for a VAS current of 6mA you can expect a maximum (symmetrical) output of ±6mApk
That would lead to an R98 current range of 0mA to 12mApk.
On that basis you could increase R98 to 33r
When the 12mApk from the VAS does arrive that would apply 396mVpk to the base of the protection transistor. It would be just turning on.
This is on the assumption that the gains of the drivers are closely matched. If the driver gains are not matched then the drivers can inject, or remove, current from the VAS to sink route.
As far as small signal transistor go only a small amount of base current is needed. This can be seen from Figures 1. and 3. see https://www.fairchildsemi.com/datasheets/BC/BC547.pdf Variations in gain between devices of the same type can be covered by a small blanket Ib.
The ratio of collector current to emitter current is always less than one.
Alpha (hFB) = Beta(hFE) divided by Beta +1
For a hFE of 50, 50/51 is 98% and 200 is 200/201 or 99.5% or the current at the collector - the remainder small % being base current. Power transistors would have the lower values of hFE but you can be confident there will be less of a spread in hFE with small signal transistors and you will not need to be driving them to a point of saturation.
As far as the clamp action is concerned I think activation of this is not an issue affecting performance - rather the mere presence of the clamp transistor is in question because the collector sits at the point of entry of the two pole compensation network.
This may be doing very little but you are seeking very low distortion. Consider then the presence of parasitic effects on the signal base currents to the Vas.
The stopper resistor values are unequal 220R in the upper circuit half seems to be enough 150 R in the lower gives a less good result. Try increasing the values there. The Iq trimpot bypass capacitor value is quite low at 47nF.
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It is not clear what suffix your 2SC3423 Vas transistor is however even with the highest hFE version the current gain would not be particularly high in comparison with a small signal type - albeit you need a medium power device for the operating current entailed with your circuit.
This raises the question whether or not the LTP is able to provide enough drive to the Vas transistor. Speed in balancing the negative feedback against the input signal also has to be maintained. Looking at the LTP and the CCS components R4 in the input CCS is 47k. I think this is too high and you should think about reducing this.
This raises the question whether or not the LTP is able to provide enough drive to the Vas transistor. Speed in balancing the negative feedback against the input signal also has to be maintained. Looking at the LTP and the CCS components R4 in the input CCS is 47k. I think this is too high and you should think about reducing this.
actually, i am using 330R and 220R for R12 and 13.
how would you calculate the value for C5 ? (you can assume 100R for V2)
At this stage I think the amplifier should be tested with a square wave with a capacitor in parallel with the test resistor to see what the result is on an oscilloscope. This hands on should give an indication.
There will be some parasitic capacitance around the 2SC3423 amplified by Miller effect - one pole. Another is formed by the low pass action of the stopper resistors in series with the MOSFET gate capacitance.
The capacitor from the collector of 2SC3423 to -ve inverting terminal is to bypass the output stage should some untoward signal be reflected by the load.
The pole of this calculable roughly from the reactance of the 4.7 pF capacitor in parallel with the feedback resistor of 36k. Ohms. If this was to be the dominant pole the value should be near enough to 10 times larger.
There was a circuit by Linsley-Hood with the same general layout to this -including the 4.7 pF capacitor. In this the Vas device was a VN1210M MOSFET and there were other refinements to do with phase correction.
The circuit was sensitive to the board layout. In this respect the dominant pole may have been due to the pole formed by the stopper resistors and the MOSFET gate capacitance.
With a 2SC3423 in place of the VN1210M (low gate capacitance) this introduces another dynamic with the Miller effect. I have already mentioned the question of the LTP current which is in line with using VN1210M.
I think the best direction is to stick with the 2SC3423 and sort out any compensation questions as necessary. This should be able to be done by experimenting with capacitor values.
The Linsley-Hood circuit is available on The Class A amplifier pages hosted by Rod Elliot.
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You are dealing with a 10 kHz signal and the capacitor will have increasing reactance with fall in frequency.
With a 47nF the high pass filter - 3dB frequency is 33.8 kHz.
If you want to reduce this to 10 kHz or below the capacitor has to be increased in value.
There are various RC calculators on line that go into formulae and do the work for you. Here is a link to one I looked up (Sample)RC Low-pass Filter Design Tool - Result -.
The nearest standard capacitor value value to bring the -3dB point in range of the test signal is 220nF (7.2kHz) but 470nF (3.3 kHz) would be OK if there is space on the board - you would possibly need a type in a tombstone package. I have seen 1 uF values in that package.
There is a body of opinion that low ESR electrolytic capacitors are as good as plastic dielectric ones - these allow more generous capacitance in a small package.
The choice is yours to decide.
I tried all combinations of LTP I from 1mA to 6mA and VAS I from 5mA to 30mA, there was no significant change in the clipping performance at 10KHz.
I assume you left R4 at the value of 47k in the LTP CCS.
There should be a voltage drop of 0.6 volts across R3 equivalent to the base emitter drop of Q1. With R3 at 330R this indicates a current of 1.8 mA.
There is a feedback loop around Q1 and Q2. Ideally there should be equality in the current in the sense resistor R4 and in the load (LTP etc).
The balance can be made lop-sided by selecting the wrong value for R4.
I recommend you check the voltage drop across this and see what sort of balance you have from Ohms law. You should also look at the CCS for your Vas similarly.
what mechanism is at work here??
the placement of a cap (2n2) from Q3e to Q4e cleans up the clipping artifacts quite well......why?... is this the way to go?
the placement of a cap (2n2) from Q3e to Q4e cleans up the clipping artifacts quite well......why?... is this the way to go?
This is the signal transfer path between Q3 and Q4. The 220R series resistors increase the linearity at the expense of some loss of gain in the LTP stage. The 2n2 capacitor bypasses redresses the loss at high frequency.
Increasing high frequency improves the phase of the forward signal and the feedback but this is not where the story ends.
Read posts 72 and 73 again and take your queues from the Linsley-Hood circuit from ETI of July 1984 where you appear to have derived your circuit.
OK, i will revise C5 to 470nF, and recheck the effects of adding a cap from Q3E to Q4E.
if the clipping is now ok, why not?
if the clipping is now ok, why not?
It may be OK but it is important to know what the stability is like - if you have not done calculations you are taking pot luck.
You can take that of the question by square wave testing at 10 kHz the step rise being equivalent to much higher frequencies where there could be problem in handling out of phase return signals with a speaker load.
After all the trouble and expense you would not want to spoil the ship for the want of a penny worth of tar -the cost of a capacitor change for C4.
You will also need a test capacitor in parallel with an 8R load - try that with 100nF for starters.
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