Hi all,
I've got a power trans, marker 700V/90mA and I rectify half wave, using 5R4, got 460VDC.
A man said that, max current is 180mA when I rectify half way like that. But in my knowledge, I think that it's just 90mA.
His theory: he said that, this trans has 90mA for full wave, so in case half wave, to preserve energy, max current will be multied 2, means 180mA
My theory: I know that max current of trans belongs to gauge of wire. So, gauge is constant then max current is constant too. So, in this case, max current is 90mA, not 180mA.
Pls, could you tell me who is right, and explain why is it. In fact, I really wanna his theory is right, so I can use this one to provide for some power tube like 6L6 or 6B4G...
ThX & Rgds,
Cominup
I've got a power trans, marker 700V/90mA and I rectify half wave, using 5R4, got 460VDC.
A man said that, max current is 180mA when I rectify half way like that. But in my knowledge, I think that it's just 90mA.
His theory: he said that, this trans has 90mA for full wave, so in case half wave, to preserve energy, max current will be multied 2, means 180mA
My theory: I know that max current of trans belongs to gauge of wire. So, gauge is constant then max current is constant too. So, in this case, max current is 90mA, not 180mA.
Pls, could you tell me who is right, and explain why is it. In fact, I really wanna his theory is right, so I can use this one to provide for some power tube like 6L6 or 6B4G...
ThX & Rgds,
Cominup
Voltage
Voltage is .707 of peak with full wave rectification and no filter capacitor. Or .3535 of RMS with half wave and no filter cap. With half or full wave rectification and a filter cap, you should be getting around 1000V. Therefore, either the circuit is not as described, something is defective or the loading is very heavy.
And if the steady current draw is 90mA, with a filter cap, peak current will be a couple amps. A diode rated at a minimum of 200mA continuous should be used for a 90mA load.
Voltage is .707 of peak with full wave rectification and no filter capacitor. Or .3535 of RMS with half wave and no filter cap. With half or full wave rectification and a filter cap, you should be getting around 1000V. Therefore, either the circuit is not as described, something is defective or the loading is very heavy.
And if the steady current draw is 90mA, with a filter cap, peak current will be a couple amps. A diode rated at a minimum of 200mA continuous should be used for a 90mA load.
Hi,
why did you miss out that crucial info first time around?
Can you separate the centre tap into two windings? 0-350 & 0-350?
you have options;- full rectify or bridge with centre common.
If two windings then add in // outputs with full rectification and double current or bridge without centre common again at double current.
why did you miss out that crucial info first time around?
Can you separate the centre tap into two windings? 0-350 & 0-350?
you have options;- full rectify or bridge with centre common.
If two windings then add in // outputs with full rectification and double current or bridge without centre common again at double current.
cominup said:
What he's saying is this, the transformer wire is rated at 90ma continuous. Assume you have a center tap and it's grounded, you now have two secondary outputs with respect to the center tap. With a 5AR4 for example, you half wave rectify each of the two secondaries for full wave between both of them. With 90ma continuous rating, you pull that much current for only half of the AC sine wave cycle and zero for the other half of the cycle for each secondary. So (90ma + 0ma)/2 cycles = 45ma average. Remember that a AC sine wave is periodic (1 cycle) from 0 to 2pi. When you rectify it, you are periodic (1cycle) from 0 to pi or half of the original, thats where the 2 cycles come from. Your 2nd cycle is blocked for a half wave and conducts nothing. So he's saying that you can pull 180ma for each *half* cycle and average 90ma on each *full* cycle on each secondary output. So with both secondaries working as half wave rectified outputs together you get 180ma. Also keep in mind that each secondary is out of phase with the other one so one conducts while the other is blocked. Each secondary sees an *average* of 90ma continuous while the load pulls 180ma.
This is NOT the case if you don't use the center tap, in which case you revert back to 90ma.
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