Question about Pass a40 and speaker loads

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OK, I just can't seem to find this basic information anywhere concerning the Pass Labs a40 amp. The writeup Nelson has about indicates 40-45wpc into an 8 ohm load, and he demonstrates that it will even drive a load as difficult as 1 ohm.

My question is: what happens to the power rating as the speaker load changes? Does 40w into 8 ohms become 80w into 4 ohms? I was planning on using a Plitron torroid rated at 625VA with dual 24v secondaries, each rated at 13A - so the design shouldn't be current limited...

I'm scoping out speaker kits, and found a nice one from BESL but its sensitivity is only 87dB and is rated at 4 ohms. I'm wondering if this amp will be sufficient to drive this speaker.

Thanks for your help!

As no one else has replied to your query, I suppose it's down to me, even though I am not a Pass amp 'expert'.

Assuming I have the correct diagram in front of me, the A40 is a push-pull Class-A amp using BJTs. Unlike single-ended amps which have a fixed maximum current (so the available power reduces into low impedance loads), the push-pull output stage will source or sink higher currents as the load impedance decreases and will therefore deliver more power into 4ohms than into 8ohms.

However, once the load current exceeds twice the quiescent current, the amp will cease to operate in Class-A, as one or other of the output transistor pairs will cut-off and the amp will be working in Class-AB.

The exact increase in power output into lower load impedances depends on several factors, not the least the size of the power supply components. It is possible that you will get near to 80W into 4ohms(Class-AB) but you will still only achieve 20W into 4ohms for full Class-A working.

You're shaking lose a few cobwebs...

Hi Geoff, Thanks for your reply. I've been looking all over for my notes and calculations from the design paper, and just found them at home...

It looks like the a40 will remain in class A operation up to 80wpc into 8 ohms (peak). Above 80w (looks like up to 100w max) it operates in class AB. If max dissipation is ~100wpc into 8 ohms (30v rails), Nelson recommends doubling amperage requirement for a margin of safety (to handle 80w peaks), thus requiring a 350VA transformer that delivers 7.3A on each secondary. I was planning on using a 625VA transformer with 13A on each secondary. Will this allow the amp to safely double its output into 4 ohms? I'm not sure if I've calculated this correctly or not...


I'm not really sure what the exact spec is for bias. I haven't given any thought to modifying this parameter of the design... This is the area I feel least secure with, so I'm not really sure how to answer you. My non EE background begins to emerge around here..

Nelson mentioned in the design for the a40 "No, you cannot easily modify the design for higher power". I'm not necessarily looking for higher power, just curious about its ability to double output into a more demanding load. I'm not trying to turn a 40w amp into a 100w amp...

[Edited by Eric on 04-20-2001 at 01:00 PM]

I've now had a look at the article on the Pass site. The quiescent current for this amp is 1.6A. As it has a push-pull output stage, the maximum current that can be supplied to the load whilst remaining in Class-A is 3.2A. This will provide 40Wrms into 8ohms and 20Wrms into 4ohms.

As I said in a previous posting, additional power is available from this design by the output moving into Class-AB working. With the 30v supply rails, I would expect a maximum output of between 40 and 45Wrms into 8ohms and between 75 and 85Wrms into 4ohms.

If you want more Class-A power (ie over 20Wrms) into 4ohms you will need to increase the quiescent current. The quiescent current for 40Wrms into 4ohm will need to be 2.25A. This will increase the power dissipation in the output devices and so larger heatsinks will be necessary. It might also be necessary to increase the number of parallel output devices used from two to three. This depends upon the ratings of the darlingtons you intend to use and the size of heatsinks.

We're getting closer

Thanks, Geoff! You are helping me to understand these relationships a little better! I was wondering if you could share the calculations you use to figure the changes in output that result from changing the bias current - that would help me figure out some of this on my own...

I already have my heatsinks. The design calls for 0.25w/c of heatsinks per channel. What I have is the equivelant of 0.1675w/c per channel (one 0.67c/w for each of 4 output transistors per channel), so it looks like a little bit of over-engineering pays off!

The output transistors are Motorola MJ11015 and MJ11016 which are rated at 200 degrees centigrade. Their internal bias resistor is 40 ohms instead of 25 ohms from the original output devices that are not available anymore. I'm not sure what their total wattage dissipation is, though; haven't gotten my hands on a data sheet for them yet.

Overall, I think that my heatsinks and the Motorola's have sufficient ratings that I could push up the bias current and still keep everything within safe operating ranges.

Determine your required output power, say 40Wrms, and load resistance, say 4ohm.

The rms load current for 40Wrms into 4ohm is the square root of the rms power divided by the load resistance i.e sqrt(40/4) which is 3.16Arms.

The peak load current is sqrt(2) times the rms current i.e. 1.414 x 3.16 or 4.47Apeak.

For a push-pull output stage operating in Class-A, the quiescent current needs to be half the peak load current i.e 2.235A (or 2.25A by the time I had rounded it off).


Hope this explains it.

See If I'm doing this right

Geoff: Thanks for the formulas! I think I'm coming closer to understanding this: In order for this amp to double its rated power (40wpc into 8 ohms) to get 80wpc into 4 ohms, I not only need a transformer capable of delivering enough current, but I *also* have to change the bias of the amp by hand.

I've been running through some calculations. It looks like to drive 80w into a 4 ohm load will require a quiescent current of 3.16Amps [sqrt(80/4)*sqrt(2)/2]

In order to deliver this current (per channel), I need a transformer that can deliver 3.16A x 30v rail x 2 (for both positive and negative banks of push-pull) = ~190w per channel. Double for margin of safety, torroid needs secondaries that will deliver 384w on a 30v rail or ~ 13Amps per secondary winding. I have a Plitron 625VA torroid with 24V secondaries rated at 13A. Looks like a match (just barely)...

Next, can the output devices handle that amount of dissipation? It looks like each channel will have to dissipate ~ 200 watts with 0.1675c/w of heatsinking. Taking into account thermal resistance, heatsinking is effectively 0.3175c/w per channel. Multiplied by 200 watts we see thermal rise of 63c degrees. Add in 25c ambient, yeilds 88c final junction temperature. The MJ11015 &16s look like they are rated to 200c junction temp.

So, overall, assuming I did this correctly - this looks possible, but a little bit of a stretch...

Maybe I'd be better aiming for 40w into 4 ohms as a compromise between the original design and being able to double power into increased resistance?

Thanks for your help!

Are you sitting comfortably? .... then I'll begin.

You want a Class-A amp that will deliver 40Wrms into 8ohms and 80Wrms into 4ohms. The supply rail voltage is determined by the power requirements for 8ohm loading and is 30V. The quiescent current is determined by the power requirements for 4ohm loading and is, as you have correctly calculated, 3.16A.

Transformer. A good 'rule of thumb' for Class-A amplifiers is to make the transformer VA rating equal to five times the power rating of the amplifier. In this case, you would need 400VA for each channel, or 800VA in total. 625VA is not all that far off and will possibly suffice.

Bear in mind however that the rectifier diodes and transformer secondary winding carry a high current pulses, the magnitude of which depends, amongst other things, on the conduction angle of the diodes, the size of the filter capacitor and the current draw from the power supply. During diode conduction, the diodes and transformer secondary winding must supply the load current (6.32A max.) in addition to the current required to recharge the filter capacitor. The calculations to determine the total current are complex (to say the least!) because many factors have to be taken into account, including the resistance of the transformer primary and secondary windings, the diode 'on' resistance, the load resistance etc. etc., hence the use of a 'rule of thumb' figure.

Transistor temperature. The supply rail voltage is 30V. The quiescent current is 3.16A. The ouput stage has two transistors in parallel so the current in each transistor will be 1.58A and the power in each transistor will be 47.4W. You say you have a 0.67 degC/W heatsink for each transistor. The thermal resistance (junction to case) for the MJ devices is 0.87 degC/W (Manufacturer's figure) and the thermal resistance (case to heatsink)of the mounting washer/insulator will be (at best) about 0.6 degC/W.

The total thermal resistance is therefore 0.67 + 0.87 + 0.6 or 2.14 degC/W. With a transistor dissipation of 47.4W this gives a transistor junction temperature of 101 degC above ambient, or 126 degC at an ambient of 25 degC. The maximum junction temperature for the MJ devices is 200 degC, but in the interests of longevity it is best to limit the junction temperature of a BJT to a maximum of 100 degC.

I suggest that you aim for 40Wrms into both 4 and 8 ohm loads by using a quiescent current of 2.25A. This way, the transformer should be adequately sized and your existing heatsinks will give a maximum transistor junction temperature of 97 degC (at an ambient of 25 degC). This will give you some margin for error, such as the thermal resistance case to heatsink being higher than assumed (which it often is) or the supply rail voltage being higher than design due to mains voltage fluctuations.

Some additional thoughts...

Thanks for all of your help and patience with me! Is this why this type of design referred to as a "constant current" ? It seems that the bias current does not change under varying loads, so you're stuck with making the best of the available current, no matter what the speaker load is.

The A40 is not a 'constant current' amp as it has a push-pull output that will continue to operate outside the Class-A limits. The supply/load current will vary from zero to twice the quiescent current in Class-A mode and even higher when driven into Class-AB. The power availability is limited by the supply rail voltage and by the capabilities of the power supply. Taking the previous figures of 30V rails and an Iq of 2.25A, the limits will be: 40W into 8ohm (Class-A) and 40W into 4ohm (Class-A) or about 80W into 4ohm (Class-AB). Even higher power would be available into a 2ohm load, the limit being set by the power supply and by the output transistor capability/protection circuit.

A 'constant current' amp has a single ended output stage which will provide no more current into low impedance loads than that catered for in the design (i.e. Iq or sometimes up to twice the Iq for special designs). Single ended amps are limited to Class-A operation and cannot move into Class-AB if presented with a low impedance load (unlike most push-pull output stages).

Thanks Again

Thanks for the explanations, Geoff! My mentality started off with "just follow the instructions, and you'll have a nice amp". I'm quickly moving into the "gotta know more!" phase.

I've read Randy Sloan's book, but there are many topics here that he does not address. Most of his designs seem to lean toward Class-B stuff.


[Edited by Eric on 04-24-2001 at 08:13 AM]
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