. Oh and you want the regs to work at minimum supply voltage and not overheat at maximum supply voltage!!!!
Finally I think the 7mA you quote for the opamp is the quiescent current with no load current flowing. The load current is determined by the feedback loop and the output load in parallel and ouput voltage to determine the output current. this needs to be added to the quiescent current to give PSU loading.
However a small help is that the output current comes on alternate half cycles from first the plus then the neg rails and average power from each of the rails is approx half total power. Gives time for the reg to cool slightly.
PS I will not dispute the other contributors advice if they say you should limit power to 600mw or 1000mw.
Andrew,
Thanks. I'm mulling that advice over. I'm certainly not discounting it. My only thought is that it has worked for Rane in several successful products. I'll post the relevant portion of the Rane schematic in a few moments. Theirs is a bit more filtered than mine, but is essentially the same. I'd be interested in your take on it.
PS- the Rane schematic is publicly posted on their site, so I think this will be "fair use."
Thanks. I'm mulling that advice over. I'm certainly not discounting it. My only thought is that it has worked for Rane in several successful products. I'll post the relevant portion of the Rane schematic in a few moments. Theirs is a bit more filtered than mine, but is essentially the same. I'd be interested in your take on it.
PS- the Rane schematic is publicly posted on their site, so I think this will be "fair use."
Here's the Rane (only difference being the phone connection power plug instead of the RJ45 style Rane plug):
The LED is backwards and the LED resistor is too low in that (should be 5.6K), but I won't bother reposting the correction.
An externally hosted image should be here but it was not working when we last tested it.
The LED is backwards and the LED resistor is too low in that (should be 5.6K), but I won't bother reposting the correction.
The LED is the right way around and the resistor is a suitable value, it'll set the current through the LED to 22ma, 5.6k would set it to 5ma.
And now the real reason for my post, are you the same Lyle Caldwell from TB?
And now the real reason for my post, are you the same Lyle Caldwell from TB?
Tim__x said:
Yup, that's me. Rane has the LED and resistor the other way around, but I guess it doesn't matter with AC. As to the resistor, I thought I need to find the spec on the LED at digikey. I was using the calculations for an LED at mouser, but the digikey seems to want more current.
Hi,
if Rane's circuit works then it prompts me to wonder if that doubler is a tripler or even a 4 times? You really do need about 40v before the regs.
if Rane's circuit works then it prompts me to wonder if that doubler is a tripler or even a 4 times? You really do need about 40v before the regs.
Andrew,
From another technical forum:
"The DC voltage measured over the cap just before the regs in the RANE are at least 18 * (square root of 2) -1.2v = +24.25v This goes for the - side as well -24.25v just under 50 volts to this."
So your advice is spot on, and the circuit does more than double the voltage. Thanks for leading me to a better understanding of this.
From another technical forum:
"The DC voltage measured over the cap just before the regs in the RANE are at least 18 * (square root of 2) -1.2v = +24.25v This goes for the - side as well -24.25v just under 50 volts to this."
So your advice is spot on, and the circuit does more than double the voltage. Thanks for leading me to a better understanding of this.
- Status
- Not open for further replies.