Thank you, resuming low ESR, high ripple & high temperature.
There is a relation between the current draws & the amount of capacitance?
I ask because normally for tube preamps drawing 60mA I used two caps 100uF each & was enough but when used the same PSU for an OTL headphone amp drawing 140mA I had to increase to 660uf (3 caps 220uF).
There is a relation between the current draws & the amount of capacitance?
I ask because normally for tube preamps drawing 60mA I used two caps 100uF each & was enough but when used the same PSU for an OTL headphone amp drawing 140mA I had to increase to 660uf (3 caps 220uF).
There is a relation between the current draws & the amount of capacitance?
The more current drawn, the more ripple results, for a given capacitance.
For DC current, I = C x delta(V)/delta(t).
Decide what % ripple is needed for your given supply voltage,
and then you know delta(V). Also, the supply refresh rate delta(t)
is 1/120 seconds (for 60Hz).
Correct capacitance value. Lowish ESR. Adequate ripple curent rating (if used as a reservoir cap). Good toleration of temperature.
Lifetime is a good thing to look at too.
for a
The law of the capacitor: C = Q/V, where Q is the charge on one of the plates.
Then rearranging, we have Q = C x V.
Differentiate both sides wrt time: dQ/dt = (C x dv/dt + v x dC/dt)
We know that dQ/dt = i, the current. Normally C is considered to be a constant, so dC/dt = 0.
Then we have i = C x dv/dt
For DC we can use for dv = delta(v), the p-p amount of ripple voltage
between rectifier recharge pulses. For a FWR this is 1/120 second for 60Hz,
or 1/100 second for 50Hz. The i is now I, the amount of constant DC current drawn.
Then C = I / (ripple voltage/recharge time), or C = I / (delta(v) / delta(t) )
So we have C = I x delta(t) / delta (v), which is the capacitance needed
for the desired amount of ripple delta(v), neglecting the ESR.
Thanks for support rayma.
C = i / (delta (V) / delta (t) )
???
The law of the capacitor: C = Q/V, where Q is the charge on one of the plates.
Then rearranging, we have Q = C x V.
Differentiate both sides wrt time: dQ/dt = (C x dv/dt + v x dC/dt)
We know that dQ/dt = i, the current. Normally C is considered to be a constant, so dC/dt = 0.
Then we have i = C x dv/dt
For DC we can use for dv = delta(v), the p-p amount of ripple voltage
between rectifier recharge pulses. For a FWR this is 1/120 second for 60Hz,
or 1/100 second for 50Hz. The i is now I, the amount of constant DC current drawn.
Then C = I / (ripple voltage/recharge time), or C = I / (delta(v) / delta(t) )
So we have C = I x delta(t) / delta (v), which is the capacitance needed
for the desired amount of ripple delta(v), neglecting the ESR.
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Please could you do the maths for a known current, for example 100mA?
How much maximum ripple voltage do you want on the rectifier input capacitor?
This would typically be less than 5% of the DC supply voltage.
Also, what's the rectified DC supply voltage?
For 8.5V ripple, 100mA DC current, and 60Hz operation, we have:
C = I X delta(t) / delta(v)
and then C (in farads) = 0.1A x (1/120 Seconds) / (8.5V), so C = 98uF.
Use a 100uF, 400V capacitor. This will be a problem for a tube rectifier.
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This is for the 1st input mains capacitor or the total capacitance of the mains PSU?
This is only for the input capacitor, just after the rectifier.
The subsequent dropping resistors and capacitors can reduce
the residual ripple that is left, by RC filter action.
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Mine calculations: 0.14A x (1/120 seconds) / 8.5V = 16.1476uF
You dropped a decimal somewhere, since that comes to about 137uF.
What about if a bigger cap is used when rectifier is all SS?
That will be ok if the inrush current is tolerable, but add a discharge resistor to ground for safety.
Wich value? Resistor connected between + & -? P.S. I have now 220uF
Add a resistor to ground somewhere in the supply, to discharge it in a minute or less for safety.
A reasonable value would be around 200k. Take into account any upstream series resistors,
because the voltage will drop since it shunts current to ground.
You can consider using a high temperature cap (105C) versus a standard temperature rated cap (85C). Even though the high temp caps don't have particularly long life ratings every reduction in temperature of 10C is said to double capacitor life. So with the difference (20C) you are looking at four times rated life assuming your temperatures themselves were below 85C before or as designed.
Higher voltage rating will typically result in better ripple voltage rating, but be conservative here because (electrolytic power supply) caps work best when close to the voltage rating. So maybe a 63V when a 50 is specified, but no higher.
Higher voltage rating will typically result in better ripple voltage rating, but be conservative here because (electrolytic power supply) caps work best when close to the voltage rating. So maybe a 63V when a 50 is specified, but no higher.
Add a resistor to ground somewhere in the supply, to discharge it in a minute or less for safety.
A reasonable value would be around 200k. Take into account any upstream series resistors,
because the voltage will drop since it shunts current to ground.
That's a bleeder resistor, I used in the Valve Itch phono preamp.
What's better spec first the ripple or the esr?
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