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Old 7th November 2011, 01:50 AM   #1
johnr66 is offline johnr66  United States
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Default Why is DC out less than AC in?

I have a 6.3V, 1.5 A transformer. It is a two lead affair into a full wave bridge. With an 8 ohm load resistor connected to the output to test, the transformer's secondary reads 6.7 volts and the DC out reads 6.1 volts with 6,800uf filter cap.

I'm using a true RMS meter, so I'm not sure what I'm missing.
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Old 7th November 2011, 02:37 AM   #2
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Can you link the datasheet for the diodes forming the bridge?

Last edited by twest820; 7th November 2011 at 02:45 AM.
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Old 7th November 2011, 02:39 AM   #3
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You're missing (from open circuit peak voltage (6.7vac x 1.41)) two diode drops, and the resistance of the transformer multiplied by the rms current into the fullwave rectifier. keep in mind that the RMS current drawn by your capacitive filter will exceed the average current drawn by the 8 ohm load.
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Old 7th November 2011, 02:45 AM   #4
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The 6.7V is with the test load attached, so the output should be closer to 8V with two Vf drops for typical rectifier diodes.
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Old 7th November 2011, 06:23 AM   #5
Elvee is offline Elvee  Belgium
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Quote:
Originally Posted by johnr66 View Post
I have a 6.3V, 1.5 A transformer. It is a two lead affair into a full wave bridge. With an 8 ohm load resistor connected to the output to test, the transformer's secondary reads 6.7 volts and the DC out reads 6.1 volts with 6,800uf filter cap.

I'm using a true RMS meter, so I'm not sure what I'm missing.
The 6800µ//8Ω clips the secondary voltage heavily, meaning it is more a trapeze than a sine. This alters (decreases) the peak to rms ratio.
The peak voltage is now ~=8V, and after the rectifier you get your 6.1V.
If you connect an oscilloscope, you'll be able to see the details.
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Old 7th November 2011, 10:40 AM   #6
DF96 is offline DF96  England
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You will be getting nearly 1V pk-pk of ripple so your 6.1V will be varying between about 6.6V and 5.6V over each half-cycle. 6.6 +1.4 (two diode drops) is 8V. A 10VA transformer may have 20% regulation, so an effective secondary resistance a bit under an ohm. This will make the transformer output have its peaks lopped off by the charging pulse, which will be several amps for a short period. Get hold of Duncan's PSUD2 and model it.
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Old 7th November 2011, 11:42 AM   #7
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Hammond agrees with your measured results:

www.hammondmfg.com/pdf/5c007.pdf

At full load Vdc = ~ 0.9 Vac with a full wave bridge rectifier and capacitor input.
As per DF96, the I2R losses are significant.
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Old 7th November 2011, 11:45 AM   #8
AndrewT is online now AndrewT  Scotland
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6.3Vac and 1.5Aac is the maximum continuous AC rating of your transformer. That is another way of telling you that feeding a resistor load the maximum power available is ~9.5W.

When you feed a capacitor input filter, you must de-rate the transformer output to ~750mAdc of continuous output current. That is the maximum current cureent into a resistive load.

750mA into 8r0 requires ~6Vdc after the rectifiers.
The effective DC voltage available is 6+1.4V = 7.4V, but that 1.4 is dissipated internally in the PSU. The transformer is being asked to deliver an effective 5.6W of DC power. That is too much. It is going to get hot.

The previous posters have correctly drawn your attention to the internal transformer and rectifier and smoothing capacitor losses that cannot be ignored.

To complete your experimentation. Remove the 8r0 resistor and measure the emf available at the smoothing capacitor. Is it as predicted by the open circuit voltage of the transformer and the Vdrop of the diodes when passing virtually zero current?
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Last edited by AndrewT; 7th November 2011 at 11:47 AM.
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Old 7th November 2011, 12:04 PM   #9
johnr66 is offline johnr66  United States
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Okay, so it has to do with the way the filter cap draws current only at the peak. I tried a 1000uf cap and the DC dropped to 5.9v. I'm using a GBL04 bridge I found in my parts drawer. Just to be clear, the secondary RMS AC voltage (6.7v) was measured with the load on the output using a decent quality meter. I can't be clear on how well this meter handles DC with a ripple.

Last edited by johnr66; 7th November 2011 at 12:31 PM.
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Old 7th November 2011, 12:17 PM   #10
AndrewT is online now AndrewT  Scotland
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the DC voltmeter samples the voltage and averages it.
The ripple will be effectively ignored by that sampling and averaging.
The DC reading will be the average DC and superimposed on that average is the ripple. That ripple may not be symmetrical, but that is unimportant.

Measure the DC using a high voltage setting on your AC voltmeter.
Is the AC value shown as near zero volts AC?
re-set your AC voltmeter to the next ACV scale down.i.e from 600Vac to 199.9Vac.
Does the value still read near zero volts AC?
If it is still very low, then turn down to the 19.99Vac scale.
If it is still very low go down to the 1.999Vac scale.
This scale should be sensitive enough tho read the effective average ripple voltage superimposed on the DC voltage. The peak to peak ripple voltage is ~3 times the average ripple voltage. A scope can confirm this.

If the AC scale starts to read a high voltage as you move down the scales then stop and disconnect. Your meter may be passing high DC currents which can damage your meter.
This can be solved by adding a high voltage plastic film capacitor to your probes. This DC blocks the AC voltmeter input circuit and provided the capacitor passes the lowest frequency you are interested in, it will give a useful value that you can use to check operation is similar to prediction.
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