power supply calculation?

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Tdale
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power supply calculation?

Post by Tdale »

How do I calculate the voltage of a power supply?

Let's say I have a rectifier bridge, with a cap to ground, a resistor and then another cap to ground, just like the normal filtering section of a power supply. How do I calculate the voltage of B1, B2 etc?

The relay power supply in the picture attached, is a good example. Is there a formula for calculating the voltage across the relays?

Tommy
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doctord02
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Re: power supply calculation?

Post by doctord02 »

You have to know the AC voltage input to the bridge recto before you can do anything else...
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Tdale
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Re: power supply calculation?

Post by Tdale »

Ok. Let's say I tap from the 6,3V heater supply? How do I calculate from there?

(By the way, is that 6,3 0 6,3 or is it 3,15 0 3,15) normally

TOmmy
jazzyjoepass
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Re: power supply calculation?

Post by jazzyjoepass »

Based on your info: 6.3V-0-6.3V

If you are doing a Full Wave Bridge (ignoring the center tap), then it is 1.414 x (6.3+6.3)

If you are using the normal Full Wave rectifier making use of the center tap, it is 1.414 x 6.3 V

But both will get you a rectified 120Hz waveform.
Half wave will get you a 60Hz.


MIke.
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Tdale
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Re: power supply calculation?

Post by Tdale »

Ok.. but is that the voltage that comes out of the FWR or is it the B+ voltages that can be tapped from each cap/resistor?

Usually, the different B+'s have different voltages, it's those I'm interested in calculating, like in the picture attached. (B1, B2, B3 etc.)
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jazzyjoepass
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Re: power supply calculation?

Post by jazzyjoepass »

I remember some article in bluesguitar.org regarding calculating the B+ voltages.

BTW, if the B+ are not connected up to the rest of the circuit, there's no current following (except minor currents thru the capacitors), thus the voltages are all around the B+1 voltage - slightly lower as you go down each B+.

Hope this helps... :cry:
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LeftyStrat
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Re: power supply calculation?

Post by LeftyStrat »

Ohm's law to the rescue!

amps = volts / ohms

or

volts = amps * ohms

So you need to know the current draw across the resistors in order to calculate voltage drop.

You can guestimate using something with similar topology. Take a look at the Kelly schematic. The drop from B+4 to B+5 is:

282 - 267 = 15

15 volts / 9100 ohms = .0016 amps

So if you're going to have a similar topology connected to that node, use that current draw and plug in your resistor value. For example, 100k:

.0016 * 100000 = 160

So you would have 160v drop across a 100k resistor. If you haver 460 at b+4 you'd have 300 at b+5 assuming a similar current draw.
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Tdale
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Re: power supply calculation?

Post by Tdale »

I realise now that I have sort of misread the power supply schematics a bit. I searched for "voltage divider" and discovered that the filter section of the power supply is actually just a number of resistors in series, and that's why they produce different voltages when you measure between each of them to ground.

http://www.tpub.com/neets/book1/chapter3/1-35.htm

The link shows it quite good. I sort of never realised that the voltage is measured across the resistors.

I suppose the reason why there usually is one pretty large resistor, and several small, is to get more equal voltages in the first part, and then some lower voltages in the end.

Are the caps there just to take AC out of the circuit, or do they affect B+ voltages`?

Tommy
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LeftyStrat
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Re: power supply calculation?

Post by LeftyStrat »

Yes, the caps are there to remove any AC ripple. Here's another good doc for the different types of rectifiers and the expected AC output:

http://www.hammondmfg.com/pdf/5c007.pdf
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Bob-I
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Re: power supply calculation?

Post by Bob-I »

Tdale,

Sorry I didn't see this before, it's my schematic.

The calculation for the DC Voltage using a bridge rectifier is VAC * .9. I measured 6.6VAC on the heaters, a tad high but in tolerance. I'm getting about 5.8VDC so the formula is not far off.

Mike (jazzyjoepass) is not correct in that this is a non-center tapped supply. It's simply 6.3VAC across the 2 leads.
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