I quite agree. You get a much better law by using the active-passive system.
The downside is what I call 'the overlap penalty' whereby headroom is reduced if the active part is amplifying while the passive part is attenuating.
The only complete solution to this is switched resistor ladders, which solves the pot problem but requires something like a 60-way 4-pole switch.
Such things exist but they are NOT cheap.
Happy new year!
May I ask you something else? I simulated circuit Fig. 4.19 from Audio Power Amp Design book with LTSpice.
According to Fig. 4.10 THD would have to be 10 times less. But the simulation is the other way around. There it is about 10 times lower, if R2 and R3 are zero. As the current increases to 9.7mA, the values get even better. What's wrong here?
Fig 4.19
Freq., current, R2/R3, R6, Total Harmonic Distortion
1khz (9.7mA, 0, 68): 0.000043%
1khz (6.1mA, 0, 68): 0.000053%
1khz (9.7mA, 100, 68): 0.000502%
1khz (6.1mA, 100, 68): 0.000507%
20khz (9.7mA, 0, 68): 0.000443%
20khz (6.1mA, 0, 68): 0.000573%
20khz (6.1mA, 100, 68): 0.009438%
20khz (9.7mA, 100, 68): 0.009438%
May I ask you something else? I simulated circuit Fig. 4.19 from Audio Power Amp Design book with LTSpice.
According to Fig. 4.10 THD would have to be 10 times less. But the simulation is the other way around. There it is about 10 times lower, if R2 and R3 are zero. As the current increases to 9.7mA, the values get even better. What's wrong here?
Fig 4.19
Freq., current, R2/R3, R6, Total Harmonic Distortion
1khz (9.7mA, 0, 68): 0.000043%
1khz (6.1mA, 0, 68): 0.000053%
1khz (9.7mA, 100, 68): 0.000502%
1khz (6.1mA, 100, 68): 0.000507%
20khz (9.7mA, 0, 68): 0.000443%
20khz (6.1mA, 0, 68): 0.000573%
20khz (6.1mA, 100, 68): 0.009438%
20khz (9.7mA, 100, 68): 0.009438%
Last edited:
In general, the fact that SPICE distortion results are not dependable because of the approximations in component modelling.
SPICE is never, never, a substitute for measurement.
The short answer is, "Yes". The power supply as specified is 'universal' as it can be adjusted to provide higher voltages to power JFET and legacy 2520 (old pro audio format) circuits.
An 18Vac transformer will give 24Vdc to 28Vdc after rectifying and smoothing.
Allowing for low mains and high load current let's start with 24Vdc as our worst case.
Allow 2Vdrop through the regulator, allow 0.5V for the half ripple (1Vpp ripple)
That leaves a guaranteed 24-(2+0.5) = 21.5Vdc as the highest regulated voltage that is free from drop out during worst case operation.
For 15Vdc operation the extra 21.5-15 = 6.5 volts will be dropped across the regulator in addition to the 2Vdrop already taken account of.
Some regulators perform a bit better when there is excess voltage above their specified drop out. Do we have any evidence that 8.5Vdrop gives better performance than 4Vdrop ?
Now look at normal operation where the available DC voltage is at 26Vdc.
Required output is 15Vdc. Allow a normal regulator Vdrop for the 1.5V drop out voltage and 0.3V for the half ripple. The regulator now has an excess of 25-(15 +1.5+0.3) = 9.2V
The result is an excess of dissipation in the regulator. Say the circuit is drawing 150mA and the reg Vdrop is 11V, then the dissipation 1.65W
That is extra ventilation inside the equipment.
An 18Vac transformer can easily drive a 18Vdc regulator and in most circumstances would drive a 20Vdc regulator.
For 15Vdc, there is no need to go above a 15Vac transformer.
Last edited:
No. Assuming transformer spec' 15Vac rating is for rated current output then 15Vac on the secondary translates to typically 19.8Vdc ((Vac * sqrt(2)) - 1.4V) for full wave rectification. (1.4V is loss on 2 x diodes so approximate).
see Andrew T post above.
I took the time to make that post and you can't even be bothered to read it !
you can check by measuring the the Vdrop across the regulator when you apply 216Vac to the primary.
You can also check the Vripple with a scope.
Your DMM measures the average voltage, so half the ripple is above the DMM voltage reading and half the ripple is below the DMM voltage reading.
Say you have 22Vdc (average by DMM) at the input to the regulator when mains is at 240Vac.
Now reduce to 216Vac and take a new measurement. it will be around 19.8Vdc
The scope will confirm the Vripple at the input. Let's say you measure 0.8Vpp on the scope screen.
Half that is below the DMM reading, so your regulator sees 19.8 - (0.8/2) = 19.4Volts as the minimum voltage
You want an output of 17Vdc so you have 2.4V across the regulator (the DMM will measure 19.4+0.4 -17 = 2.8Vdrop)
Look up your Datasheet and see what drop out they specify for the current demand you are satisfying.
Last edited:
what is the total secondary voltage? 15Vac or 30Vac?
A 15VA 15Vac secondary has an AC current rating of 1Aac.
This must be de-rated when feeding a capacitor input filter.
The maximum continuous DC current from the PSU is just over half the AC rating.
If you want the transformer to run cool then you need to keep the continuous demand to <= 50% of that derated DC current i.e. ~25% of the AC rating.
Thus a 15VA dual 15Vac transformer will run cool if you draw a continuous DC current of ~130mAdc
Thanks, I will do more research, I haven't built the power supply yet, I'm running the preamp off an existing dual 15 volt supply, I'm trying to decide whether to use a 15 or 18 volt transformer, and whether to keep the supply at +/-15V instead of raising it
It's a dual secondary transformer, so 30Vac (or 36Vac) total.
The preamp contains 15 LM4562 op-amps so minimum 150mA quiescent current. I will measure actual current draw
The output current of each opamp ADDs to the quiescent current drawn from each supply rail.
If the loads on each opamp can only draw 1mA, or 2mA worst case, then quiescent plus a bit is a good enough estimate.
If any/some opamps are driving a current hungry load, then you MUST allow for that addition. Feedback and capacitance add to the current loading.
eg.
a 2Vac output with a 1k0 feedback resistor will ADD 2.83mApk to the quiescent current draw, even though the output load of 10k only appears to add 0.283mApk
If the loads on each opamp can only draw 1mA, or 2mA worst case, then quiescent plus a bit is a good enough estimate.
If any/some opamps are driving a current hungry load, then you MUST allow for that addition. Feedback and capacitance add to the current loading.
eg.
a 2Vac output with a 1k0 feedback resistor will ADD 2.83mApk to the quiescent current draw, even though the output load of 10k only appears to add 0.283mApk
Last edited:
That's true, but it is the peak voltage. It doesn't matter. RMS is still 15V.
Of course it matters. The voltage rails for the circuit are (obviously) DC.
15Vac rectifies to give > 19Vdc and that is the rectified DC voltage at the input to a subsequent voltage regulator.
It will have some 2 x Mains frequency ripple on it. Magnitude of the ripple is a function of load current and capacitance.
This is not a subjective opinion
