I was wondering if I might ask for assistance regarding Hammond's ‘Design Guide For Rectifier Use’, and helping me to choose a power transformer for an amplifier. I have watched Uncle Doug's guides to rectification recently and thought I had it sussed, then started looking for a transformer and re-entered the land of confusion.
Here's the Hammond guide:
https://www.hammfg.com/electronics/transformers/rectifier
If I were to take one transformer as an example and work through with it 369BX: https://www.hammfg.com/part/369BX
The A.C. High Voltage Secondary (RMS) is given as 300V C.T. for this transformer. My first assumption here is that this means 300-0-300.
Working through the examples of HALF WAVE Resistive Load, FULL WAVE Resistive Load and FULL WAVE Bridge Resistive Load, here’s how I interpreted the guide:
HALF WAVE Resistive Load
V D.C. = 0.45 X Sec. V A.C. (All voltages RMS)
So I would interpret this as
V D.C. = 0.45 x 300V (from one of half of the secondary windings when centre tapped)
Giving 135V DC RMS with a lot of ripple
FULL WAVE Resistive Load
V D.C. = 0.45 X Sec. V A.C. (All voltages RMS)
So I would interpret this as
V D.C. = 0.45 x (300V + 300V) (from both halves of the secondary windings when centre tapped)
Giving 270V DC RMS and a lot less ripple
FULL WAVE Bridge Resistive Load
V D.C. = 0.9 X Sec. V A.C. (All voltages RMS)
So I would interpret this as
V D.C. = 0.95 x 600V (skipping the centre tap and having both halves of the secondary windings)
Giving 570V DC RMS
Is this anywhere near correct? I'm not sure I found Hammond's guide and docs all that straightforward, but that's mainly because it's all new to me and I'm not 100% sure I'm on the right path.
Here's the Hammond guide:
https://www.hammfg.com/electronics/transformers/rectifier
If I were to take one transformer as an example and work through with it 369BX: https://www.hammfg.com/part/369BX
The A.C. High Voltage Secondary (RMS) is given as 300V C.T. for this transformer. My first assumption here is that this means 300-0-300.
Working through the examples of HALF WAVE Resistive Load, FULL WAVE Resistive Load and FULL WAVE Bridge Resistive Load, here’s how I interpreted the guide:
HALF WAVE Resistive Load
V D.C. = 0.45 X Sec. V A.C. (All voltages RMS)
So I would interpret this as
V D.C. = 0.45 x 300V (from one of half of the secondary windings when centre tapped)
Giving 135V DC RMS with a lot of ripple
FULL WAVE Resistive Load
V D.C. = 0.45 X Sec. V A.C. (All voltages RMS)
So I would interpret this as
V D.C. = 0.45 x (300V + 300V) (from both halves of the secondary windings when centre tapped)
Giving 270V DC RMS and a lot less ripple
FULL WAVE Bridge Resistive Load
V D.C. = 0.9 X Sec. V A.C. (All voltages RMS)
So I would interpret this as
V D.C. = 0.95 x 600V (skipping the centre tap and having both halves of the secondary windings)
Giving 570V DC RMS
Is this anywhere near correct? I'm not sure I found Hammond's guide and docs all that straightforward, but that's mainly because it's all new to me and I'm not 100% sure I'm on the right path.
Your first assumption is not correct. 300V C.T. means 150-0-150.
But if it were 300-0-300 than I think your calculations are right, but I'm not sure if you interpreted the "HALF WAVE Resistive Load" correctly. You calculate with only one half of the total winding. That would be correct if you would only use half a winding, which is possible but a bit unusual.
All your examples are for resistive loads, so not of much use in audio electronics.
The factors being used for the voltages on the Hammond site (0.45, 0.9 and 1.41) are not the factors you wil experience in real life, especially if you would use tube rectifiers.
An example: When using an Amplimo 28070 toroid (with the two 115 V secondaries in series to give 230 V at 0.22 A maximum) with a bridge rectifier made out of 4 x 1N4007, and a capacitor of 220 uF after the bridge, I get about 307 V while the current draw is 113 mA. So the 'real life factor' in this case is 307/230 = 1.33. And probably this 'real life factor' is even a bit lower in reality because the mains voltage at my place is a bit higher than 230 V, while the primary is for 230 V.
The diffences are caused by resistance in the windings (even the resistance of the primary winding has some influence) and the rectifier(s).
If I use tube rectifiers, I prefer to study the datasheets. In Philips datasheets you see curves for a couple of different secondary voltages and different resistances. That gives me a good enough indication of the range of DC voltage I can expect after rectification.
But if it were 300-0-300 than I think your calculations are right, but I'm not sure if you interpreted the "HALF WAVE Resistive Load" correctly. You calculate with only one half of the total winding. That would be correct if you would only use half a winding, which is possible but a bit unusual.
All your examples are for resistive loads, so not of much use in audio electronics.
The factors being used for the voltages on the Hammond site (0.45, 0.9 and 1.41) are not the factors you wil experience in real life, especially if you would use tube rectifiers.
An example: When using an Amplimo 28070 toroid (with the two 115 V secondaries in series to give 230 V at 0.22 A maximum) with a bridge rectifier made out of 4 x 1N4007, and a capacitor of 220 uF after the bridge, I get about 307 V while the current draw is 113 mA. So the 'real life factor' in this case is 307/230 = 1.33. And probably this 'real life factor' is even a bit lower in reality because the mains voltage at my place is a bit higher than 230 V, while the primary is for 230 V.
The diffences are caused by resistance in the windings (even the resistance of the primary winding has some influence) and the rectifier(s).
If I use tube rectifiers, I prefer to study the datasheets. In Philips datasheets you see curves for a couple of different secondary voltages and different resistances. That gives me a good enough indication of the range of DC voltage I can expect after rectification.
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Lovely jubbly, good to know where I'm going wrong.
In the example of half wave rectification I'd seen, you only seemed to use half a winding with one diode to get your DC, whereas full wave involved using both windings, and then combining the two out of phase DC currents. I'm not sure I understand how else to use half wave rectification without it just becoming full wave.
Also, thank you for pointing out that I should be looking at capacitive loads rather than purely resistive, is that owing to always using a circuit with capacitive filtration and the subsequent charging/discharging of said capacitor?
Thanks rdf, will look at the duncan sim.
In the example of half wave rectification I'd seen, you only seemed to use half a winding with one diode to get your DC, whereas full wave involved using both windings, and then combining the two out of phase DC currents. I'm not sure I understand how else to use half wave rectification without it just becoming full wave.
Also, thank you for pointing out that I should be looking at capacitive loads rather than purely resistive, is that owing to always using a circuit with capacitive filtration and the subsequent charging/discharging of said capacitor?
Thanks rdf, will look at the duncan sim.
There are two main ways to make full-wave rectification. A CT winding and two diodes; and a single winding with 4-diode bridge.
There are several mistakes on that chart, which Hammond did not create. They make transformers, they carefully avoid saying how to use them.
98% of home audio will use Capacitor Input filter, because caps are cheap today.
There is a tradition (from costly-cap days) of Choke Input filters. These have advantages, but often over-stated, and today chokes cost more than capacitors.
Un-filtered DC is only for electroplating and rude battery charging.