Sime filter calculation help needed.

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An externally hosted image should be here but it was not working when we last tested it.

Nuuk's class a buffer

Would someone please help me figure out what R3 is doing on the output?
Also how does R3 fit into the highpass filter on the output?
How would R3 be incorporated in calculations if you wanted to change the output cap.?

Also in another category, how would one calculate the gain for a circuit like this?
 
No. I was trying to highlight that C1, R3 & R4 are not a filter.

C1 blocks DC from getting to what is being driven and also protects the circuit against a DC fault on the driven circuit. The corner frequency will depend on the input impedance of what is being driven.

R3 is as I said previously. It's impedance is so low it's negilgible for anything else.

R4 is simply to provide a charge/drain path for C1 when the circuit is not connected to anything. It will be swamped by the input impedance of what is being driven.
 
I think I got the bit about the funtcion of the 68R resistor. Also it carries less than 0.0068V for every volt over the outpu, so I'm willing to discount it...

If you don't mind bearing with me me a little longer until I have enough info to make it usefull, you could mabe check my understanding...?


So lets assume the next stage being powered has an input inpendance of also 100k, would I parallel that with the 100K of R4 to get 50K as part of the highpass filter?

Would that be a less than ideal load? as it demands more current? In other words
 
Many of your questions a can be answered by spending a few minutes with SwitcherCAD. It's free, and it can save you a lot of time, if you invest a little time learning to use it.

I_F
 

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Nordic said:
So lets assume the next stage being powered has an input inpendance of also 100k, would I parallel that with the 100K of R4 to get 50K as part of the highpass filter?

Yes that's right.

Nordic said:
Would that be a less than ideal load? as it demands more current? In other words

No it's fine. Any load you hang on the output is going to draw more current. The trick is to stop it being too heavy a load, which drags down the output noticeably and/or causes the output stage to get too hot.
 
Thanks, I found a basic tutorial which had me analyse a basic resistor divider equivalent as well as an emitter follower (allthough not a constant current one, it was fine to teach the basics) where changes in the output load made a big diffirence on the output impendance of the resitor based version. On the NPN version goin from 10k to 100k (examples I used) changed the impendance by only a few hundred ohms.

It seems Q3 is used in place of the 2 series diodes one would normaly have in its position between the base of Q2 and V-.

Transistors seem mighty interesting, and actualy not that hard to learn as I'm learning basic valve based electronics at the same time... so far the circuits I have a grip on, are very similar in approach accross both mediums...
 
Hi,
the output of your buffer will see the 2uF AND the DC blocker fitted in the receiver. You need to take acount of BOTH DC blockers.
The output also sees the 100k in parallel to the input impedance of the receiver. Again you need to take account of both.

You have two filters in series (cascaded) and the maths gets quite complicated if the two caps and/or the two resistors are similar in value. The sym will help a lot here.

However it is quite normal to ensure the 100k is many times the input impedance of the receiver and then you can virtually ignore it's effect.

Now it becomes an easy hand calculation.
let's set the 100k to 1M0 for this example.
R1=2uf
receiver Zin =50k & DC block=4u7F.

The high pass filter time constant is 50k * caps in series= 50k * 2*4.7/[2+4.7]=70mS. F(-3db) = 1/2 Pi RC = 2.27Hz.
I usually aim for the time constant to be a full decade below the response of the speaker. for 20Hz use 80 to 100mS. Your sym will show the phase error when the filter is set a decade below the required response. Try setting the filter one octave below your required response and see the effect on phase!!! a single pole filter will be about -1db at one octave down. Many adopt this standard as acceptable, but it is VERY audible (bass light).

Since most power amps have Zin between 10k and 100k and the DC blocker between 1uF and 10uF, I find the best compromise is to set the source DC blocker to 10uF polypropylene (you only need to buy two for stereo) and can power as many amps as you have speaker drivers. Each power amp can have a smaller DC blocking cap selected to complement the filter response, input Z and that good quality 10uF. I find that Zin=43k to 51k and 2u2F in the power amp does well for me.
 
Hi Richie,
That posted schematic shows a two transistor constant current sink.
I don't think it operates as a mirror nor do I think it can be equated to a Wilson mirror.

The Vbe of Q3 sets a (nearly fixed) voltage across R1. This voltage will vary with current through Q3 but will be in the range of 600mV to 650mV for a sensible Ic.
The 600mV will force a fixed curent through r1 (120r) of about 5mA and will hold fairly steady at that value for all output currents to the load.
The feedback from Q2 does a good job of controlling that fixed current independantly of the current from R2.
Variations of voltage on the rails will vary the current through r2 and that in turn WILL vary the voltage on Q3. This in turn changes the fixed current of the CCS. That mechanism of variable rail voltage to variable sink current is poor PSRR and can be improved by adding a capacitor across the CE of Q3. This helps remove load induced modulation of the supply rails and if of good quality will also remove some rail noise.

Hopefully, this explanation will help some understand the mechanisms at play.
 
Hi Nordic,
that CCS you refer to is a fixed voltage reference (the two diodes) to keep the emitter resistor at constant current.
Many favour this style for it's alledgedly better sound quality.
This may in part be due to the lack of NFB and that is anathema to those same protagonists. But, they might be right. I don't have the ears nor the patience to repeat this level of experimentation.

BTW.
the diode voltages tend to be slightly higher than the Vbe of the transistor. The voltage across the emitter resistor is more correctly expressed as [2*Vdiode - Vbe], this could be in the range of 700mV to 800mV depending on devices and operating currents.
 
An externally hosted image should be here but it was not working when we last tested it.

THE CONSTANT-CURRENT EMITTER-FOLLOWER

Fig 4 Emitter-follower with constant-current source replacing emitter resistor. (ef2)

The simple emitter-follower can be improved by replacing the sink resistor Re with a constant-current source. The voltage across a current-source does not (to a first approximation) affect the current through it, so if the sink current is large enough a load can be driven to the full voltage swing in both directions.
The current source Q2 is biased by D1,D2. One diode cancels the Vbe drop of Q2, while the other sets up 0.6V across the 100R resistor. This establishes the quiescent current at 6mA. The 22K resistor in turn biases the diodes. It works fine as shown if the supply rails are regulated, but might require filtering if they are not. The 22K value is non-critical; so long as the current through the diodes exceeds the Ib of Q2 by a reasonable factor (say ten times) there will be no problem.
The linearity of this emitter-follower is still degraded by increasing loading, but to a much lesser extent: see Fig 5 below.

It seems we are getting some lines crossed with the next step in the evolutionary chain

An externally hosted image should be here but it was not working when we last tested it.

THE PUSH-PULL EMITTER-FOLLOWER

Fig 6 Circuit of push-pull emitter-follower. Quiescent current still 6mA as before, but the load-driving capability is twice as great. (ef3)

This is an extremely useful and trouble-free form of push-pull output; I have used it many times in preamplifiers, mixers, etc. I derived the notion from the valve-technology White cathode-follower, described in a long-ago Wireless World. I'm afraid I have no idea who Mr White was.

When the output is sourcing current, there is a voltage drop through the upper resistor R3, so its lower end goes downwards in voltage. This is coupled to the current-source Q2 through C, and tends to turn it off. Likewise, when the current through Q1 falls, Q2 is turned on more. This is essentially a negative-feedback loop with an open-loop gain of unity, and so by simple arithmetic the current variations in Q1,Q2 are halved. In other words, this stage can sink twice the current of the constant-constant source version described above, while running at the same quiescent current. The effect of loading on linearity is once again considerably reduced.

Only one resistor and one capacitor have been added to the constant-current version.

This configuration needs fairly clean supply rails to work, as any upper-rail ripple or disturbance is passed directly through C to the current-source, modulating the quiescent current and disrupting the operation of the circuit. A suitable value for C in the circuit as shown is 47uF. Note that the biasing resistor is not shown, to aid clarity.
 
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