A couple of things are missing. The 33.94 volts from the secondary is close to correct, but it doesn't take into account the voltage drop across the diode bridge. Since the path is through 2 diodes, the drop will be approximately 1.5 volts. That doesn't leave enough drop across the regulators at 32 volts out. You need 2 or 3 volts there, so better to aim for 30 volts out. The circuit also needs larger filter capacitors in parallel with the 0.1uF capacitors. Something like 3300uF to 6800uF, 50WVDC across each + & - rail.
The revised modification
Hi There,
Thanx, Sofa Spud. You were right. I checked some other Lm3876 projects on line, and they all have large caps coming off the rectifiers. One guy had 2200s and another had two 8000 per side. So I've left the size indeterminate for now. I'll see what my electronic genius friends can come up with on that score.
If anyone else has ideas, just holler. As I said, I'm really a beginner. I more or less just copy parts from other peoples designs, and hope it works. So here is the latest and greatest schematic.
Thanx
The Happy Hippy
Hi There,
Thanx, Sofa Spud. You were right. I checked some other Lm3876 projects on line, and they all have large caps coming off the rectifiers. One guy had 2200s and another had two 8000 per side. So I've left the size indeterminate for now. I'll see what my electronic genius friends can come up with on that score.
If anyone else has ideas, just holler. As I said, I'm really a beginner. I more or less just copy parts from other peoples designs, and hope it works. So here is the latest and greatest schematic.
Thanx
The Happy Hippy
Check your programming resistors.
R1, R2 = 120, 2700 ohm will be closer to 30V output.
With 10 mA programming current the 2700 ohm will also need to be a 0.6 or 1 W unit.
Use the 'standard' 240 ohm for R1 to avoid this. 220, 4700 ohm units are common resistor values and will give you a more realistic 28V output. Use 1/4 W for R1 and 1/2 W for R2.
Note that 30 V output may be too close to the input voltage. The regulator will drop out of regulation as the transformer output sags and the ripple increases under load.
Size your big filter capacitors to avoid the ripple during maximum load to cause the regulator input voltage to drop below the output + 2 or 3 volts.
For 60 Hz mains the required C is aprox. 5.I/(2.pi.120.Vr).
...where I is maximum load, Vr is maximum allowed ripple.
E.g. for 1 V ripple at 1.5 A use 10 000 uF.
R1, R2 = 120, 2700 ohm will be closer to 30V output.
With 10 mA programming current the 2700 ohm will also need to be a 0.6 or 1 W unit.
Use the 'standard' 240 ohm for R1 to avoid this. 220, 4700 ohm units are common resistor values and will give you a more realistic 28V output. Use 1/4 W for R1 and 1/2 W for R2.
Note that 30 V output may be too close to the input voltage. The regulator will drop out of regulation as the transformer output sags and the ripple increases under load.
Size your big filter capacitors to avoid the ripple during maximum load to cause the regulator input voltage to drop below the output + 2 or 3 volts.
For 60 Hz mains the required C is aprox. 5.I/(2.pi.120.Vr).
...where I is maximum load, Vr is maximum allowed ripple.
E.g. for 1 V ripple at 1.5 A use 10 000 uF.
There's no negative version of the LM338 sadly unless someone knows better.
I recently used up a few bits box LM317 and LM337's making up a general purpose handy dual rail variable +-30v PSU. I wanted over 2A so I just paralleled the devices using 0.1R output pin resistors to help balance currents (it works). I lost a little of the regulation performance, but I didn't want to use a parallel transistor to boost current as I didn't think that would be short circuit proof without a lot of extra circuitry. A fan and a transformer tap switch that can halve the input DC for low output voltages keeps temperatures nicely below the LM317 shutdown temperature.
What happens if?
OK So my question is, what happens if I leave it set for 30V, and the input drops below that? Remember, I'm new to this. And I do not understand the need for a 1/2 W resistor. Can you exlpain that more?
Thanx
The Happy Hippy
OK So my question is, what happens if I leave it set for 30V, and the input drops below that? Remember, I'm new to this. And I do not understand the need for a 1/2 W resistor. Can you exlpain that more?
Thanx
The Happy Hippy
The regulator will need to recieve aproximatly 3.5v DC More than the output voltage to maintain regulation ...... so with 30v DC output you will need at least 33.5vDC at the input ......
If the load on the transformer drops the DC voltage below 33.5vDC then the regulator will stop maintaining regulation which negates the purpose of the regulator and might cause other problems .......
That resistor needs to draw 10ma to maintain regulation , 10ma at 30v is about 300mW so a 1/4w resistor is too small , if you only have 1/4 resistors you can use 2 resistors in paralell that are double the Value or 2 resistors in series that are half the value ......
If the load on the transformer drops the DC voltage below 33.5vDC then the regulator will stop maintaining regulation which negates the purpose of the regulator and might cause other problems .......
That resistor needs to draw 10ma to maintain regulation , 10ma at 30v is about 300mW so a 1/4w resistor is too small , if you only have 1/4 resistors you can use 2 resistors in paralell that are double the Value or 2 resistors in series that are half the value ......
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I'm still trying to figure out why you are trying to use a small-current regulator in a power amp.
If it's to do it just to do it, that's great! I'm not going to try to stop you.
But if you are just trying to make a good PSU for a LM3876, there is no need to make it that complicated.
If you are trying to throw away a volt or two in order get your rails a bit lower, use a single diode bridge and wire your transformer as a center-tap.
If it's to do it just to do it, that's great! I'm not going to try to stop you.
But if you are just trying to make a good PSU for a LM3876, there is no need to make it that complicated.
If you are trying to throw away a volt or two in order get your rails a bit lower, use a single diode bridge and wire your transformer as a center-tap.
Hi,
the short answer is no, no and no. Regulators do not increase current
capability, quite the opposite. Currents from regulators in each rail don't
add up. Regulators do not smooth the "flow" of electricity i.e. current,
they smooth the voltage or voltage ripple only. 1.5 amps is nowhere
near enough to provide 56W into 8R, that needs 2.65A RMS (or
+/- 1.3A RMS) and you need double that for 4 ohm resistive loads.
It simply won't work without additional pass transistors, and you are
complicating what should be a very simple supply for the LM3876.
Bridge rectifier and capacitors is all you need, if you want regulators
you need 3A regulators in each rail, that is pointless as PSSR is 120dB.
For real 8ohm not 8R speakers you'd probably want 5A peak per rail.
Easy with a standard supply, not so with current limited regulators.
rgds, sreten.
I said as much back in post #5.
LM317 doesn't need 3.5V overhead.
The "T" package (TO-220) is max'ed at 3.4 amps in the datasheet. (NatSemi, July2004)
@happy hippy, you're being told again that you don't need a regulated supply for your chipamp.
I honestly don't know what would happen if the regulator input fell below the output setting; it would probably shutdown and give 0 volts out.
This particular regulator works by creating a constant current between output and adjust pin (thru a resistor). That resistor sets the amperage. By placing another resistor in the path to ground, the desired output voltage is set. Using Ohm's Law - V=I*R.
So there is current through the resistor, and that develops a voltage across the resistor. Volts times amps equals power. Since this application puts about 30 volts across this resistor, it doesn't require much current to get relatively high power dissipation. Hence the 1/2W power rating suggestion.
Resistors are basically (but not entirely) specified by value, tolerance, and power rating.
LM317 doesn't need 3.5V overhead.
The "T" package (TO-220) is max'ed at 3.4 amps in the datasheet. (NatSemi, July2004)
@happy hippy, you're being told again that you don't need a regulated supply for your chipamp.
I honestly don't know what would happen if the regulator input fell below the output setting; it would probably shutdown and give 0 volts out.
This particular regulator works by creating a constant current between output and adjust pin (thru a resistor). That resistor sets the amperage. By placing another resistor in the path to ground, the desired output voltage is set. Using Ohm's Law - V=I*R.
So there is current through the resistor, and that develops a voltage across the resistor. Volts times amps equals power. Since this application puts about 30 volts across this resistor, it doesn't require much current to get relatively high power dissipation. Hence the 1/2W power rating suggestion.
Resistors are basically (but not entirely) specified by value, tolerance, and power rating.
happy hippy, to provide a little more (non-engineer) explanation... different circuits and applications have different power supply needs. You're attempting to get as smooth a DC as possible by filtering and regulating the voltage from the transformer/diode bridge. The regulator isn't actually necessary in your case IMO. An amp has what is called a Power Supply Rejection Ratio (PSRR). Variations in the supply rails cause a change in circuit operation, and a good amp will reject those variations. In effect the amp makes it only a very slight change at its input. The PSRR tells you how well that is accomplished. I didn't look at the LM3876 datasheet, but I gather from an earlier post that its PSRR is 120dB. A good figure, and it means in all likelihood you don't need voltage regulators, you only need to knock out most of the ripple voltage with capacitor filtering after the diode bridge.
If you are using that 24V transformer, you will be staying under the max voltage ratings for the LM3876 with the 33.94V transformer secondary output. Nothing to worry about there.
If you are using that 24V transformer, you will be staying under the max voltage ratings for the LM3876 with the 33.94V transformer secondary output. Nothing to worry about there.
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happy hippie:
The only advantage of a regulator in this application is that it allows you to operate with supply rails close to the maximum design of the amp chip.
Without regulation the supply voltage surges and sags somewhat as the load varies, risking chip damage during no-load conditions.
This only applies when you try to run with the highest supply voltage as possible.
In your case, however, there is very little to be gained from a regulated supply (and a lot to loose if you consider the extra complications and thermal management requirements).
Just use your 2x24VAC transformer, rectifier bridge(s) and big filter caps, the amp chip will take care of the rest.
The only advantage of a regulator in this application is that it allows you to operate with supply rails close to the maximum design of the amp chip.
Without regulation the supply voltage surges and sags somewhat as the load varies, risking chip damage during no-load conditions.
This only applies when you try to run with the highest supply voltage as possible.
In your case, however, there is very little to be gained from a regulated supply (and a lot to loose if you consider the extra complications and thermal management requirements).
Just use your 2x24VAC transformer, rectifier bridge(s) and big filter caps, the amp chip will take care of the rest.
Happy Hippy,
When the voltage at the input drops below the minimum voltage required for the regulator to hold the output at the set voltage (say roughly 32V for your 30V example) then the regulator output voltage will basically follow the input less this 2V difference (30Vin 28Vout, 20Vin 18Vout)
This happens on every turn on and turn off, as well as during input sagging.
So basically the output voltage will be a few volts below the input voltage as each ramps upto the set voltage.
Hope this helps
-Antonio
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