howto: replacing the ALWSR pre-regulator by a VBE
- Thread starter teddy_pardo
- Start date
ron
Tweaker
Sure, just solder the smaller npn to the bigger pnp which is the easy bit, the harder more fiddly part I did was to cut the larger legs of the PNP and solder ptfe covered wire to the stumps so you can reverse the pinouts of the emitter and collector to the holes on the super reg. a 470R resistor is also soldered across the base and collector of the PNP.
I'll try and post a diagram to make it easier
mjrod99@comcast
pfm Member
teddy_pardo
Trade: Teddy Pardo
I'm not sure about the advantage of R2 in this circuit. It is important in switching circuits where switch off speed is important. But OTOH it reduces the hfe of the combined transistor at low currents.
martin clark
pinko bodger
R2 damps any tendency for the pair to oscillate at VHF - due to the gain in the 'loop' and the strays associated with conencting the two transistors together.
The effect on Hfe is negligible; since the dropout voltage is so low, you can afford the tiny extra current through R5
The effect on Hfe is negligible; since the dropout voltage is so low, you can afford the tiny extra current through R5
teddy_pardo
Trade: Teddy Pardo
R2 damps any tendency for the pair to oscillate at VHF - due to the gain in the 'loop' and the strays associated with conencting the two transistors together.
The effect on Hfe is negligible; since the dropout voltage is so low, you can afford the tiny extra current through R5
Is R2 connected at the right place? I was refering to the resistor mentioned in wikipedia:
"As with a Darlington pair, a resistor between 100Ω and 1kΩ is usually connected between the emitter of Q2 and the collector of Q1 to improve the turn-off time of Q2 - this helps to improve the performance of the pair at high frequencies."
Is it a mistake in the diagram, or does it have another function?
For the wikipedia link click here
martin clark
pinko bodger
No the diagram really is correct, even if it looks wrong
The wikipedia description is badly written... but correct about location: the resistor ends up between the collector and emitter of the driver.
The wikipedia description is badly written... but correct about location: the resistor ends up between the collector and emitter of the driver.
ron
Tweaker
When I first drew the schematic and showed it Martin before building it I did place the resistor in the wrong place.
Anyway its bit of a fiddle with changing the pinouts of the pass transistor to fit the pcb.
Martin suggested the Sziklai pair earlier on and I'm surprised nobody else tried it, I quite like this one and in my case makes the sound really clear, it may not suit everyone
Anyway its bit of a fiddle with changing the pinouts of the pass transistor to fit the pcb.
Martin suggested the Sziklai pair earlier on and I'm surprised nobody else tried it, I quite like this one and in my case makes the sound really clear, it may not suit everyone
teddy_pardo
Trade: Teddy Pardo
No the diagram really is correct, even if it looks wrong
The wikipedia description is badly written... but correct about location: the resistor ends up between the collector and emitter of the driver.
Martin - the wikipedia explenation and the diagram above don't look the same. In the diagram above R2 is between the base and collector of Q2, while in wikipedia it's between the base and emitter of Q2 (it's not connected to the driver at all). Here is another example.
ron
Tweaker
Martin - the wikipedia explenation and the diagram above don't look the same. In the diagram above R2 is between the base and collector of Q2, while in wikipedia it's between the base and emitter of Q2 (it's not connected to the driver at all). Here is another example.
Hi Teddy,
This is the first one I drew out, all I can is changing that resistor to how Martin advised works a treat here
martin clark
pinko bodger
Teddy, don't worry about it - both connections work, it's just not quite as straigthtforward as a darlington
As I said- it's worth a go if you need an absolute minimum of drop-out voltage. I do in my CD2, where a BC547/MJE15031 pair with R2=470ohms where ron has it in post #63 drops just 0.8v at 100mA.
As I said- it's worth a go if you need an absolute minimum of drop-out voltage. I do in my CD2, where a BC547/MJE15031 pair with R2=470ohms where ron has it in post #63 drops just 0.8v at 100mA.
teddy_pardo
Trade: Teddy Pardo
Anything between 50K-100k
Hi Jarthel,
The VBE replaces the PRE-regulator, hence it precedes the SR.
Connection guidelines are mentioned on post #2. See also the picks.
Final V is set by the ALWSR.
You must have "enough" V before the VBE (there may be a formula for predicting V drop here but I don't know it ) and good heatsink.
Regards,
M
The VBE replaces the PRE-regulator, hence it precedes the SR.
Connection guidelines are mentioned on post #2. See also the picks.
Final V is set by the ALWSR.
You must have "enough" V before the VBE (there may be a formula for predicting V drop here but I don't know it
) and good heatsink.Regards,
M
kubota1000
Active Member
Hi Jarthel,
The VBE replaces the PRE-regulator, hence it precedes the SR.
Connection guidelines are mentioned on post #2. See also the picks.
Final V is set by the ALWSR.
You must have "enough" V before the VBE (there may be a formula for predicting V drop here but I don't know it
Regards,
M
with a darlington, is the circuit still the same? can I place the VBE circuit together with the filter caps and then a wire from this PCB to the ALWSR PCB?
I am somewhat confused how can a VBE replace the LM317T pre-regulator. the LM317 is configured to output 2.5V and in this photo: http://i27.photobucket.com/albums/c156/princesnuggles/cdp-fet.jpg, the output of the VBE is way above 2.5V.
any explanation on how would this work?
thank you.