Increasing current capacity for capacitance multiplier circuit

[Potato Cakes]

Hello, everyone!

I have been using a capacitance multiplier circuit for a couple of years now with a very high degree of success in drastically reducing noise when using low cost SMPS with various audio circuits (tube microphones, compressors, preamps, etc). So much so that I have been gradually removing linear, transformer based PSUs in gear I have built and replacing them with this circuit.

A little while a go I ran into an issue when I was using this with an eight channel 1290 style preamps build that the BD140 would get very hot and melt the solder used to connect it to the board. I added some small heat sinks that that seemed to solve the issue. But if I wanted to increase the current capacity beyond what the BD140 can do, would I simply just use a higher amp rated transistor in place of the BD140? The BC550 did not seem affected as the load increased.

Thanks!

Paul

 

[gyraf]

Dissipated heat in watts is (always) just input-output voltage times the current. A watt heats like a watt heats - the exact amount is easy to calculate from the degC/W rating of your pass element housing with-or-without thermal coupling to a heatsink

You need more current or decide to drop more voltage? More watts dissipated as heat.

Your capacitance multiplication follows the gain (beta) of the transistor, you may want to use a darlington here. or even a mosfet.

Most often I use IRF840's or the like for HV rail denoising

 

[ccaudle]

if I wanted to increase the current capacity beyond what the BD140 can do, would I simply just use a higher amp rated transistor in place of the BD140?

There are two considerations:

• maximum current rating of the transistor, which is derated with temperature, it is not a static value
• maximum temperature rating

Calculating basic thermal behavior is a lot like Ohms law, with power dissipation as the "voltage" source, the temperature difference between the device and ambient as the "voltage" difference, and the thermal resistance of the die to package, package to PCB and package to heat sink, and heat sink to air as the analog to the circuit resistance.
The thermal resistance is given in degrees C per Watt, so you multiply power dissipation by thermal resistance, and that increase in temperature is relative to the ambient temperature, so would add to e.g. 25C for room temperature use, or maybe 35C or 40C if you want to handle operating outside in the summer time in the shade, or even higher if you have to handle operating outdoors in direct sunlight (since sunlight shining on a chassis would cause the base temperature of the heatsink to rise even before it started dissipating any power).

This link is one of the first I found about the topic, but probably all of the large transistor manufacturers have similar app notes:
https://www.rohm.com/electronics-basics/transistors/tr_what7
https://resources.pcb.cadence.com/b...emperature-transistor-temperature-operability

So you need to know power dissipation but also the thermal resistance of the package to air, and the total resistance of die to package to heatsink to air if using a heatsink. Heatsink datasheets should include that value, but it is good to design in some margin, because the data was presumably gathered using optimal installation technique in terms of proper thermal paste, spreading a very thin even layer, proper torque of any mounting screws, etc.

the BD140 would get very hot and melt the solder used to connect it to the board

Most transistors are rated for a maximum junction temperature of 125degC, and most solder melts around 250degC, so I would change the transistors as well, you significantly shortened their lifespan if they are still working properly (and they may be working, but at reduced beta, you would probably have to pull them out and test all the parameters to be sure).

 

[Potato Cakes]

I am aware of calculating heat dissipation (watts). I was getting about a 1V drop in the circuit with about a 1.4A load, which I calculated as 1.4W. The BD140 has a rating of 12.5W, albeit with a heat sink. But even without a heatsink this should still handle the 1.4W I had calculated, which apparently is wrong. My question if I simply use a higher rated transistor for the BD140 with the proper thermal management, does that allow for more current or is there something else I am not calculating correctly?

Thanks!

Paul

 

[JohnRoberts]

Without an effective heatsink or forced air, 1 1/2W can get hot.

JR

 

[Potato Cakes]

Dissipated heat in watts is (always) just input-output voltage times the current. A watt heats like a watt heats - the exact amount is easy to calculate from the degC/W rating of your pass element housing with-or-without thermal coupling to a heatsink

You need more current or decide to drop more voltage? More watts dissipated as heat.

Your capacitance multiplication follows the gain (beta) of the transistor, you may want to use a darlington here. or even a mosfet.

Most often I use IRF840's or the like for HV rail denoising

I just need more current.

 

[JohnRoberts]

Higher current devices routinely come in larger packages. The BD140 theta JA (thermal resistance junction to ambient) is 100'/W so 1.5W dissipation will raise the junction temperature 150' C above ambient, not cool.

That 12+W dissipation is with the junction held at 25'C (room temperature). Get thee a heatsink and probably a physically bigger device.

JR

 

[Matador]

BD140 is rated at 1.5A: you need more than 1.5A? TIP41 is rated at 6A, with a similar specification for hfe as the BD140. Obviously the heatsink needs to be sorted first if you want to draw more current at a similar VCE drop.

 

[JohnRoberts]

They don't spec the theta JA for the TIP41 but claim 2W dissipation JA. Since they also spec 150'C max junction temp this works out to roughly 75'/W JA thermal resistance. So larger package dissipate more heat to ambient. Clip on heatsinks are fairly cheap and easy.

JR

 

[Matador]

I've had great luck with the Aavid / Boyd 10C/W aluminum heatsinks for TO-220 style packages:

https://www.digikey.com/en/products/detail/boyd-laconia-llc/531102B02500G/1216359

An insulating mica between the package and the heatsink, with a dollop of thermal paste, you can dissipate a few watts in perpetuity with the HS only getting warm to the touch.

 

[JohnRoberts]

Yup, heat sinks are not rocket science and air is a poor thermal conductor. A mature well inspected subject.

JR

 

[ccaudle]

But even without a heatsink this should still handle the 1.4W I had calculated

You misread the datasheet. It clearly says maximum dissipation of 1.25W at ambient of 25C or less, and that is meant literally, i.e. if the room is 25C but the transistor is inside a closed case, the ambient at the transistor case is no longer 25C, it will be higher because the transistor is warming the air. Even with forced air movement you cannot meet spec in a 25C room while dissipating 1.4W.

 

[DaveDC]

The BD140G does not seem to be a particularly good choice as a pass transistor. I did a QSpice DC simulation with a 15 volt input source and a constant 2 amp load. The BD140G design output 11.8 volts and the transistor dissipation was 6.35 watts. Substituting a 2SA2222SG for the pass transistor with no other changes, the output voltage was 13.9 volts, and the transistor dissipation was 2.1 watts, or about one third of the BD140G. The 2SA2222 has a much higher minimum hfe.

You'll still need to do the heatsink calculations as others have said. Section 4.14 of the National Semiconductor Audio Handbook gives a very good primer on thermal and heatsink design. Its Appendix A1 also has a good section on power supply design