Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronic Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

Register Log in

The purpose of the transistor and the opamp in this power supply

Zoldar71

Newbie level 5
Joined
Jan 17, 2021
Messages
9
Helped
0
Reputation
0
Reaction score
0
Trophy points
1
Activity points
52
Hi,

I'm trying to get my head arround the power supply part of a schematic of a guitar effect pedal.

What is the function of the transistor and the opamp in this circuit? I guess it must be something to do with decoupling the circuit from the 9v power input, but I can't figure out why to use the transistor to create the 8v (or 8.36v in fact) and an opamp to create the 4v.

Schermafbeelding 2021-01-17 171652.png

+v is 9v
TL072 is powered from the +8v point (the opamp shown is half of the chip, the other one is used in the signal chain).

Cheers,
Martin
 

KlausST

Super Moderator
Staff member
Joined
Apr 17, 2014
Messages
18,945
Helped
4,215
Reputation
8,433
Reaction score
4,160
Trophy points
113
Activity points
124,552
Hi,

You say "power supply circuit".
I can't see a power supply circuit.
I see a 4V reference generator. It most probably is used as analog signal_gnd reference. (To avoid a symmetric power supply).

The transistor circuit is a gyrator and acts like an inductance.
The Opamp just is a voltage follower.

The whole circuit generates low pass filtered (about) V_Batt/2 with low noise. (Rather complicated, but not bad).

Klaus
 
  • Like
Reactions: d123

    Zoldar71

    points: 2
    Helpful Answer Positive Rating

    d123

    points: 2
    Helpful Answer Positive Rating

Zoldar71

Newbie level 5
Joined
Jan 17, 2021
Messages
9
Helped
0
Reputation
0
Reaction score
0
Trophy points
1
Activity points
52
Hi Klaus, thank you.

I called it power supply because it is not part of the signal chain, and it is used to generate the 8v which powers the rest of the schema. But you are probably right that it is not a power supply. I understand the 4v ref voltage and it is indeed a ground refrence for the rest of the circuit.

I looked up gyrator. Quite interesting. Now I also understand your remark that it is basically al low pass filter.

I know the opamp is a voltage follower, but why use it here. What is the advantage of delivering the 4v reference voltage from an opamp?
 
Last edited:

treez

Advanced Member level 5
Joined
Sep 22, 2008
Messages
7,679
Helped
568
Reputation
1,139
Reaction score
543
Trophy points
1,393
Location
cambridge
Activity points
76,865
The opamp will keep it at 4V...whereas if it was just a potential divider giving 4V, then the voltge would vary with the load.
In other words, the opamp has low output resistance.
 

    Zoldar71

    points: 2
    Helpful Answer Positive Rating

KlausST

Super Moderator
Staff member
Joined
Apr 17, 2014
Messages
18,945
Helped
4,215
Reputation
8,433
Reaction score
4,160
Trophy points
113
Activity points
124,552
Hi,

8V indeed is a supply, but I focussed on the 4V output

I know the opamp is a voltage follower,
What's the job of a voltage follower? Low current, high impedance input --> low impedance output. This means it stabilizes the outpout voltage.

Klaus
 

    Zoldar71

    points: 2
    Helpful Answer Positive Rating

Zoldar71

Newbie level 5
Joined
Jan 17, 2021
Messages
9
Helped
0
Reputation
0
Reaction score
0
Trophy points
1
Activity points
52
Hey Claus and Treez. Thanx a lot. It's clear to me now. Now I continue investigating the rest of the circuit.
 

danadakk

Advanced Member level 3
Joined
Mar 26, 2018
Messages
758
Helped
148
Reputation
295
Reaction score
149
Trophy points
43
Activity points
3,631
The Transistor and its base C is acting as a C multiplier presenting the
9V input withe very large C and causing a slow ramp up of V to OpAmp
follower NI input and its supply.

The 47 uF on base is essentially multiplied by beta of transistor.




1610993513665.png

The TL072 has a crappy output CM range so circuit not exactly a homerun
in its design.


Regards, Dana.
 

Zoldar71

Newbie level 5
Joined
Jan 17, 2021
Messages
9
Helped
0
Reputation
0
Reaction score
0
Trophy points
1
Activity points
52
Oh cool graphs. Don't understand the X1 out en X1 inp. In my spice simulation they are more or less 4 volts. On your graps they are much higher.
 

danadakk

Advanced Member level 3
Joined
Mar 26, 2018
Messages
758
Helped
148
Reputation
295
Reaction score
149
Trophy points
43
Activity points
3,631
Note the R divider keeps the inputs, more or less, inside the CM range
of the OpAmp inputs as the power to the opamp is ramped up.

Regards, Dana.
 

Zoldar71

Newbie level 5
Joined
Jan 17, 2021
Messages
9
Helped
0
Reputation
0
Reaction score
0
Trophy points
1
Activity points
52
Note the R divider keeps the inputs, more or less, inside the CM range
of the OpAmp inputs as the power to the opamp is ramped up.

Regards, Dana.
I understand, but on the graph you provided X1-out climbs to 7v and not 4v as I expected because of the 10k voltage dividers. This is what puzzles me.

Out of curiosity, what tool do you use?
 

betwixt

Super Moderator
Staff member
Joined
Jul 4, 2009
Messages
14,715
Helped
4,810
Reputation
9,637
Reaction score
4,590
Trophy points
1,393
Location
Aberdyfi, West Wales, UK
Activity points
125,344
What confuses this is the pulse input. The circuit is basically a gyrator followed by a voltage divider. Normally it would be powered by a battery or similar steady voltage but here it seems to be driven by a pulse.

As I understand it the transistor stage is the gyrator, it serves to roughly pre-regulate the voltage. If its emitter voltage drops under load, C2 holds the base voltage up hence Vb-e tries to increase and the transistor conducts more. Effectively it makes the capacitor seem to have a much larger value but at the expense of losing about 0.6V in the transistor B-E junction.

The op-amp is a unity-gain buffer stage to isolate the load. The voltage is set by R3/R2 which being equal value makes it half the supply. Without the amp, any load in parallel with R2 would affect the division and the voltage would drop. C4 is, I think, there to filter noise from the potential divider but with a pulsed input it will be slow to charge, hence the slow rise in output voltage.

Brian.
 

    Zoldar71

    points: 2
    Helpful Answer Positive Rating

danadakk

Advanced Member level 3
Joined
Mar 26, 2018
Messages
758
Helped
148
Reputation
295
Reaction score
149
Trophy points
43
Activity points
3,631
I understand, but on the graph you provided X1-out climbs to 7v and not 4v as I expected because of the 10k voltage dividers. This is what puzzles me.

Out of curiosity, what tool do you use?
I also noticed that, but got distracted and did not re-address the issue. My apologies.
The supply ground of the OpAmp was not connected. DUH !

Note SIM is a bit incomplete in that the input diode and C1 100 uF cap are essentially
not in circuit since I am driving that node with a V source. I should put some generator
R in the circuit to get a more complete sim. I tried it with 10 ohms, that did not alter the
results in any significant way.

Here is updated sim (note the Vcesat of Q1 curve that affects max V sent to OpAmp) -

1611055882386.png

SIM is SIMetrix. Most folks are using LTC Spice these days. SIMetrix was default SIM
used at Analog Devices I gather, and when they bought out LTC they settled on
their spice. I find SIMetrix much more user friendly than LTC in general.


Regards, Dana.
 
Last edited:

    Zoldar71

    points: 2
    Helpful Answer Positive Rating

Zoldar71

Newbie level 5
Joined
Jan 17, 2021
Messages
9
Helped
0
Reputation
0
Reaction score
0
Trophy points
1
Activity points
52
What confuses this is the pulse input. The circuit is basically a gyrator followed by a voltage divider. Normally it would be powered by a battery or similar steady voltage but here it seems to be driven by a pulse.

As I understand it the transistor stage is the gyrator, it serves to roughly pre-regulate the voltage. If its emitter voltage drops under load, C2 holds the base voltage up hence Vb-e tries to increase and the transistor conducts more. Effectively it makes the capacitor seem to have a much larger value but at the expense of losing about 0.6V in the transistor B-E junction.

The op-amp is a unity-gain buffer stage to isolate the load. The voltage is set by R3/R2 which being equal value makes it half the supply. Without the amp, any load in parallel with R2 would affect the division and the voltage would drop. C4 is, I think, there to filter noise from the potential divider but with a pulsed input it will be slow to charge, hence the slow rise in output voltage.

Brian.
Thanx Brian. I'm very new to the gyrator. I'm reading up on this. Very interesting stuff. The setup in this diagram is different from the other schemas I see on the net. Most of them are build around an op amp, and the ones with transistors have a different setup. Like this one for instance.
Schermafbeelding 2021-01-19 165812.png
But I guess it boils down to the same.

Since it is the power stage of an guitar effects pedal it is powered by a relatively stable 9 volt source. But a lot of the cheap 9v adapters may have a bit of noise on the output voltage. I don't know why danadakk put a pulse in the schema.
--- Updated ---

I also noticed that, but got distracted and did not re-address the issue. My apologies.
The supply ground of the OpAmp was not connected. DUH !

Note SIM is a bit incomplete in that the input diode and C1 100 uF cap are essentially
not in circuit since I am driving that node with a V source. I should put some generator
R in the circuit to get a more complete sim. I tried it with 10 ohms, that did not alter the
results in any significant way.

Here is updated sim (note the Vcesat of Q1 curve that affects max V sent to OpAmp) -

View attachment 167044

SIM is SIMetrix. Most folks are using LTC Spice these days. SIMetrix was default SIM
used at Analog Devices I gather, and when they bought out LTC they settled on
their spice. I find SIMetrix much more user friendly than LTC in general.


Regards, Dana.
Thanx Dana
 

danadakk

Advanced Member level 3
Joined
Mar 26, 2018
Messages
758
Helped
148
Reputation
295
Reaction score
149
Trophy points
43
Activity points
3,631
One of the areas gyrators drew lots of attention to, back in the day,
was filters. It allowed capacitors to be transformed into inductors,
thereby eliminating inductors in filters of yore. Was used, for awhile,
in ladder filters where floating L's were needed, eg. L's not terminated
to a supply rail.

A lot of that interest faded with DSP.

I would be curious if anyone on forum still using them and reasons why....


.



Regards, Dana.
 

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Top