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Old 21st April 2010, 08:35 PM   #21
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But the load regulation is not as good as with NFB ?
I did a simulation of the discrete part,
with ideal supply and reference instead of LM317.
Simulated load was active sine source with expected Ri and 1 Vpp.

Last edited by Bernhard; 21st April 2010 at 08:40 PM.
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Old 21st April 2010, 09:31 PM   #22
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load regulation is very not bad - 0.1%
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Old 21st April 2010, 10:18 PM   #23
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Quote:
Originally Posted by Nazar_lv View Post
[snip]Its not a shunt type regulator and dont have cap. multiplication, look carefully.
Advantages of this schematic is
-no electrolitycs and it problems
-low and stable(resistive) output impedance
-Perfect transient response (Series regulator(317\337) is completely detached from load)
-High ripple rejection and low noise
-No dependence noise-output current, output impedance-output current
-no global nfb

Disadvantages
-Efficiency
Ohm's law R=U\I ... 17ohms
So, Nazar, I still don't get it. The shunt (it is a shunt, no doubt) pulls a constant current from the regulator, the same current a 17 ohms resistor would pull.
All these advantages you list are the result of a resistor load on the regulator?
The shunt (transistor or resistor) does nothing to the Zout, the Zout is the Zout of the 317, (which is something in the order of 0.02 ohms up till 10kHz when it starts to rise), in parallel with 17 ohms and the B-E junction. It still rises with frequency, therefore is still inductive.
Can you please explain?

When you change the load current from, say, 50mA to 100mA, how much does the final output voltage change? That is the indication for the Zout.
I expect it to be very much larger than 1 ohms, but I may be wrong.

jd
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Last edited by jan.didden; 21st April 2010 at 10:21 PM.
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Old 21st April 2010, 10:22 PM   #24
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Quote:
Originally Posted by Bernhard View Post
But the load regulation is not as good as with NFB ?
I did a simulation of the discrete part,
with ideal supply and reference instead of LM317.
Simulated load was active sine source with expected Ri and 1 Vpp.
Bernhard, what did you see in the sim?

jd
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Old 22nd April 2010, 09:23 AM   #25
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AFAIK the simplified diagram of the circuit is as follow

VT1 is represented as an (ideal: infinite gain, bandwidth, ...) OP Amp

VBias = 1.25 - Vbe
where
1.25 = Vref of LM317

Vbe = Vbe of VT1

So we get that the current flowing in R2 is constant and his value is:

Ir2 = VBias / R2

Vout = Vin - Ir2 * R2 which is constant (regardless the current flowing in the output load)

The output resistance is 0 (zero) due to the fact that current and voltage are constant

We observe that:

- The current supplied by the LM317 is constant regardless the load current
- VT2 try to sink the current not flowing into the load


If we take into account the finite gain of VT1 (non ideal Op Amp) we guess that output resistence of the circuit is

Rout = R2 / AVvt1

where AVvt1 is the (finite and frequency dependent) gain of VT1

We must also take into account the Vbe modulation of VT1 due to the change of the emitter current of VT1

A simulation is in order :-)
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Old 22nd April 2010, 09:32 AM   #26
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Quote:
Originally Posted by diy_audio_fo View Post
AFAIK the simplified diagram of the circuit is as follow

VT1 is represented as an (ideal: infinite gain, bandwidth, ...) OP Amp

VBias = 1.25 - Vbe
where
1.25 = Vref of LM317

Vbe = Vbe of VT1

So we get that the current flowing in R2 is constant and his value is:

Ir2 = VBias / R2

Vout = Vin - Ir2 * R2 which is constant (regardless the current flowing in the output load)

The output resistance is 0 (zero) due to the fact that current and voltage are constant

We observe that:

- The current supplied by the LM317 is constant regardless the load current
- VT2 try to sink the current not flowing into the load


If we take into account the finite gain of VT1 (non ideal Op Amp) we guess that output resistence of the circuit is

Rout = R2 / AVvt1

where AVvt1 is the (finite and frequency dependent) gain of Vt1

We must also take into account the Vbe modulation of Vt1 due to the change of the emitter current of VT1

A simulation is in order :-)
I think your analysis is right on target. This is what shunts do: absorb the variations in load current, thereby presenting a constant voltage to the load looking as if it comes from a very low (ideally zero) Zout. As you noted, the limit on reaching zero Zout is the 'loop gain' of the shunt.

But it can only work within the range of the shunt current which must be low for the type of transistor used. Do we know the actual value of R2? I thought about 17 ohms for a 35mA shunt current but I think Nasar indicated a lower dissipation in the shunt so consequently a lower shunt current and a higher R2 value.

jd
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Old 26th April 2010, 09:32 AM   #27
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diy_audio_fo good, but VT1 and VT2 together forms superbeta transistor.
Quote:
Nasar
Nazar

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Old 26th April 2010, 10:11 AM   #28
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Hi Nazar,

Is the purpose of the no. of 0.1 caps at the output to reduce
impedance & is the last cap place close to the pins of the IC


TQ
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Old 26th April 2010, 10:46 AM   #29
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Quote:
Originally Posted by Nazar_lv View Post
diy_audio_fo good, but VT1 and VT2 together forms superbeta transistor.
Nazar

Practical use DAC Picardian chord v1.2
Click the image to open in full size.
Nazar,

Is there a point to this post or is it only used to advertise your stuff?

jd
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Old 26th April 2010, 01:06 PM   #30
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Quote:
Is the purpose of the no. of 0.1 caps at the output to reduce
impedance
Hi, yes, on HF
Quote:
is the last cap place close to the pins of the IC
all caps as close as possible
Quote:
Is there a point to this post or is it only used to advertise your stuff?
Quote:
Do we know the actual value of R2? I...
If there is no need i delete my previous post
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Last edited by Nazar_lv; 26th April 2010 at 01:09 PM.
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