"idont know if u noticed that am paralleling the 2 transformers together just to get 24-0-24 secondries since icant get a trans.
in here rated as the same sec. volts and current.. so icant simply
build a 2 separated PSUs as u suggested.."
I'm not sure I understand. You have two transformers? So for each transformer you make a supply, each outputing 0V and 30V (max). Neither is split rail. My comments are based on the schematic you posted.
"and about getting more current using the pass transistor as u
mentioned ,ill need to get a higher rated trans. and amnt sure i
can but ill try .. and i think ineed to add more filter capacitors
right?"
No extra filter caps are needed. When you say "trans." do you mean transistor or transformer? Yes in either case you need the transformer to be able to supply the current you want and you need the pass transistor current and power rating to be adequate. Tell me the current you want and I will suggest some common parts that you are most likely to find in most places...except maybe Egypt!
in here rated as the same sec. volts and current.. so icant simply
build a 2 separated PSUs as u suggested.."
I'm not sure I understand. You have two transformers? So for each transformer you make a supply, each outputing 0V and 30V (max). Neither is split rail. My comments are based on the schematic you posted.
"and about getting more current using the pass transistor as u
mentioned ,ill need to get a higher rated trans. and amnt sure i
can but ill try .. and i think ineed to add more filter capacitors
right?"
No extra filter caps are needed. When you say "trans." do you mean transistor or transformer? Yes in either case you need the transformer to be able to supply the current you want and you need the pass transistor current and power rating to be adequate. Tell me the current you want and I will suggest some common parts that you are most likely to find in most places...except maybe Egypt!
traderbam said:"idont know if u noticed that am paralleling the 2 transformers together just to get 24-0-24 secondries since icant get a trans.
in here rated as the same sec. volts and current.. so icant simply
build a 2 separated PSUs as u suggested.."
I'm not sure I understand. You have two transformers? So for each transformer you make a supply, each outputing 0V and 30V (max). Neither is split rail. My comments are based on the schematic you posted.
"and about getting more current using the pass transistor as u
mentioned ,ill need to get a higher rated trans. and amnt sure i
can but ill try .. and i think ineed to add more filter capacitors
right?"
No extra filter caps are needed. When you say "trans." do you mean transistor or transformer? Yes in either case you need the transformer to be able to supply the current you want and you need the pass transistor current and power rating to be adequate. Tell me the current you want and I will suggest some common parts that you are most likely to find in most places...except maybe Egypt!
1st, am using 2 x (12-0-12) transformer in a parallel connection
to give 24-0-24 secondries (deal with it as one transformer),
so am able to be within the output range of say +-1.25 > 32V.
as shown in the schematic.
my transformer are 3A rated, so if iwant to modify the schematic
so the regulator can provide up to say 3A also,, dont i need more
filter caps.??
take alook at this:
http://ourworld.compuserve.com/homepages/Bill_Bowden/page12.htm
BTW what about the LM317K TO-3 package?
can handle more current?
Yep, the article shows clearly how to do things. So what I'm suggesting is that you build two of the "LM317 with pass transistor" circuits. Make them identical, and include the pot for voltage adjustment as in your original circuit.
The two transformers thus have their primaries in parallel and their secondaries completely separate. You don't use the centre taps. What you then get is two identical, floating psus that can be connected in series like a couple of batteries.
As for smoothing caps, the caps get recharged every 10ms (if your mains is 50Hz) by the transformer. At 3A of current, the voltage drop will be (1/C)*3 volts/s or (1/C)*0.03 V/10ms. Your schematic shows about 5000uF so the drop will be 200*0.03=6 volts. This is too much drop, so you need more C as you point out. Eg: for a 1V ripple you need 30000uF. In any case, for 1.5A you still need 15,000uF for 1V ripple so you may be a little short in your schematic.
I was wrong! Careless.
The two transformers thus have their primaries in parallel and their secondaries completely separate. You don't use the centre taps. What you then get is two identical, floating psus that can be connected in series like a couple of batteries.
As for smoothing caps, the caps get recharged every 10ms (if your mains is 50Hz) by the transformer. At 3A of current, the voltage drop will be (1/C)*3 volts/s or (1/C)*0.03 V/10ms. Your schematic shows about 5000uF so the drop will be 200*0.03=6 volts. This is too much drop, so you need more C as you point out. Eg: for a 1V ripple you need 30000uF. In any case, for 1.5A you still need 15,000uF for 1V ripple so you may be a little short in your schematic.
I was wrong! Careless.
traderbam said:Yep, the article shows clearly how to do things. So what I'm suggesting is that you build two of the "LM317 with pass transistor" circuits. Make them identical, and include the pot for voltage adjustment as in your original circuit.
The two transformers thus have their primaries in parallel and their secondaries completely separate. You don't use the centre taps. What you then get is two identical, floating psus that can be connected in series like a couple of batteries.
As for smoothing caps, the caps get recharged every 10ms (if your mains is 50Hz) by the transformer. At 3A of current, the voltage drop will be (1/C)*3 volts/s or (1/C)*0.03 V/10ms. Your schematic shows about 5000uF so the drop will be 200*0.03=6 volts. This is too much drop, so you need more C as you point out. Eg: for a 1V ripple you need 30000uF. In any case, for 1.5A you still need 15,000uF for 1V ripple so you may be a little short in your schematic.
I was wrong! Careless.
look, ive never seen any PS. schematic thats based on the LM3x7
regulators that have more than 2200uF filter caps!
in my schematic am using 4400uF filter caps which i think is more
than enough upto 1.5A, but to be honest idont know an exact
accurate equation or methode to determine the proper filter caps value!
so what do u think ya all?!
how to properly determine the filter caps value with respect to the
max load current?
There is an equation that describes the discharge voltage over time of a capacitor into a resistance. It is:
v(t) = Vo * (1-e^(-t/RC))
where:
v(t) is the voltage of the capacitor at time t
Vo is the voltage of the capacitor at time t = 0
t is the discharge time, in seconds
C is the capacitance of the input filter capacitor, in Farads
R is the load resistance, in Ohms.
For a power supply, you can calculate R by noting that Rload = Vload/Iload. Thus, if you are putting out 15 volts and 1.5 Amps, the load resistance is 10 Ohms.
Note that for a 50 Hz power source, t will be 10 ms, assumming full wave rectification. For 60 Hz, it will be 8.33 ms. The value of Vo will be the voltage to which the input filter capacitors are charged, which should also be equal to their no load voltage.
Thus, for a 50 Hz system where the regulator is outputting 15 volts and 1.5 A and where the filter capacitors of 4400 uF charge to 20 volts at no load, the following equation will tell us to what voltage the filter capacitors will drop before being recharged:
v(t=10 ms) = 20 * (1 - e^(-.01/.0044*15/1.5))
I do not have my scientific calculator with me so I can't find the answer to this. The ripple voltage one would expect would be the difference between Vo and the calculated value of V(t). If you want less ripple, increase C to a higher value and resolve.
One could solve the capacitor equation in terms of C, which would involve natural logarithms. I think the result is:
C = -t/(R* ln (1 - v(t)/Vo))
So, if you want to see what capacitance is needed to have a one volt ripple, assumming the same parameters as earlier:
C = -.01( 15/1.5 * (1 - 19/20))
v(t) = Vo * (1-e^(-t/RC))
where:
v(t) is the voltage of the capacitor at time t
Vo is the voltage of the capacitor at time t = 0
t is the discharge time, in seconds
C is the capacitance of the input filter capacitor, in Farads
R is the load resistance, in Ohms.
For a power supply, you can calculate R by noting that Rload = Vload/Iload. Thus, if you are putting out 15 volts and 1.5 Amps, the load resistance is 10 Ohms.
Note that for a 50 Hz power source, t will be 10 ms, assumming full wave rectification. For 60 Hz, it will be 8.33 ms. The value of Vo will be the voltage to which the input filter capacitors are charged, which should also be equal to their no load voltage.
Thus, for a 50 Hz system where the regulator is outputting 15 volts and 1.5 A and where the filter capacitors of 4400 uF charge to 20 volts at no load, the following equation will tell us to what voltage the filter capacitors will drop before being recharged:
v(t=10 ms) = 20 * (1 - e^(-.01/.0044*15/1.5))
I do not have my scientific calculator with me so I can't find the answer to this. The ripple voltage one would expect would be the difference between Vo and the calculated value of V(t). If you want less ripple, increase C to a higher value and resolve.
One could solve the capacitor equation in terms of C, which would involve natural logarithms. I think the result is:
C = -t/(R* ln (1 - v(t)/Vo))
So, if you want to see what capacitance is needed to have a one volt ripple, assumming the same parameters as earlier:
C = -.01( 15/1.5 * (1 - 19/20))
It is actually simpler than Jeff suggested, since his formula
assumes the capacitor is discharged by a resistor. In this
case it is not appropriate to assume the load is resistive.
If the design specification is that the PSU should be able
to source 1.5A, as in your case, it is more appropriate to
model the load as a 1.5A current sink, since the load current
is typically not correlated to the mains frequency.
We know that the charge Q in a capacitor is Q = C*V, where
V is the voltage over it. We also know that if we charge a
capacitor with a constant current I the charge is Q=I*t, where
t is the time we charge it. Putting this together we get
C*V = I*t
we can differentiate this to get the voltage change over time
C*dV = I*dt
I is given by our design specification, that is, the max current
the PSU must be able to supply to the load. dt is the time
between charging phases. To simplify this we may assume that
charging the capacitors takes no time. If we have 50Hz mains
and use full-wave rectification, we can thus set dt = 10ms and
get
dV = (I*dt)/C = (1.5*0.01)/C
If C is 1000uF, we get dV = 15V
If C is 10000uF, we get dV = 1.5V, which is more reasonable.
Note that this is an overestimation, since we overestimated
dT, so the voltage drop will be somewhat lower in practice.
I would not go below 4700uF, which would give a loss of
about 3V. Maybe even that is too much, but that depends on
other factors such as max. output voltage, min. voltage drop
over the regulator and over the bridge etc.
assumes the capacitor is discharged by a resistor. In this
case it is not appropriate to assume the load is resistive.
If the design specification is that the PSU should be able
to source 1.5A, as in your case, it is more appropriate to
model the load as a 1.5A current sink, since the load current
is typically not correlated to the mains frequency.
We know that the charge Q in a capacitor is Q = C*V, where
V is the voltage over it. We also know that if we charge a
capacitor with a constant current I the charge is Q=I*t, where
t is the time we charge it. Putting this together we get
C*V = I*t
we can differentiate this to get the voltage change over time
C*dV = I*dt
I is given by our design specification, that is, the max current
the PSU must be able to supply to the load. dt is the time
between charging phases. To simplify this we may assume that
charging the capacitors takes no time. If we have 50Hz mains
and use full-wave rectification, we can thus set dt = 10ms and
get
dV = (I*dt)/C = (1.5*0.01)/C
If C is 1000uF, we get dV = 15V
If C is 10000uF, we get dV = 1.5V, which is more reasonable.
Note that this is an overestimation, since we overestimated
dT, so the voltage drop will be somewhat lower in practice.
I would not go below 4700uF, which would give a loss of
about 3V. Maybe even that is too much, but that depends on
other factors such as max. output voltage, min. voltage drop
over the regulator and over the bridge etc.
the error amp does the de-rippling
that's the purpose of the regulator and the reason you don't need high capacitance -- if you place 1000u or 2000u cap ahead of the regulator, the regulator has less work to do but it's an optimization problem, not a philosophical one !
But it is important to distinguish noise from ripple.
you will get "tens of millivolts" of noise with an LM317/337 -- squeaking this down to negligible levels takes work --- Wenzel Associates application note on "finnessing regulator noise" has a good description of how this is done with a little feedback, and can get the noise down to very small levels -- comparable with voltage references.
it gets beyond the point of this discussion, but there are switching regulators and gate drivers with a few hundred uV (100uV in one case) of switching noise, problem is that they are in the "several dollar range" while an LM317 can be had for about $0.25 even for the hobbyist if they shop carefullly.
that's the purpose of the regulator and the reason you don't need high capacitance -- if you place 1000u or 2000u cap ahead of the regulator, the regulator has less work to do but it's an optimization problem, not a philosophical one !
But it is important to distinguish noise from ripple.
you will get "tens of millivolts" of noise with an LM317/337 -- squeaking this down to negligible levels takes work --- Wenzel Associates application note on "finnessing regulator noise" has a good description of how this is done with a little feedback, and can get the noise down to very small levels -- comparable with voltage references.
it gets beyond the point of this discussion, but there are switching regulators and gate drivers with a few hundred uV (100uV in one case) of switching noise, problem is that they are in the "several dollar range" while an LM317 can be had for about $0.25 even for the hobbyist if they shop carefullly.
"that's the purpose of the regulator and the reason you don't need high capacitance"
Yes, but it depends how high a regulated voltage you want at the output. For a 24Vac xformer you should get about 34Vdc peak. The highest, regulated output voltage will be 34V - ripple voltage - regulator drop-out voltage. At 3A using 5000uF pre-reg you get 6V ripple. If the drop-out is 3V(?), then the max regulated output at full current will only be 34-6-3=25V. And a little more will be lost due to the currrent sense of the bootstrap circuit.
Yes, but it depends how high a regulated voltage you want at the output. For a 24Vac xformer you should get about 34Vdc peak. The highest, regulated output voltage will be 34V - ripple voltage - regulator drop-out voltage. At 3A using 5000uF pre-reg you get 6V ripple. If the drop-out is 3V(?), then the max regulated output at full current will only be 34-6-3=25V. And a little more will be lost due to the currrent sense of the bootstrap circuit.
well, about the filter capacitance.. ive read achapter in a good
book (its about linear IC circuits) in a chapter dealing with the
powersupplies. that i only need about 1000uF of filter capacitance
and it recommends 2000uF per 1A !!
so am in confusion
anyway ive found a good brand of 50V 10,000uF capacitors ,so
ithink ican put about 30,000uF into each rail so when mesured
ill get about 1.6V ripple at 5A ,ithink its acceptable..
and sure to get the 5A output ill use the pass transistor (2N2955)
and substitute the 2 transformers with a more powerful ones (maybe 10A ratings)..
so anything iforgot before buying the new items?!
book (its about linear IC circuits) in a chapter dealing with the
powersupplies. that i only need about 1000uF of filter capacitance
and it recommends 2000uF per 1A !!
so am in confusion
anyway ive found a good brand of 50V 10,000uF capacitors ,so
ithink ican put about 30,000uF into each rail so when mesured
ill get about 1.6V ripple at 5A ,ithink its acceptable..
and sure to get the 5A output ill use the pass transistor (2N2955)
and substitute the 2 transformers with a more powerful ones (maybe 10A ratings)..
so anything iforgot before buying the new items?!
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