Hi Eric,
Thanks a stack. Here's a pic (not great) of the layout I'll be using.
Kevin
An externally hosted image should be here but it was not working when we last tested it.
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Not necessarily.
A Capacitance Multiplier (CM) is specifically a ripple attenuator that has a low voltage drop.
Normal CMs drop sufficient voltage to eliminate the ripple.
We measure the average voltage at our supply rails. For this example I will use 30Vdc.
That supply will have a ripple voltage. Let's assume it is 400mVpp when the client circuit is drawing operating current.
The voltage on the supply rail varies from 30.2Vdc to 29.8Vdc.
The average is 30Vdc and the ripple is 0.4Vpp.
For the CM to work properly, the output must not fall below the trough of the ripple minimum, i.e. not below 29.8Vdc. When the voltage before the CM is above that value the CM absorbs the excess voltage.
A CM works properly when the half ripple voltage below the average level is attenuated so that the dips are not significant. i.e the Vdrop appears to be about the half ripple voltage, i.e. 0.2V in the example given above.
IT DOES NOT need to be 3 to 4 V to be defined as a "properly working CM".
I made a simulation of the power supply I will build. The simulation only shows "half" of the power supply. The V+.
Sorry for the Swedish-English
.
The 15 Ohm is simulationg the load from the amplifier later. About +-21Volt and 1,4 A.
Soon to find out if it also is working properly in parctice.
Sorry for the Swedish-English
.The 15 Ohm is simulationg the load from the amplifier later. About +-21Volt and 1,4 A.
Soon to find out if it also is working properly in parctice.
post246 is not a Capacitance Multiplier.
The active device is an double EF stage.
The output is two Vbe drops below the RC filter voltage.
That RC filter voltage wanders all over the place, as output current changes.
The Vbe drops also wander, though not as badly, as output current changes.
The RC filter needs an extra resistor to make a CM. add R4 parallel with C3. The two series connected resistors make a resistive resistor ladder. That is what makes a CM.
Then you add a capacitance (C3) to the lower leg of the ladder, that gets multiplied by the current gain of the EF stage.
You can use a PNP instead of an NPN to reduce the voltage lost across the active devices.
This is what LDO regulators do to reduce the Vdrop of the regulator.
The active device is an double EF stage.
The output is two Vbe drops below the RC filter voltage.
That RC filter voltage wanders all over the place, as output current changes.
The Vbe drops also wander, though not as badly, as output current changes.
The RC filter needs an extra resistor to make a CM. add R4 parallel with C3. The two series connected resistors make a resistive resistor ladder. That is what makes a CM.
Then you add a capacitance (C3) to the lower leg of the ladder, that gets multiplied by the current gain of the EF stage.
You can use a PNP instead of an NPN to reduce the voltage lost across the active devices.
This is what LDO regulators do to reduce the Vdrop of the regulator.
Hi,
Actually according to ESP, post 246 is a CM.
Please read this from ESP, according to the author R4 (12k) as you mention is not even necessary in the circuit therefore the addition of R4 does NOT create a CM...
Enjoy : Capacitance Multiplier Power Supply Filter
BR,
Eric
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A CM needs a resistor ladder to define the voltage proportion that is available at the active device.
Somewhere around 90% to 95% is usual.
i.e. upper resistor is 1k5, lower resistor is 20k.
The voltage at the junction will be at an average of 20/(20+1.5) * input voltage = 93% of Vin.
To get the multiplier, you now add a capacitor to that junction. Try 330uF
The RC filter giving the average is 1k5||20k (~1k4) & 330uF = 0.46s (a slow averaging).
For an average input voltage of 30Vdc with lots of ripple the ladder has a defined reference voltage of 27.9Vdc, which slowly tracks the changes of Vin.
If you omit the 20k then the voltage at the active device is all over the place. Because the current flowing to the active device determines the Vdrop across the 1k5.
Somewhere around 90% to 95% is usual.
i.e. upper resistor is 1k5, lower resistor is 20k.
The voltage at the junction will be at an average of 20/(20+1.5) * input voltage = 93% of Vin.
To get the multiplier, you now add a capacitor to that junction. Try 330uF
The RC filter giving the average is 1k5||20k (~1k4) & 330uF = 0.46s (a slow averaging).
For an average input voltage of 30Vdc with lots of ripple the ladder has a defined reference voltage of 27.9Vdc, which slowly tracks the changes of Vin.
If you omit the 20k then the voltage at the active device is all over the place. Because the current flowing to the active device determines the Vdrop across the 1k5.
I received my boards and parts in March 2012. If I suspect fakes transistors (MJ15003), what is the best source for genuine Motorola parts? Aside from the different markings, is there a conclusive way to test the parts to know for sure? I haven't hooked up my transistors yet and would rather not if they are faulty/fake parts. I have no problems purchasing genuine parts to improve the odds of a successful build.
Thx. - Rick
With my PCBs I also got 4 MJ15003.
I connected my first built amp with two Lab power supply set at + - 20 V DC, and with current limiting. The VR2 trimpot was first set for minimum current.
The Lab power supply limited the current to 2 A ! And I had to disconnect the amp.
After much work I found out that one MJ15003 was blown.
I the checked the other 2, not used MJ15003, and found that the Emitter Pin was lose !
Right now I have the first amp running at the work bench with the 2 only left Mj15003. Still using the Lab power supply and the current adjusted to "only" 550 mA.
Seem to be stable. The output DC adjusted to about +1mV.
After one hour in operation both the current and DC offset is stable.
Tomorrow I will do some measurement and also deceide if I shall use some old faithful 2N3055s instead.
I connected my first built amp with two Lab power supply set at + - 20 V DC, and with current limiting. The VR2 trimpot was first set for minimum current.
The Lab power supply limited the current to 2 A ! And I had to disconnect the amp.
After much work I found out that one MJ15003 was blown.
I the checked the other 2, not used MJ15003, and found that the Emitter Pin was lose !
Right now I have the first amp running at the work bench with the 2 only left Mj15003. Still using the Lab power supply and the current adjusted to "only" 550 mA.
Seem to be stable. The output DC adjusted to about +1mV.
After one hour in operation both the current and DC offset is stable.
Tomorrow I will do some measurement and also deceide if I shall use some old faithful 2N3055s instead.
As it turned out the transistors that came with my boards are Toshiba 2SC3280 and not the MJ15003. I understand that Toshiba quit manufacturing them in 2000, so these must not be original Toshiba's! I really suspect these devices as they look very used, so I'm leery of moving forward using these devices. Would Siliconray have sold used parts as new? I hope not. I like the TO-247 package over the TO-3, so I'm wondering if there's another replacement device in the TO-247 package that would work with this amp design? -Rick
I need some output transistors for my build on the Siliconray PCBs, how were the MJL21194 transistors you ordered. They are top of my list, especially as I can get some easily, but I wonder whether they'll be too fast and will oscillate? Thanks
Ray