oh? so a 'normal' power supply won't do then?
Guess I'll have to find a circuit and build one of these 'dual-rail' supply thinggies 😕
Guess I'll have to find a circuit and build one of these 'dual-rail' supply thinggies 😕
Two diode Bridges are not needed
Can someone explain why again that the two bridge supply has gained favour in this Chip Amp forum? One will do for either a center-tapped tranny or two non-center-tapped trannies. This fellow needs a simple logical straightforward design and that two bridge design is not.
Can someone explain why again that the two bridge supply has gained favour in this Chip Amp forum? One will do for either a center-tapped tranny or two non-center-tapped trannies. This fellow needs a simple logical straightforward design and that two bridge design is not.
IIRC, If you use a transformer with 2 secondary windings and a bridge on each secondary, you'll end up with less DC in the transformer which means less heat / better regulation etc etc. Of course, you lose twice as much in diode losses.
I used such a configuration for its versatility: it can be combined into a +/- rail supply, or two independent single-rail supplies. works wonderfully.
the absolute bare minimum requires two resistors: Rfi (20k) and Ri(1K?). Of course, you need the speaker (RL) to make any sound, 🙂.
I have done such a test up successfully. what else you may need depends on a lot of other factors. I would, to be safe, use two small decoupling caps on the power supply rails. In my actual gainclone, I used also two 3300uf caps (one on each rail) as my power supply is pretty skinny in terms of capacitance.
I have done such a test up successfully. what else you may need depends on a lot of other factors. I would, to be safe, use two small decoupling caps on the power supply rails. In my actual gainclone, I used also two 3300uf caps (one on each rail) as my power supply is pretty skinny in terms of capacitance.
Nuuk said:I suggest some reading may be in order!
nuuk, there may be a typo somewhere here. you wrote:
As an example lets say that you want to use a nice blue LED (Farnell order code 233-493) with a PIV of 4.5 volts and If of 30mV and your voltage rail is 26 volts. So you will require a series resistor of the value (26-4.5)/0.0030 = 7167. The nearest standard value to this is 7500 (7K5) and it won't do any harm to operate the LED at just below it's maximum voltage.
you probably wanted to say "If of 3ma", based on the subsequent calculation. 30ma will also work.
however you are seriously over-engineering the serial resistor by about 7 times: it dissipates about 0.07w in your example (or 1w if you were running it at 30ma).
I used a 10k resistor in on a 36v rail.
This is not the main reason here. Just read what Rb is in the datasheet, very good described. This resistor is to avoid possible damage at low voltage conditions.theChris said:Rb is to reduce DC offset
MC said:heh, I guess I'll get them all as soon as I can figgure out what Cs is suppose to be
Cs - Supply Capacitance. in general, analog circuits use bypassing capcacitors located close to the IC. most professional circuits at least. Cs is typically an electrolytic capacitor between 1000uF and 10000uF with the preferance being to 1000uF for the gainclones. sometimes a 100nF or 1uF mylar (or some other cap) is used as close to the IC pins as possible. the idea to to improve the high frquency preformance of the capacitor combo.
My testclone is at heart just 2 100nF caps and 2 resistors on the lm1875 ICs with hot glue preventing it from touching the heatsink which is a few mm from the pins on 2 sides... located further away from the ICs is a 4700uF capacitor on each rail (the PSU caps). this gives a little over 1000uF per IC per rail voltage. there is no volume control or power switch... or case...
there is a difference buffer followed by an active crossover before the lm1875s.
there is a bit of DC offset, and the lack of sheilding means the signal wire needs to be further from the transformer. otherwise the amp works fine.
the lack of a volume control or power switch is annoying.
also, you do realize that you need a transformer and full-wave bridge if you plan to run this circuit from AC?
Nuuk said:Yes, that should be 30mA. Thanks for pointing it out, I'll change it.
if it is indeed 30ma, you may need to redo the calculation and upsize the resistor to match.
Oops, see what you mean, it should be 0.03mA not 0.003mA! Damn those decimal points! 😀
Thanks again.
Thanks again.
Nuuk said:Oops, see what you mean, it should be 0.03mA not 0.003mA! Damn those decimal points! 😀
Thanks again.
are you sure your LED will lit up at 0.03ma or 0.003ma? that's a very small current, even for a LED, 🙂.
it is like either 3ma (a little on the low-side) or 30ma (on the high side). In your example, a 7k, 1/4w serial resistor will do that job for 3ma; or a 750ohm, 1w serial resistor for 30ma.
I would personally go with the first one.
Hey - what are you doing up at this time of day? 😉
The If is 30mA although that is the MAX figure. Perhaps it would be better to run it at say 20mA so I'll alter the text.
The If is 30mA although that is the MAX figure. Perhaps it would be better to run it at say 20mA so I'll alter the text.
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