Just need a quick double-check on this;
I'm planning on using a Hammond 187C12 as a small transformer for a line level device power supply - freeDSP Aurora to be specific. Since this is center-tapped, I am going to put 1 diode on each windings (as opposed to the 4 for bridge rectification) for full-wave rectification where the center tap is the common (0 VDC). Maybe a two-diode device like this.
The transformer has (12.6 / 2) = 6.3 VAC RMS on each winding, so my rectified and smoothed output voltage (assuming high enough capacitance) should be (6.3 * 1.414 - 0.4) = 8.5 VDC.
Does that make sense? Did I make a mistake somewhere? Should I use bridge rectification anyway (will there be less power loss overall)? Also, what's the maximum C I should limit myself to, for this transformer?
I'm planning on using a Hammond 187C12 as a small transformer for a line level device power supply - freeDSP Aurora to be specific. Since this is center-tapped, I am going to put 1 diode on each windings (as opposed to the 4 for bridge rectification) for full-wave rectification where the center tap is the common (0 VDC). Maybe a two-diode device like this.
The transformer has (12.6 / 2) = 6.3 VAC RMS on each winding, so my rectified and smoothed output voltage (assuming high enough capacitance) should be (6.3 * 1.414 - 0.4) = 8.5 VDC.
Does that make sense? Did I make a mistake somewhere? Should I use bridge rectification anyway (will there be less power loss overall)? Also, what's the maximum C I should limit myself to, for this transformer?
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Your math assumes zero conduction angle (i.e. the diodes conduct infinite current for zero time). In reality, you'll likely have something like (6.3-0.4)*1.35 = 8.0 V out at nominal mains (7.2 V at low mains), assuming your 0.4 V diode drop holds up.
There's no maximum capacitance for a transformer. A larger capacitance will result in lower conduction angle, thereby, higher ripple current. It may be that the RMS value of the ripple current exceeds the RMS current rating of the transformer at some capacitor value, but that's pretty unlikely. Mostly what you'll find is that the extreme current spikes will end up coupling into sensitive circuits, resulting in hum.
I'd have to know your target output voltage and load current to provide better advice. I also suggest that you set up a simulation in PSUD2.
Tom
There's no maximum capacitance for a transformer. A larger capacitance will result in lower conduction angle, thereby, higher ripple current. It may be that the RMS value of the ripple current exceeds the RMS current rating of the transformer at some capacitor value, but that's pretty unlikely. Mostly what you'll find is that the extreme current spikes will end up coupling into sensitive circuits, resulting in hum.
I'd have to know your target output voltage and load current to provide better advice. I also suggest that you set up a simulation in PSUD2.
Tom
You need 12 volt dc for the dsp, secondary in series - rectifier- caps - and a fixed regulator like a 7812 , that would give you 12 v dc, is clean dc critical?
Am I correct that that the two secondary each hold 6.3v ?
Am I correct that that the two secondary each hold 6.3v ?
Yeah, you'd always use that if you had no center-tap, but with the center tap I can (in theory) get away with 2 diodes for half the diode voltage drop. At the same time, the transformer is used more efficiently with a full-wave bridge, so I don't know...
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Thanks Tom. The Aurora takes between 6.5 to 12 VDC, with 7 VDC "typical" and power consumption around 6-8 W (give or take). Others have successfully used ~9ish VDC, a stable 12 wouldn't be an issue either I think. This Hammond seems to be the cheapest I can find on Mouser/Digikey that would do the trick.
EDIT: Never mind, I had some unrealistic filtering happening. There are also 10 VAC CT options as well, but between that and 12.6 VAC CT that should cover both the FWCT as well as FWBR.
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Not knowing the source impedance is giving me a headache. I think I might just use an SMPS module with a CRC or pi filter downstream.
Figure small transformers are 20% regulation, "un-sag". This part is rated 12.6V 1A. So it can power a 12.6 Ohm resistor. The 20% un-sag implies this is through a 2.52 Ohm resistance and the no-load voltage is 15.12VAC.
> get away with 2 diodes for half the diode voltage drop. At the same time, the transformer is used more efficiently with a full-wave bridge, so I don't know...
Don't get stuck on small issues in DIY! There is not enough power here to justify much brain-strain. Even with "bad" design it will cost a half-cent an hour to pay the electric bill. 2,500 hours a year costs less than a 6-pack of beer or one sad bottle of wine.