Linear Regulator Amplifier Experiments

[Lman]

Tom C,
Thanks for clearing that out. I thought it could be me since english is not my native and I am not perfectly confident in my language skills sometimes.
Randy Fromm is a pleasure to watch the way he explains things. That video even hints at how the linear amplifier works.

Regards
Lman

[Lman]

Hey Lman,

My Question is...what exactly does determine (besides the input source) how much current is put to the load. I suppose that's my next area of discovery. I suppose one factor would be the darlington pair and how hard it is driven...I think I may try a pulse width signal to drive the darlington pair.

One other quick thing, when John refers to the "impedance of the battery" is he talking about batt voltage or actual resistance? I'm pretty sure it's actual resistance but I am wondering how he senses that. I assume it would be some kind of sensitive voltage divider op amp circuit that when the impedance rises because of a load being attached to the output (I assume that would happen b/c the impedance of just about any load would be greater than that of a 12V batt) it would divert the current to run the load and float the battery.

Hi Joster,

Consider my answer as my understanding of the things. I might be wrong somewhere.
What determines the amount of current going to the load is namely the amplifier. Let us think for a minute how an audio amplifier works. It sees the impedance of the speaker/s (the load) and adjusts itself. I think that is simple Ohm's Law. If you have an 8 Ohm speaker how the audio amp knows how much current to deliver ? If you have two 8 Ohm speakers in parallel how does the amplifier sees them ?
After John has posted the video with the audio amplifier, the light bulb in my head turned on.
He mentioned in the posts Class A amplifier, single ended, Emitter Follower.
In Class A the transistor is driven between its on point and saturation carrying the whole 360 degrees of the sine wave. The transistor in Emitter Follower configuration will give us power gain plus it is easier for impedance matching.

You are right that the Darlington pair is a factor but in this circuit it is more like a filtered power supply for the Linear amplifier. And it is another Emitter Follower.

When John refers to the impedance of the battery he means exactly this and not the voltage. And yes the impedance of the battery could be sensed through a voltage divider and op-amp. This is exactly how the linear regulators do it. Check the block diagram of a linear regulator given in the datasheets. And I think this is why John pointed me out in his responses to Servo Control. And it is one way to do it.

Hope that answers your questions.

I am awaiting to see what diagrams you wanted to post and than I will get to something you seem to skip.

Regards
Lman

[Joster]

Linear Reg SS SG Input.jpg

Hey Lman, Thanks for replying...here she is hope it works

[Joster]

Linear Reg SS SG Input Aug 11 2014.jpg
this should be better

[Joster]

here is another one ive been messing withbeta-multiplier-w_feedback.jpg

[Lman]

Joster,

That is not correct again. If the Cap is kept at 30V and you want to pass let say 2 amps to the battery that R4 resistor must be really big.
John B. commented on similar circuit I posted here :
link
You can see this circuit given in the datasheets of the regulators for increasing the current of the regulator but as I have researched and read the NPN darlington is much more stable. NPN is the way.

But let's get back to the Beta Multiplier.
Now I suppose you have learned how zener voltage regulator works. Take a look at the following link and pay attention to figure 3, and figure 7.

http://www.bristolwatch.com/ele/zener_power_supply.htm

Study that circuit and read carefully about the calculations. The Resistor should be calculated to supply just enough current so that the pass-transistor opens and the Zener could work properly. Once the transistor is open it will amplify that current based on the Beta. Imagine you have a Darlington arrangement with total gain of 1000. This means that every 1 mA on the base will become 1A on the output. So, the calculations are important. And passing the predetermined amount of current and voltage the transistors in the darlington should stay in their SOA (Safe Operating Area). So you should look for a transistor which is capable of passing the desired current at 30V DC.

Regards
Lman

[Joster]

thanks a ton for the link...you find the best info i guess I'm just not as clear on what Im looking for...anyway so figure 7 is what I'm after for sure!....

trying to figure out how to calculate Ib. I must have to have a zener across the filter cap to keep it at 30V then from there I can use Ohms law to get my resistance value if I know the zener on the base of the beta multiplier is say 15V and I know I want 1ma for Ib......R = Vfiltercap - V(zener on base)/Ib = 15V/1ma = 15kohms...I'll try and simulate this....but that will only get me 15V-0.6V = 14.5 V on Vout...do I not want about 24V on Vout since I am charging a battery and need a higher voltage at that point??? but then I will need a 30V zener on the base of the darlington pair but then where does that leave the zener diode that is in paralell across the filter cap....head spinning

[Lman]

Joster,

I suggest to start with the calculations from the maximum load current.
You do not need a Zener across the filter cap. You need only the zener on the base of the beta multiplier. Actually there are two filter caps so far. One would collect the spikes and the other is indeed across the zener but on the base of the multiplier.
Watch again the series-pass video, read through the John's post. I admit I was doing it again and again the moment I would grasp something
My head was spinning for quite a while since I was going trough all the links and trying to filter the information to what fits the guidance of JB.
Now I have it a more clear but still have my doubts about some things.
I just share what I have learned so far to save the time of others who would be interested in that. After all that is why we are all here, right ? Learning, sharing, helping !

Regards
Lman

[Joster]

i should be shooting for 30V-0.6V from emitter to ground correct?? Which means I need a 30V zener on the base....if i used a 15v the highest I could ever charge the battery would be 14.4V which isnt bad but I need 15V

[Joster]

i wonder how the spike accumulator stays at 30V without a zener? i cant calculate the resistor value if I don't know what the voltage will be on the collector??