Hi Guys,
A little confused need some tutorial. Assuming that I've a 16 vdc supply rated at 200 ma, I connect a 10H choke across which is rated at 120 ma with a dcr of 210 ohm which are the figures to use to calculate voltage drop at output of choke ?
Thks
A little confused need some tutorial. Assuming that I've a 16 vdc supply rated at 200 ma, I connect a 10H choke across which is rated at 120 ma with a dcr of 210 ohm which are the figures to use to calculate voltage drop at output of choke ?
Thks
The input supply's current capability doesn't mean anything w.r.t. the choke. But you need to make sure that the load doesn't draw more than the choke's capability of 120 mA. Otherwise it's core will saturate, as yet been said, and it's inductance will decrease dramatically (more or less...).
Best regards!
Best regards!
Unclear what is meant by connecting choke "across". Is the choke in series or shunt? If series, before or after the reservoir capacitor?
What is the current draw from the supply? All we have been told is that the supply can do 200mA; this does not mean that it will be doing 200mA, as that depends on the load.
What is the current draw from the supply? All we have been told is that the supply can do 200mA; this does not mean that it will be doing 200mA, as that depends on the load.
Aaaah here you are DF96
Sorry my bad should have been clearer. It’s a CLC, current draw around 90ma. Basic ohms law seems inaccurate for this application
Thks
Sorry my bad should have been clearer. It’s a CLC, current draw around 90ma. Basic ohms law seems inaccurate for this application
Thks
If the choke is the L in the CLC then you just use Ohm's Law to calculate the DC drop across the choke.
In addition, there may be voltage drop from the discharge of the reservoir capacitor but this may already be taken care of if you know how the supply varies with curent without the final LC.
In addition, there may be voltage drop from the discharge of the reservoir capacitor but this may already be taken care of if you know how the supply varies with curent without the final LC.
Thks Df96 calculated via ohms law but figures doesn’t look right. Will build the ps to verify
Thks
Thks
A 90mA (0.090A) load, through a 210 Ohm choke, makes a 18.9V drop.
In light of your 16V supply, this can't be true. More likely your load will wind up near 7V 40mA or something, and not as happy as you would like.
At this _low_ voltage and high current, a 10 Henry choke seems wrong. 16V 90mA is 178 Ohms. 10 Henries at 100Hz is 4,780 Ohms. WAY more than you should need for smoothing in a 178 Ohm circuit.
Conceptually you could un-wind the choke, re-wind it but in four exact-equal (turns count) sections. Put those in parallel. Now you have 0.625H at 13 Ohms DC, 392 Ohms at 100Hz. The 178 Ohm load suggests > 50uFd, 470uFd is cheap-enough. 470uFd is 3.4 Ohms at 100Hz. The 392r choke against the 3.4r capacitor makes 115:1 or 41dB buzz reduction. In fact it will not be quite this good; because of parasitics, it is hard to get 40dB reduction in a single stage. But if re-winding is possible, job done.
Otherwise look for <1H and <15 DCR. A full choke range will include such a part. Same core, different wire gauge and turns count.
In light of your 16V supply, this can't be true. More likely your load will wind up near 7V 40mA or something, and not as happy as you would like.
At this _low_ voltage and high current, a 10 Henry choke seems wrong. 16V 90mA is 178 Ohms. 10 Henries at 100Hz is 4,780 Ohms. WAY more than you should need for smoothing in a 178 Ohm circuit.
Conceptually you could un-wind the choke, re-wind it but in four exact-equal (turns count) sections. Put those in parallel. Now you have 0.625H at 13 Ohms DC, 392 Ohms at 100Hz. The 178 Ohm load suggests > 50uFd, 470uFd is cheap-enough. 470uFd is 3.4 Ohms at 100Hz. The 392r choke against the 3.4r capacitor makes 115:1 or 41dB buzz reduction. In fact it will not be quite this good; because of parasitics, it is hard to get 40dB reduction in a single stage. But if re-winding is possible, job done.
Otherwise look for <1H and <15 DCR. A full choke range will include such a part. Same core, different wire gauge and turns count.
Thank you PRR. I build a supply to test & dummy loaded it with a 4R resistor. Voltage drop from around 17v to 12.5v. Tried using PSUD but it's still a learning process for me. Now I wish that I have a scope. Lol
Easy: 0V DC.
Choke input: 16VDC (or whatever it supplies under load)
Choke output: 0V *always* , by definition, since it´s grounded.
No problem at all because that choke will pass a maximum of 76 mA, no matter what, even if output is a direct short to ground 😱
This does not add up. You better measure the parts and draw-out what you are really doing.
What do you see that is wrong PRR ? Im actually doing it as an experiment to self learn. Did it yrs back when building tube preamp but now testing to see if it works on low voltage applications.
Thks
Thks
If the load is 16V at 90mA, why test with a 4 Ohm load?
If the choke is 210 Ohms, and you load with 4 Ohms, "drop ...17v to 12.5v" makes no sense. We expect a 210r:4r divider to drop to 0.02 of input, or to 0.3V, not 12V.
Please DRAW it up, verify values (the choke may not be as specified), and do some rough math.
We are assuming this is pretty-smooth DC. If it is raw rectified AC, then simple DC math does not get the whole story.
You said earlier that you have a CLC filter with a current requirement of 90mA. Your picture does not show the second C. It seems to me that you have the choke and resistor in series after the first and only capacitor. You need to add a capacitor to ground after the choke.
However with a choke dcr of 210 ohms, you will not be able to maintain a steady voltage from the supply as the choke will drop too much voltage. For your desired 90mA, the choke will drop 210 ohms x .090V = 18.9V. This is not even possible as the transformer and capacitor cannot even provide it. A more reasonable choke dcr would be less than 1 ohm so that a load demand would not reduce the output voltage significantly. A choke with low dcr and with only millihenries of inductance would suffice.
It looks like you have an 100uf input capacitor. Your supply would benefit from having a much larger capacitor to reduce ripple. 1000uf capacitors before and after the choke would help to reduce the ripple, and provide a higher rms voltage out.
PSUD2 example:
However with a choke dcr of 210 ohms, you will not be able to maintain a steady voltage from the supply as the choke will drop too much voltage. For your desired 90mA, the choke will drop 210 ohms x .090V = 18.9V. This is not even possible as the transformer and capacitor cannot even provide it. A more reasonable choke dcr would be less than 1 ohm so that a load demand would not reduce the output voltage significantly. A choke with low dcr and with only millihenries of inductance would suffice.
It looks like you have an 100uf input capacitor. Your supply would benefit from having a much larger capacitor to reduce ripple. 1000uf capacitors before and after the choke would help to reduce the ripple, and provide a higher rms voltage out.
PSUD2 example:
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