# Theory on when to use each voltage regulating method

Discussion in 'Homework Help' started by masked, Jul 16, 2010.

Jul 1, 2010
48
1
Hi,

This isn't homework; I'm just trying to teach myself electronics by doing step by step project. ( at http://forum.allaboutcircuits.com/showthread.php?t=40633 )

My goal on this step is to use 3 9v batteries to power two circuits, and regulate the voltage of one circuit to 9v. After learning about different V-regulator approaches, it seems the simplest is just to put resistors in series to make a voltage divider.
The drawback appears to be current related because the resistance and current draw of the driven circuit need to be known in order to choose my 2 resistors properly.

However, in my application I will have a 2.64mA current limiting diode right by the battery.
Because I know the 2.64mA spec, would it be appropriate to use a voltage divider for this circuit? or is there a better method to use in this case?

thanks,

2. ### JoeJester AAC Fanatic!

Apr 26, 2005
3,393
1,211
If you have a constant current application, use a constant current circuit. Look at the LM317

3. ### Ghar Active Member

Mar 8, 2010
655
72
A voltage divider only makes sense if a) you're not worried about how variable the voltage is and b) it's a low enough power that you don't care about the inefficiency.

If your original source is regulated and the load is constant then the variability of your voltage divider's output won't be terrible. However, a 9V battery will be constantly dropping in voltage as you use it up so it's your call.

As for efficiency if you need to drop the voltage significantly and you require decent current then you will be wasting far more power than you're delivering. If it's milliwatts you might not care either way.

I actually never heard of a current limiting diode but just looked it up. Thanks for the interesting device!

4. ### Wendy Moderator

Mar 24, 2008
20,772
2,540
Are you trying to regulate voltage or current?

Generally a resistor divider is used as a reference, something a straightforward amplifier uses to create a new voltage. By itself a resistor divider is not a good way to create a power supply voltage as the moment you add anything to load it the load itself is a 3rd resistor, and there goes your target voltage.

By adding an amplifier the voltage is isolated from the load. This is the bases of a voltage regulator.

Jul 1, 2010
48
1
Thanks for the replies. I'll look up the LM317 now and see if I can work that solution with the parts on hand.

I'm trying to make sure I understand the decision I'm making here, so with regard to Ghar's comments:
This circuit is just a 555 timer driving an H-Bridge (R6-754410), so I think that means that it's
a) low power (enough that I need not worry about inefficiency)
b) not worried about a little voltage variation.
...so a voltage divider may not be best, but not disastrous either. Is that right?

On Bill's comments: I'm only trying to regulate voltage for the 555 cicrcuit so that I can power it from the same 27v source used by the driven circuit on the other side of the H-bridge. The load IS effectively a 3rd resistor, but since the load is my 555 circuit that never changes, I thought that I could simply factor in the resistance when I chose my 2 resistors for voltage division, and be okay.
Is that right thinking?

I'd also considered adding a 18v zener diode to the 555 circuit thinking that it should drop the ~27v from the batteries down to ~9v well enough to avoid overload.

Again, I've got
A) a working 9v timer and H-bridge
B) a working 27v, 2.64mA circuit driven by the timer/H-bridge
...I'm just trying to understand a designer's approach to powering the timer from the existing 27v source.

thanks again,

6. ### Wendy Moderator

Mar 24, 2008
20,772
2,540
OK, this is one variation of the techniques...

7. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
515
If you have 3 separate 9v batteries, why don't you just use one of these?

8. ### Wendy Moderator

Mar 24, 2008
20,772
2,540
His first schematic shows that is what he is doing.

Jul 1, 2010
48
1
Thanks again Bill.
A week ago that schematic would have left me confused, but I think I understand it all now. It's more complex thatn I'd have come up with, but it'll be educational and give me my first excuse to use a transistor!

What's the drawback to adding 1 zener diode to drop 18v from the timer?
just the potential for the voltage to vary a bit as the batteries weaken?

As for using one of the existing 9Vs... when I posted my schematic asking about doing it that way I was told that it could mess things up. I still don't understand the mechanics of why (why not) to use that setup though.

Jul 1, 2010
48
1
Now I'm confused about the path from R1 to PNP transistor Q3.

The circuit from R3 to Q4 makes sense to me:
When the output from pin 3 is high, Q2 allows current to flow from R3 to ground, which drops the ZR3 cathode voltage below breakdown. ZR3 thus prevents current flow to the Q4 base, and Q4 prevents current flow to ground.
This makes any +voltage from Q3 collector to flow to "OUT".

When pin 3 is in low state, Q2 prevents flow to the emitter (GND) resulting in ZR3 receving breakdown voltage and supplying Q4 base with higher voltage than Q4 emitter.
This allows any + voltage from Q3 collecter to flow to GND.

What I don't get is if the Q3 emitter is attached to +30v then it seems like the Q3 base will always be at a lower V than the emitter and will always allow current to flow.
What function is performed by the circuit from R1 to Q3?

thanks,

11. ### Wendy Moderator

Mar 24, 2008
20,772
2,540
The potential problems I see is 9V are fairly weak over any amount of time, they don't have much capacity.

The design I drew will do pretty much the same thing. The output transistors have no protections what so ever, you short them they will blow. The exact values for the zeners aren't critical, figure 12V to 22V for a 27V power supply.

Your load is going to be water? To electrolyze it? You may want to put a heavy duty rheostat (the old wire wound kind) between the water and the output, start with a high resistance and move down, to limit the amount of amps.

Ideally, what are the specs you're after?

12. ### Wendy Moderator

Mar 24, 2008
20,772
2,540
When the output is low (pin 3) both Q1 and Q2 are off, since they are NPN transistors. This means Q4 is turned on because of R3 feeding it current, and Q3 gets no current and stays off.

When the output is high both Q1 and Q2 are on. Since Q3 is a PNP (reversed polarity) a ground feed through R4 turns it on, while Q4 is firmly grounded by Q2. In other words, they have flipped states. Whichever transistor is on is likely going to be saturated, which is another way of saying completely on.

The zeners ZR2 and ZR3 try to prevent shoot through, a condition where Q3 and Q4 are not both on at the same time. The larger the zeners values are the better it works.

ZR1 drops its rated voltage (22V), so the 555 only sees 10V for the power supply voltage.

The design is such that Q1/Q2 respond to a firm ground by turning completely off. Since (compared to the 30V power supply) the 555 never turns completely on (it has a max of 8.6V out) the transistors still turn on OK.

If you have read my articles The 555 Projects you will note that a conventional 555 never comes out with the full positive voltage out, but looses around 1.3V due to the internal circuitry. It is a fundamental drawback of the basic design of the 555.

Jul 1, 2010
48
1
Wow, you're fast and thorough!

Ideally my output specs are electrodes powered at ~27V with a max current of 2.64mA. Things got complex when I wanted them to reverse polarity every minute or so. The original plan was to limit current with one CRD, but a rheostat is a good option. ...although I'm starting with distilled water (big resistance), and as electrolysis progresses the water becomes more conductive, so I'd have to manually adjust the rheostat.

I had noticed the voltage drop from my 555 on the meter, but i ignored it since the output was still enough to flip a relay or a bridge IC. I'll go read the 555 series.

I think I have a basic misunderstanding of the PNP. My reading led me to believe that it was only ON when the base (from R4) was lower than the emitter (from +30v).
With that in my head, these two statements:
are still evading my comprehension. To my eyes, the low state does indeed turn Q1 off but, to me, turning Q1 off appears to leave Q3 base at 0v which is lower than the 30v emitter, so I'm imagining current flowing from emitter to collector. (or Q3 being ON).

Then in high, I see that Q1 is grounded, but I'm imagining a +.7V Q3e-Q3b leakage current trying to pass through R4 and ZR2.

what am I missing?

14. ### Wendy Moderator

Mar 24, 2008
20,772
2,540

If Q1 is not conducting Q3 gets no current, so it turns off. If Q1 is conducting Q3 gets current from ground, then turns on.

I tried to make one path at one time, and finally concluded it couldn't be done. This is why Q1/Q2 are both there, they process the signal differently.

I have another scheme for 555 circuits that are within the 555's power supply ratings, but that is off topic. It looks something like this...

Jul 1, 2010
48
1
For any newbies who had the same difficulty as me, I kept rearranging Bill's diagram until the symmetry became obvious to me and finally figured out where my thinking went wrong.

The PNP transistor base requires a NEGATIVE voltage to be turned on.
When Q1 was turned off, I kept seeing the Q3b voltage as zero and thinking that Q3 would turn ON because 0 was less than 30. Doh! In reality, there WAS no closed circuit to ground, so Q3 remains OFF.

Bill - thanks SO MUCH for walking me through this and making a diagram for me!

I also like how Q3 doesn't waste any energy when turned off because of the open circut at Q1. It worried me that Q4 didn't behave as nicely when turned off, because it allowed all the current through R3 to run to ground.

At the risk of taking this further off topic...
Can we make this circuit more efficient by changing Q4 to a PNP so that it's default state is still ON with a new connection to ground, but that it will waste less current when OFF because we ground a smaller current from the new PNP transistor Q5?

16. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,990
1,115
Unfortunately your modification well not work properly.
When the output of a 555 is high both Q1 and Q2 are ON.
Q5 and Q3 is also ON.
So, on emitter of a Q4 we have almost 30V.
Q4 is ON because there is a path for Q4 base current to gnd through:
Vcc ---> emitter-collector Q3---> emitter-base Q4--->ZR3--->R3--->GND.
And so we have a beautiful short-circuit, because Q3 on Q4 ore both ON an the same time.

• ###### 10.PNG
File size:
6.4 KB
Views:
11
Last edited: Jul 18, 2010
17. ### Wendy Moderator

Mar 24, 2008
20,772
2,540
Also, there is a very good reason I used an opposite polarity transistor on the output. The drop of my configuration is around 0.1V for each transistor, in other words you get a true rail to rail output. If I had used the common collector mode you are trying for I drop more voltage (around 0.6V), which means it gets hotter and is no longer rail to rail (ignoring the fact the design is a dead short).

Transistors are relatively cheap, why not go with a design as drawn? Many cases they make what is called a complimentary device, a PNP with basically the same specs as it's NPN counterpart. Transistors 2N2222A and 2N2907A are just such a case, as is 2N3055 and 2N2955. In this case it doesn't matter much, since it is a digital output.

Using the PNP/NPN was not an accident, it is core to the design.

Jul 1, 2010
48
1
Both good points. Thanks!

There was no particular reason to monkey with the design - just tinkering to learn. (in this case , to learn about short circuits and better output.)

My thought had been that Q4 would be OFF and prevent the short circuit.
The idea being that when high output turned on Q5, that the + voltage it allowed through R7 and ZR4 would equal the voltage through ZR3 and R3. The equal voltage would provide a differential of 0v across Q4e-b and would turn Q4 off.

Last edited: Jul 18, 2010
19. ### Wendy Moderator

Mar 24, 2008
20,772
2,540
No problem, if you print the schematic out you can draw the current paths. (Visions of transistors roasting in my head).