chrispsound
Well-known member
Also I forgot to mention that I am using the string of zeners instead of the regulator tubes. Thanks again. ChrisP
P2P Redd 47 - a few questions
- Thread starter letterbeacon
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That makes a serious difference. How many resistors are there between the ca. 400V unreg voltage and the Zeners? What is the Zener voltage?
If there is only one 1K res and the zener voltage is ca. 310V, then you're dissipating 90v in it, resulting in 8W of dissipation.
If there is only one 1K res and the zener voltage is ca. 310V, then you're dissipating 90v in it, resulting in 8W of dissipation.
chrispsound
Well-known member
Can't see th eschemo. Way too small.
Drawing 45mA from the second tube seems largely exaggerated anyway.
Can you check the voltages without the tubes?
Drawing 45mA from the second tube seems largely exaggerated anyway.
Can you check the voltages without the tubes?
chrispsound
Well-known member
Sorry about the schematic, it is on page 7 of this thread posted by earthsled. For the purposes of testing I am just using the power supply(schematic is on my original post) with just 2 capacitors a 1k 5w resistor and a 470k bleeder and I am not using zeners. Since my toroid PT has many taps I am now using the 220vAC tap. My power supply is now at 300vDC unloaded after rectification and filtering. With the E88CC pulled I am getting about 297v at pin 1 and 6 and no voltage on any other pin at tube socket except heater ac at pin 4 and 5. I am getiing 299 volts at the top of the plate resistor. I think my tube may be bad? Thanks for the help so far.
With current pulled through ECC88's plate resisor voltage will drop to around 160V meassured directly at the plate pin. Check if you properly connected cathode resistor and the rest of the circuit. Your current draw is around 2x higher than it should be! I think you have a problem with preamp/last tube circuit, not PSU.
chrispsound
Well-known member
I am getting proper voltage drop across the plate resistor, the problem I am having is excessive current draw so the plate resistor is getting way hot, I quadruple checked everything else and the 200R cathode resistor is in properly. I did notice that my cathode voltage is around 10 volts were the original schematic says 3.7 volts.AZ999 said:
It seems to indicate the second triode is shorted. Check the plate voltage.
chrispsound
Well-known member
According to the original schematic, plate voltage should be 138 volts and cathode should be 3.7 volts. Voltage pre plate resistor should be 290 volts with the tubes in, if I am reading the original schematic correctly. Is this correct? What should voltage be with tubes removed? Is it possible that I damaged the tube by sending too high plate voltage?
With the tube removed, the plate voltage should be equal to B+ and the cathode voltage zero.
chrispsound
Well-known member
I ordered another tube, is there a way to check the tube pins with the tube out with a multi-meter to determine if the tube is shorted internally? Thanks again abbey road d enfer for all your responses, if you are ever in Southern California let me know, I will buy you a beer(or whatever else you like to indulge in). 
ChrisP
ChrisPYes. Just check in ohms mode the resistance between catode and plate. Should be infinite.chrispsound said:
You may want to check your wiring. A short happens so easily.
chrispsound
Well-known member
If there was a wiring short wouldn't it still be drawing current with the tube out?
99% correct; there's always the exception where engaging the tube pushes something.chrispsound said:
earthsled
Well-known member
Sorry to bring up this question again, but I just found some conflicting info for the two-diode full wave current calculation on this page: http://www.sowter.co.uk/rectifier-transformer-calculation.htm
So, if 60mA current is needed (for a 2 ch REDD.47), should the transformer be rated at 60mAx1.414 or simply 60mA? I don't want to get a current rating too high due to the flux issues discussed earlier.
Thanks!
chrispsound
Well-known member
http://www.analogmetric.com/goods.php?id=78 This is the 100mA version, I have the 50mA version of this one, has a 18v winding that I am going to use for for phantom. It seemed to work well with their 6SN7 pre-amp that I ordered as well, but I am a newb and have no way of testing quality, but voltages were all close to spec and the transformer did not get warm. They seem a little shady but everything arrived from Hong Kong in a timely manner and in one piece. Built like a tank with a nice mounting bracket and a lot of leads for prototyping. ChrisP
This is confusing because the actual mean value given by this valculator is correct, and that's what should be taken into account for the DCR related losses, but in terms of power, which is your concern I believe, the law of conservation of energy prevails. Since the DC voltage is 1.4 times the ACrms, the AC rms amps is 1.4 times the DC amps. If 380V 20mA is your starting point, you need 7.6W for B+. You need some safety margin, but a factor 2, as suggested earlier, may be somewhat excessive.earthsled said:
The mains xfmr in the V72 may have 20% safety margin, no more.
You do need a 100mA version.
earthsled
Well-known member
Does the same formula hold true for both bridge (four diodes), and full-wave (two diodes with CT)?
Excellent! A 20% safety margin for current was exactly what I was figuring.
Thanks for your help!
The same power formula in both cases. So, for the bridge, the voltage is 380/1.4 and the current is 60mx1.4.earthsled said:
For the full-wave CT, the voltage is 2x380/1.4 and the current is 1/2.60mx1.4. The actual mean current used to evaluate resistive losses is slightly higher.
There are two dimensional design parameters: the power rating, which ensures that there's enough energy transfer possible, and the current rating, which basically governs the gauge of the wire. Transformers are spec'd for resistive loading (sine current) where the power dissipation is r.I² (r is the DCR of the winding, I is the rated rms current).
With a rectifier, current is not a sinewave anymore
The problem with the calculations on the Sowter website is that they don't explain these differences, and anyway they don't mention a very important factor, which is the fact that when hitting the reservoirs, there are large current spikes, which govern the actual instant DCR loss and thence the actual output voltage. The DCr must be several times smaller than whatever of the heat dissipation or average loss calculations infers. Now it depends on the application. For a low-level app such as a preamp, the voltage can be allowed to sag, in particular with class A operation where current does not vary much, but in power applications both average and peak losses are a problem.
Finally, looking at it more deeply, I can't figure out where the 1.61 coefficient comes from. Golden number? :-\
Maybe someone who's still fresh from school would chime in...?
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