According to the IRF datasheet for the 640 there's a junction to heatsink thermal impedance of 1.5 K/W. The heatsink has a sink to ambiet impedance of 6.3 K/W best case with natural convection. So if you're dissipating 6.6W then you'll have a junction temperature of 51°C above ambient.
If the heatsinks are in an inclosure, even one with ventilations holes above and below the heatsink, then the ambient will be some what higher than room temperature. Assuming 40°C is usually a safe enough bet (30°C on a hot day + 10°C heating within the case). The in case heating will be much higher if you don't have adequate ventilation. This gives a junction temperature of 91°C. Which is warm, but no where near the 120°C maximum operating temp (The datasheet states 150°C but it's reasonably widely accepted that 120° is as warm as you want silicon to get if you need long term reliability).
This is assuming you're using thermal paste, but no isolator. If you use a mica isolator you can be pretty sure you'll add around 1 K/W of thermal impedance, pushing your end junction temp up to around 98°C.
Short answer is it'll work, but it'll get warm. I'd consider upgrading to the 2" ones if you have room in your case, though it's not necessary.
Mike.
If the heatsinks are in an inclosure, even one with ventilations holes above and below the heatsink, then the ambient will be some what higher than room temperature. Assuming 40°C is usually a safe enough bet (30°C on a hot day + 10°C heating within the case). The in case heating will be much higher if you don't have adequate ventilation. This gives a junction temperature of 91°C. Which is warm, but no where near the 120°C maximum operating temp (The datasheet states 150°C but it's reasonably widely accepted that 120° is as warm as you want silicon to get if you need long term reliability).
This is assuming you're using thermal paste, but no isolator. If you use a mica isolator you can be pretty sure you'll add around 1 K/W of thermal impedance, pushing your end junction temp up to around 98°C.
Short answer is it'll work, but it'll get warm. I'd consider upgrading to the 2" ones if you have room in your case, though it's not necessary.
Mike.
ok, that makes sense. 
but looking at the somewhat higher thermal resistance of the irf610 (3.5 K/W J-C), it looks like ambient temp is going to be around 108C... which is a bit high for my taste... (what do you think?) i would use the 2" sink, but it only gives 1 K/W less than the smaller one, and i'm trying to keep the case dimensions slim and trim.
there is also the possibility of milling or constructing (from milled parts) adequately sized heatsinks, or even integrating the case top panel (which will be 1/4" 6061 aluminum) into the solution, though it would require mica insulators, and i really would like to be able to test the circuit before marrying it to a chassis.
rod elliott has an interesting article on custom sinks made from milled parts, but using noncontiguous parts seems a bit too much like a hack. any opinions on methods or suppliers? or could i still safely stick with the wakefield sink?
~ brad.
but looking at the somewhat higher thermal resistance of the irf610 (3.5 K/W J-C), it looks like ambient temp is going to be around 108C... which is a bit high for my taste... (what do you think?) i would use the 2" sink, but it only gives 1 K/W less than the smaller one, and i'm trying to keep the case dimensions slim and trim.there is also the possibility of milling or constructing (from milled parts) adequately sized heatsinks, or even integrating the case top panel (which will be 1/4" 6061 aluminum) into the solution, though it would require mica insulators, and i really would like to be able to test the circuit before marrying it to a chassis.
rod elliott has an interesting article on custom sinks made from milled parts, but using noncontiguous parts seems a bit too much like a hack. any opinions on methods or suppliers? or could i still safely stick with the wakefield sink?
~ brad.
ok, i'm going with a different sink. (digikey part no. HS350-ND) interestingly enough, a sink with the same dimensions and material as the original, with a lower natural thermal resistance... wierd. oh well, as long as i sleep better at night.
i'm currently rechecking the BOM and trying to cross as many t's and dot as many i's (and j's) as possible before proceeding to the pcb design, which i should begin soon.
~ brad.
i'm currently rechecking the BOM and trying to cross as many t's and dot as many i's (and j's) as possible before proceeding to the pcb design, which i should begin soon.
~ brad.
gordy, from my experience, the szekeres is a very 'tubey' amp, so i think you'd be pleased if you chose to try it. it's more detailed than a tube amp without the harshness or overaggressiveness of a bjt amp. a good sound to complement sennheisers and to use for relaxed listening. 
i'm on break for this week, but first thing monday i'll have that estimate for you. my guess would be in the ballpark of 0.018% THD, almost completely even harmonics and relatively nothing over the 4th harmonic or so. the latest series of modifications to the design have knocked THD down a bit from the time i posted originally.
i'm also planning on simulating various changes to the shunt regulators, including increasing the CCS current and increasing the CCS BJT base resistor a bit. results and a finalized first version BOM coming monday.
~ brad.
i'm on break for this week, but first thing monday i'll have that estimate for you. my guess would be in the ballpark of 0.018% THD, almost completely even harmonics and relatively nothing over the 4th harmonic or so. the latest series of modifications to the design have knocked THD down a bit from the time i posted originally.
i'm also planning on simulating various changes to the shunt regulators, including increasing the CCS current and increasing the CCS BJT base resistor a bit. results and a finalized first version BOM coming monday.
~ brad.
more changes, results
alright, i've simulated 50mW into 30 ohms, and the results follow...
0.0488% THD:
2nd: -66dB
3rd: -107dB
4th: -116dB
5th: -119dB
past the 3rd harmonic, distortion values are roughly equivalent to (or lower than) power supply line rejection values.
on to the changes...
1. i changed R403,R503 to 0R82 vishay w/w resistors to increase ccs current to 800mA/ch. this means around 200mA/ch of constant dissipation through the regulator shunt mosfets.
2. i changed the heatsinks to HS350-ND parts from digikey and added symbols to the mosfets in the regulator schematic pages.
attached are the new bom and schematic pdf.
~ brad.
alright, i've simulated 50mW into 30 ohms, and the results follow...
0.0488% THD:
2nd: -66dB
3rd: -107dB
4th: -116dB
5th: -119dB
past the 3rd harmonic, distortion values are roughly equivalent to (or lower than) power supply line rejection values.
on to the changes...
1. i changed R403,R503 to 0R82 vishay w/w resistors to increase ccs current to 800mA/ch. this means around 200mA/ch of constant dissipation through the regulator shunt mosfets.
2. i changed the heatsinks to HS350-ND parts from digikey and added symbols to the mosfets in the regulator schematic pages.
attached are the new bom and schematic pdf.
~ brad.
I'd not paid any attention to your power supply previously. I recently designed one almost identical to it for one of my amplifiers. Based on that there are two points I noticed:
1) The 1N3514 isn't sufficiently forward biased to be operating in constant current mode. According to the datasheet I would expect you to require at least 5V to ensure CC mode but you've clamped it to ~0.7V by the base-emitter junction of Q403 (Q503). At this voltage you'd save money by using a simple 140 ohm resistor with no performance hit.
2) The transformer mentioned (TE70064) isn't going to be large enough. It only has a current rating of 1.4A and you're talking of drawing 1.6A (I'm assuming from the schematic, that you only have one unregulated supply). Additionally while it will be possible to obtain 18VDC from an 18V transformer, in reality it's cutting things pretty fine. With 5400uF of cap and a 1.6A load your unregulated voltage will drop down to 20V in the troughs which only leaves two volts for your CCS to play with. This is assuming that your transformer is large enough that there is negligible voltage sag and that your mains is at or above nominal.
Hope this helps,
Mike
1) The 1N3514 isn't sufficiently forward biased to be operating in constant current mode. According to the datasheet I would expect you to require at least 5V to ensure CC mode but you've clamped it to ~0.7V by the base-emitter junction of Q403 (Q503). At this voltage you'd save money by using a simple 140 ohm resistor with no performance hit.
2) The transformer mentioned (TE70064) isn't going to be large enough. It only has a current rating of 1.4A and you're talking of drawing 1.6A (I'm assuming from the schematic, that you only have one unregulated supply). Additionally while it will be possible to obtain 18VDC from an 18V transformer, in reality it's cutting things pretty fine. With 5400uF of cap and a 1.6A load your unregulated voltage will drop down to 20V in the troughs which only leaves two volts for your CCS to play with. This is assuming that your transformer is large enough that there is negligible voltage sag and that your mains is at or above nominal.
Hope this helps,
Mike
brad, that's very intresting and thanks for posting your results.
You previously said that the circuit was very tube-like and I guess that the 2nd harmonic response reinforces that view. It's nice to see all of the other harmonics below 100dB down. I think that is good for 50mW into 30 R. I'm certainly encouraged enough to try a source follower headphone driver like this.
miket6000, I was contemplating a shunt reg and it's genuinely helpful to read of the real-world experiences of others. I think you have just helped brad, myself, and probably about two dozen lurkers! Thank you indeed.
You previously said that the circuit was very tube-like and I guess that the 2nd harmonic response reinforces that view. It's nice to see all of the other harmonics below 100dB down. I think that is good for 50mW into 30 R. I'm certainly encouraged enough to try a source follower headphone driver like this.
miket6000, I was contemplating a shunt reg and it's genuinely helpful to read of the real-world experiences of others. I think you have just helped brad, myself, and probably about two dozen lurkers! Thank you indeed.
thanks, anything to cut BOM cost.
i'll try a 140R resistor and a JFET to see just what i can get away with in that position.but the design uses a TE70084.
do you think doubling the unregulated supply capacitance would sufficiently alleviate this problem?
glad they are encouraging.
~ brad.
it is normal to expect an 18Vac transformer to provide sufficient voltage for an 18Vdc regulated supply.
Let's start with some assumptions.
50VA 18Vac, 15% regulation transformer. Mains voltage tolerance +-6%
At maximum mains voltage the no load output voltage will be
18 * 1.06 * 1.15 * 1.414 - rectifier voltage drop ~=30Vdc at the smoothing capacitor.
At minimum mains voltage the no load output voltage will be
18 * 0.94 * 1.15 * 1.414 - rectifier voltage drop ~=26Vdc at the smoothing capacitor.
The ripple on the cap will be zero since there is zero current demand.
The regulator has 8V to 12V to use and stay in regulation.
Now draw ~0.3Amps from the regulator. The ripple will increase from zero Vpp to ~1Vpp (depends on cap value). In addition the caps will not quite reach the peak voltage from the transformer. The maximum voltage at lowest mains supply voltage will be ~ 26-0.1 at ripple max and ~26-0.1-1.0 at ripple minimum.
We now have 24.9Vdc at worst case condition at the input of the regulator.
That is still 6.9V for the regulator to work with when drawing 300mA.
At the maximum draw that this transformer can supply, about 700mA, the minimum input voltage will be about 23.5Vdc. Any decent regulator should be able to maintain regulation with 5.5V of overhead.
Let's start with some assumptions.
50VA 18Vac, 15% regulation transformer. Mains voltage tolerance +-6%
At maximum mains voltage the no load output voltage will be
18 * 1.06 * 1.15 * 1.414 - rectifier voltage drop ~=30Vdc at the smoothing capacitor.
At minimum mains voltage the no load output voltage will be
18 * 0.94 * 1.15 * 1.414 - rectifier voltage drop ~=26Vdc at the smoothing capacitor.
The ripple on the cap will be zero since there is zero current demand.
The regulator has 8V to 12V to use and stay in regulation.
Now draw ~0.3Amps from the regulator. The ripple will increase from zero Vpp to ~1Vpp (depends on cap value). In addition the caps will not quite reach the peak voltage from the transformer. The maximum voltage at lowest mains supply voltage will be ~ 26-0.1 at ripple max and ~26-0.1-1.0 at ripple minimum.
We now have 24.9Vdc at worst case condition at the input of the regulator.
That is still 6.9V for the regulator to work with when drawing 300mA.
At the maximum draw that this transformer can supply, about 700mA, the minimum input voltage will be about 23.5Vdc. Any decent regulator should be able to maintain regulation with 5.5V of overhead.
Last edited:
andrew, thanks for your advice on the unregulated section. how did you figure 700mA as a maximum current through a 50VA 18VAC trafo? wouldn't it be more like 2.8A? i'm asking because the design here draws a constant 1.6A from the supply...
also, i think the salas shunt classifies as a high-dropout regulator by most standards. typical headroom given is around 6V.
~ brad.
how did you figure 700mA as a maximum current through a 50VA 18VAC trafo? wouldn't it be more like 2.8A? i'm asking because the design here draws a constant 1.6A from the supply...also, i think the salas shunt classifies as a high-dropout regulator by most standards. typical headroom given is around 6V.
~ brad.
mike, what are your opinions on sound quality of the amp using the shunt regs? what was the application?
~ brad.
50VA 18Vac is 2.77Aac.
de-rate the transformer to ~70% for a capacitor input filter.
Now read of the maximum power available at the capacitor input filter.
Power = Voltage * current.
Max current = Max Power / voltage.
Max power ~= 50 * 0.70 ~ 35W
Max current ~= 35/18/1.414 ~1.375Adc
at this continuous output current the transformer will be running at the maximum rating that the manufacturer recommends. Most manufacturers specify a maximum VA based on maximum tolerable/reliable temperatures in the core.
Many builders and designers recommend an operating continuous current of ~ half the maximum to keep the transformer temperature much less than maximum, i.e. our 50VA 18Vac can supply a continuous 690mAdc
This is what I have adopted for a continuous load like a ClassA amplifier.
Put the other way, if you want a continuous current of 1.6Adc then select a transformer that can supply a continuous 6.4Aac, i.e. 115VA for each 18Vac secondary.
de-rate the transformer to ~70% for a capacitor input filter.
Now read of the maximum power available at the capacitor input filter.
Power = Voltage * current.
Max current = Max Power / voltage.
Max power ~= 50 * 0.70 ~ 35W
Max current ~= 35/18/1.414 ~1.375Adc
at this continuous output current the transformer will be running at the maximum rating that the manufacturer recommends. Most manufacturers specify a maximum VA based on maximum tolerable/reliable temperatures in the core.
Many builders and designers recommend an operating continuous current of ~ half the maximum to keep the transformer temperature much less than maximum, i.e. our 50VA 18Vac can supply a continuous 690mAdc
This is what I have adopted for a continuous load like a ClassA amplifier.
Put the other way, if you want a continuous current of 1.6Adc then select a transformer that can supply a continuous 6.4Aac, i.e. 115VA for each 18Vac secondary.
Last edited:
the typical CCS+Shunt regulator is a normal drop out regulator. The Salas is no different in this respect.
The Salas will regulate with a lowish drop out, although not quite into the low drop out range.
The 6V to 10V of overhead is recommended for best audio quality.
I have the DCB1 operating at ~10.3Vdc.
I chose a 15Vac transformer to achieve a 10V overhead. Not because a 12Vac would not allow regulation in worst case conditions.
My first DCB1 build would regulate with 9Vac on the secondary of the transformer or about 2Vdc above supply ripple + regulated output.
Last edited:
considering the cost of trafos, door number one will yield a cheaper BOM cost. i'm thinking about a 100VA trafo. (digikey 237-1341-ND) while it's still not quite at the 'halfway margin' mark, i think it should be more suited.
(i guess i was calling 'regular-dropout' regulators high-dropout regulators... lol)
~ brad.
(i guess i was calling 'regular-dropout' regulators high-dropout regulators... lol)
~ brad.
yet another revision!
ok, i've attached the changes to the power supply and regulators. a 2sk170 jfet is used in place of the current regulating diode, the new filter caps are 2200uF/50V FC types (x5), and the trafo is a 100VA type. crude simulations show ripple to be 1.0Vpp @ ~23Vout from the unregulated section.
~ brad.
ok, i've attached the changes to the power supply and regulators. a 2sk170 jfet is used in place of the current regulating diode, the new filter caps are 2200uF/50V FC types (x5), and the trafo is a 100VA type. crude simulations show ripple to be 1.0Vpp @ ~23Vout from the unregulated section.
~ brad.
Last edited:
On what AndrewT has said, there will always be a voltage ripple on the cap because the power supplys are drawing a constant 1.6A giving a voltage ripple of approximately dv = 1/(120*C) (Assuming 60Hz). This is a slightly pessimistic approximation, but it never hurts to make things a little better.
As for the transformer current rating, I think you've performed the same derating twice. Derating a transformer with a capacitive load is a simplification based on the fact that the output voltage will be sqrt(2) higher than the transformers rated voltage. To look at it another way, if you have a 50VA 18V transformer, and rectify the output with a large bulk cap the output voltage will be 25.4V if you draw 1.97A from this you're drawing 50VA. This will be aproximately 70% of the rating on the transformer because they rate transformers on their RMS voltage so 50VA/18V = 2.78A.
Short answer is it's not necessary to derate the transformer as much as AndrewT has and the one you had previously will be enough.
Mike
As for the transformer current rating, I think you've performed the same derating twice. Derating a transformer with a capacitive load is a simplification based on the fact that the output voltage will be sqrt(2) higher than the transformers rated voltage. To look at it another way, if you have a 50VA 18V transformer, and rectify the output with a large bulk cap the output voltage will be 25.4V if you draw 1.97A from this you're drawing 50VA. This will be aproximately 70% of the rating on the transformer because they rate transformers on their RMS voltage so 50VA/18V = 2.78A.
Short answer is it's not necessary to derate the transformer as much as AndrewT has and the one you had previously will be enough.
Mike
- Status
- This old topic is closed. If you want to reopen this topic, contact a moderator using the "Report Post" button.