V-Reg Board for Breadboard use

[chopnhack]

I salvaged a wall wart as well as the matching barrel jack from an older wireless router and wanted to make a small board to attach to my breadboard for a "benchtop" power supply until I get to that more advanced project ;-0

Should I use the 78L06 instead of the '05 if I want to make sure I have at least 5v? I have read that some PIC's when they are programmed fail flashing because of low power due to variances with USB ports. Here is the basic schematic I am thinking of using. The w.w. puts out 16.4vdc and is rated at 12v/1A max of 16W. I was thinking of using a zener instead of the 1N4000 series diode, but I read on how the changes in resistive loads that occur on the ouput can cause the zener to have to dissipate potentially too much heat. I figured I should leave as much flexibility in design to allow whatever loads I might need up to the transformers max.

Thanks for any input, all ideas welcome. Ideally, this is to be super simple so that I can build it and not over analyze it

 

[KrisBlueNZ]

That looks perfect. The diode on the input is a good idea.

No, you won't need a 78(L)06. But check the power dissipation in the regulator. If you're drawing the maximum 100 mA from the output, the input voltage (after the drop in D1) could be say 15V, so the regulator will be dropping 10V. At 100 mA, it will be dissipating 1W (P = V × I). A TO-92 package will get pretty hot - over 100 °C - dissipating a watt.

You may want to use a 7805 (TO-220 package, >1A rating), or at least a 78M05 (TO-126 package, 0.5A rating) because they're easier to heatsink.

 

[Harald Kapp]

An 78L06 will put out 6V which is above the ratings for most every 5V component. So you should use a 5V regulator.
Note that the 78L05 is rated for 140mA only. If you need more current, you need to upgrade to e.g. an 7805.

With 16.4V input voltage you'lll have a voltage drop of ~11V across the regulator. At 140mA this makes for a loss of 1.5W. You will need a heatsink to cool the regulator.
A more effective way is to use a switching regulator. There are 78xx compatible switching modules availabe, e.g. this one.

 

[cjdelphi]

Use the 7805 because you require 5v ...

But the 5v may not provide 5v maybe 4.8v, the 7806 too 5.8v not always depends on the manufacturer. .

So the trick is to adjust the "gnd" pin, it's actually a ref pin... to get a precise 5v use a potentiometer, a quick way is to use a diode to gnd.

 

[Harald Kapp]

I've never had problems with an 7805. Even at 4.8V (lower limit acc. to datasheet) this is enough for typical 5V logic which operates down to 4.75V.

If you need to adjust the output voltage, a diode from the GND pin to GND is not a good idea as the voltage drop acrosss the diode is not well defined and at 0.6V (typ. silicon diode) is too high: at the upper limit of 5.2V for the 7805 the diode would increase this to 5.8V, too much for reliable operation of 5V components).
In this case use the circuit in figure 3 of the datasheet.

 

[(*steve*)]

So the trick is to adjust the "gnd" pin, it's actually a ref pin... to get a precise 5v use a potentiometer

Actually it's not. The ground pin on a fixed regulator does not necessarily carry a fixed current. Quite often it varies with voltage. This means that a resistor here may cause the output voltage to vary with current.

a quick way is to use a diode to gnd.

That will work better because the diode's Vf changes relatively little with If

 

[KrisBlueNZ]

That's some bad advice there.

Use the 7805 because you require 5v ... But the 5v may not provide 5v maybe 4.8v, the 7806 too 5.8v not always depends on the manufacturer.

If the output voltage is outside the stated tolerance, which is 4.75~5.25V or better, then the 7805 is faulty and should be thrown out. If not, then the PIC will work fine.

So the trick is to adjust the "gnd" pin, it's actually a ref pin... to get a precise 5v use a potentiometer, a quick way is to use a diode to gnd.

Not true for fixed voltage regulators. In any case, using that "trick" can only increase the output voltage, not decrease it. And using a diode adds a poorly defined voltage of around 0.7V which is far too much. Connecting anything in series with the 0V lead could also compromise the decoupling. It's bad juju!

 

[chopnhack]

Yay! Thanks for the replies guys. I misquoted the IC, it is in fact a to-220 LM7805. A good deal on ebay, I believe, @ 10 for 0.99 cents delivered
I don't know what the max current draw will be, but I could always label it as a max of 100mA for now and see if I ever need more. For the future, I do recall somewhere seeing a schematic of a pnp transistor used to "bypass" current around the v-regulator. Can anyone explain how this works without giving away a schematic. I would like to work that out on my own with the knowledge of how it works.
Thanks again all

 

[BobK]

PICs are not at all sensitive to supply voltage. Most 5V PICS will run down to 2.5V, and the special low voltage ones down to 1.8V.

Bob

 

[Harald Kapp]

Can anyone explain how this works without giving away a schematic. I would like to work that out on my own with the knowledge of how it works.

You use a series resistor on the input side of the regulator that develops a voltage drop as the output current of the regulator increases. This voltage drop is used to control a bypass transikstor which will increasingly conduct, thus contributing to output current. If the output voltage rises above the regulator's nominal output, it will turn down the internal output transisstor, thus reducing output current and cinsequently closing the external transistor due to reduced Vbe.

Here is a link to the respective schematic - so I don't give away the schematic unless you're stuck in your analysis and follow the link.

 

[chopnhack]

You use a series resistor on the input side of the regulator that develops a voltage drop as the output current of the regulator increases. This voltage drop is used to control a bypass transikstor which will increasingly conduct, thus contributing to output current. If the output voltage rises above the regulator's nominal output, it will turn down the internal output transisstor, thus reducing output current and cinsequently closing the external transistor due to reduced Vbe.

Here is a link to the respective schematic - so I don't give away the schematic unless you're stuck in your analysis and follow the link.

Thank you Harald

So, the base and collector of a NPN transistor are in parallel to the series resistor of the circuit, the small current produced across the resistor turns on the base allowing current to flow across the collector through the emitter to the output side of the regulator.

Novice question: How is it that the voltage potential at the other side of the regulator is still only the output of the regulator? I should think it would be a product of the input in parallel with the output minus the Vdrop from the diode portion of the transistor. Does that make any sense?

 

[Harald Kapp]

Well, not quite. This won't work, but you're near.

You've asked for a circuit using a PNP transistor - your circuit uses an NPN transistor.
Apart from that, you need to apply a base-emitter voltage, not a collector base voltage to control a bipolar transistor.

 

[chopnhack]

I see the error now! Will the circuit not work with a NPN transistor or would it have to be wired differently?
Also, could you explain how the voltage potential stays regulated when there is access to the input v. pot. through the transistor ?

 

[Harald Kapp]

This circuit looks good. You may want to add a series resistor to limit base currrent.
Using an NPN transistor is possible, but requires a different circuit.

could you explain how the voltage potential stays regulated when there is access to the input v. pot. through the transistor ?

I could, but you can, too:
For the purpose of simplifying the analysis, assume we can neglect the current from In to GND of the regulator IC., such that Iin=Iout (this is physically wrong but acceptable for reasonabel Iout >> Ignd).
As long as Iout*R1 <0.6V (assuming Vbe=0.6V is the voltage where Q1 starts to conduct), where will Iout come from?
If Iout*R1 > 0.6V, how will it be distributed between the IC and Q1? And what happens to I*R1, when more and more current is delivered from Q1 instead of the IC? How will this feed back to the control voltage (Vbe) of Q1? And which are the consequences for the current through Q1?

 

[Gryd3]

I don't know if this is a good practice or not, but I've always used old computer PSUs as power supplies for tinkering with.
They output voltages of course are not exact, but most finished circuits can't rely on exact levels being provided, and it gave me options, lots of options for supply voltage.
+12, +5, +3.3, -5, -12
http://pinouts.ru/Power/atxpower_pinout.shtml

I've accidentally shorted one of my projects and the PSU immediately turned off, just needed a reset and was good to run again.

 

[chopnhack]

I could, but you can, too:
For the purpose of simplifying the analysis, assume we can neglect the current from In to GND of the regulator IC., such that Iin=Iout (this is physically wrong but acceptable for reasonabel Iout >> Ignd).
1. As long as Iout*R1 <0.6V (assuming Vbe=0.6V is the voltage where Q1 starts to conduct), where will Iout come from?
2. If Iout*R1 > 0.6V, how will it be distributed between the IC and Q1? And what happens to I*R1, when more and more current is delivered from Q1 instead of the IC? How will this feed back to the control voltage (Vbe) of Q1? And which are the consequences for the current through Q1?

HaHa, don't be so sure!

1. While Iout*R1 < 0.6V, Q1 remains low (unpowered) so all current must be flowing from In through R1 through v-regulator. Iout comes from Vdrop across R1 and v-reg (whatever internal resistance there is).

2. While Iout*R1 > 0.6V the distribution will depend on the resistances between In and Out of the Regulator as well as resistances between In the transitor and Out.

Assuming an 18Ω resistor between In and IC - and a 10V drop, 0.91A will be available to pass through the IC.
I am unsure of the internal resistance for the transistor, I looked at the data sheet but was overwhelmed by the data! I only found that the max continuous collector current of the 2N3906 was 0.2A.

The best I could find is that the Base only needs about 1mA at 0.65V to make the gate more positive to overcome the negativity of the depletion zone to allow the negatively charged electrons to flow across from the "n" through the "p" to the other "n". With those figures in mind, R2 should be valued to give 1mA. If I assume the potential after R1 is still 16.4VDC then I would need R2 to be ~16.4kΩ.

I can go no further - I don't understand or know the resistance to use for the transistor. Thanks for your help!

 

[Harald Kapp]

Don't look at the transistor as a resistor. Look at it more as a current source where Ic = gain*Ib.
The base current in this setup is approx. ib=(V(R1)-0.6V)/R2 (note that R2=16.4k may be too high resulting in too littel base current).

Look at the circuit from the perspective of the output voltage:
For Iout*R1 < 0.6V your analysis is right (and simple since Q1 is off).
For Iout*R1 > 0.6V any additional output current Delta_I tries to create an additional voltage Delta_V=Delta_I*R1 across R1. This Delta_V in turn will create a base current Delta_Ib=Delta_V/R2 (assuming Vbe is more or less constant at e.g. 0.6V, which it is not, but for the purpose of this analysis it is sufficient).
This additional base current will alllow a collector current Delta_IC=Delta_Ib*gain to flow.
A simple model for the behavior of this circuit is this:

I1 is the current through the regulator IC, I2 is the current contributed by Q1, Iout is the output curren. Therefore:
Iout=I1+I2.
I2 is (approximately and neglecting the base voltage Vbe=0.6V of Q1): I2= Ib*gain=V(R1)/R2 *gain=I1*R1/R2*gain.
Inserting this into Iout=I1+I2 results in Iout= I1*(1+R1/R2*gain) .
You can see how the output current is split between the regulator (term "1" within the parentheses) and the transistor (term R1/R2*gain within the parentheses). In practise you will have to incorporate Vbe into the equations, but that is mere handiwork.

Without any math: If the output voltage sinks below nominal. the regulator will try to increase the voltage by allowing more current through the regulator IC. this will also generate more voltage drop across R1 and consequently drive more current through Q1.
Conversely, if the output voltage rises above nominal, the regulator will allow less current through the regulator IC. This will decrease the voltage drop across R1 and consequently drive less current through Q1.

 

[chopnhack]

The base current in this setup is approx. ib=(V(R1)-0.6V)/R2 (note that R2=16.4k may be too high resulting in too littel base current).

Using the above equation I get a much larger value for R2, did I misunderstand something?
ib=0.001A, V=16.4, R1=18Ω
0.001A=((16.4vdc*18Ω)-0.6vdc)/R2
R2 = 294.6k Ω

Without any math: If the output voltage sinks below nominal. the regulator will try to increase the voltage by allowing more current through the regulator IC. this will also generate more voltage drop across R1 and consequently drive more current through Q1.
Conversely, if the output voltage rises above nominal, the regulator will allow less current through the regulator IC. This will decrease the voltage drop across R1 and consequently drive less current through Q1.

That is a great explanation Harald! Thank you I understand somewhat the balance between V-I-R but sometimes have trouble understanding how it interplays with the actual components.

 

[Harald Kapp]

did I misunderstand something?

Sorry, yes, you did.

Start with this term: 0.001A=((16.4vdc*18Ω)-0.6vdc)/R2
1) how did yyou derive this term?
2) 16.4V*18Ohm is nonsense, not a current.

When you dimension this circuit you start by defining the current where the transistor starts to take over his share of the output current.
Let's assume the regulator IC is specified for 1A (and sufficiently well cooled by a heatsink to dissipate the resulting power). A somewhat arbitrary point where the transistor could start taking over could be 1/2*Imax=0.5A. Assuming a Vbe of 0.6V for the transistor this leads to R1=0.6V/0.5A= 1.2Ω. Tweaking this value allows you to control the takeover current and to adjust for variations in Vbe of the transistor. Since my choice of 0.5A is far away from the limit of the regulator IC, we are safe here even if all tolerances work out to a higher takeover-current.

Let us now assume the circuit shall be designed for e.g. 5A output current and we want the regulator IC to contribute 0.8A at 5A output current (leaving 0.2A margin for tolerances). This means the transistor needs to contribute 4.2A in this situation. Let us further assume the transistor's gain is 50 - not unreasonable for a power transistor. A collector current of 4.2A requires a base current of 4.2A/50=84mA.
We want the currrent through the regulator IC to be 0.8A (see above). This current flowing through R1=1.2Ω develops a voltage drop of 0.96V.
The base current to Q1 is set by R2 and the voltage drop across R2. The voltage drop is Vr2=0.96V-0.6V=0.36V. The base current is 84mA (see above). From this I get R2=0.36V/0.084A=4.3Ω.

You can use e.g. a Darlington transistor to increase gain and use a higher value for R2.

 

[chopnhack]

Sorry, yes, you did.

Start with this term: 0.001A=((16.4vdc*18Ω)-0.6vdc)/R2
1) how did you derive this term?

I was using the equation you posted:
<<Harald Kapp said: ↑
The base current in this setup is approx. ib=(V(R1)-0.6V)/R2 (note that R2=16.4k may be too high resulting in too littel base current).>>

I wasn't quite sure what you were getting at with the equation. I substituted the values for V and R1 and Ib to solve for R2. At least that is what I thought the equation was for until I ended up with that large value!

When you dimension this circuit you start by defining the current where the transistor starts to take over his share of the output current.
Let's assume the regulator IC is specified for 1A (and sufficiently well cooled by a heatsink to dissipate the resulting power). A somewhat arbitrary point where the transistor could start taking over could be 1/2*Imax=0.5A. Assuming a Vbe of 0.6V for the transistor this leads to R1=0.6V/0.5A= 1.2Ω. Tweaking this value allows you to control the takeover current and to adjust for variations in Vbe of the transistor. Since my choice of 0.5A is far away from the limit of the regulator IC, we are safe here even if all tolerances work out to a higher takeover-current.

Let us now assume the circuit shall be designed for e.g. 5A output current and we want the regulator IC to contribute 0.8A at 5A output current (leaving 0.2A margin for tolerances). This means the transistor needs to contribute 4.2A in this situation. Let us further assume the transistor's gain is 50 - not unreasonable for a power transistor. A collector current of 4.2A requires a base current of 4.2A/50=84mA.
We want the currrent through the regulator IC to be 0.8A (see above). This current flowing through R1=1.2Ω develops a voltage drop of 0.96V.
The base current to Q1 is set by R2 and the voltage drop across R2. The voltage drop is Vr2=0.96V-0.6V=0.36V. The base current is 84mA (see above). From this I get R2=0.36V/0.084A=4.3Ω.

You can use e.g. a Darlington transistor to increase gain and use a higher value for R2.

I understand the relationship better. I don't feel confident enough to set up any equations to define them, but its a bit clearer. Thank you!