# Rate of load voltage decrease

Discussion in 'Homework Help' started by The Engineer, Jan 19, 2015.

1. ### The Engineer Thread Starter Member

Sep 3, 2014
35
1
I have been testing various parameters of a op amp controlled linear voltage regulator using a zener diode, and have noticed that as the load resistance decreases in linear steps, the voltage across the load seems to decrease in a non linear fashion.
looking at the graph the voltage seems to decrease by a small amount when the load resistance is lowered from 11 ohms to 9 ohms, however when resistance value is decreased below 9 ohms the voltage decreases much more rapidly and in linear fashion.

this does not follow ohms law and does not make sense to me at this point in time.
It would be great if somebody could explain this to me
Thankyou
-Con

File size:
193.9 KB
Views:
45
File size:
189 KB
Views:
76
2. ### WBahn Moderator

Mar 31, 2012
18,085
4,917
Simple. In general a voltage regulator is a highly non-linear circuit designed to behave linearly only over a particular range of output current. So to expect it to behave linearly as you increase the current beyond that limit is asking too much.

More specifically, consider the conditions under which your opamp saturates. Since you are using a 741 that will be whenever you are asking the voltage to get much closer than 2V of the rails. Now consider that you are trying to maintain a load voltage of 10.2V which means that the opamp output needs to be at about 10.9V which is only barely over 1V from the rail. So you are starting out with the opamp railed so it will hold the voltage at the load to about 0.7V below the saturation voltage.

As you increase the current demand from your transistor it you walk up its characteristic I-V curve and you need more and more base current to support the higher collector current. Since that base current has to come from a device with limited current drive capability, the saturated output voltage drops once the limit is reached and continues to drop as you demand more current from it.

It looks that the drop starts happening at about 7Ω. With about 10V across it that means that the load current is about 1400 mA. If we assume a beta of about 100, that means that the 741 is shelling out about 14 mA of base current. A look at the data sheet shows that the output short circuit current (i.e., when the output has been pulled all the way down) is typically only 25mA but can be as little as 10mA. So it's not surprising that you are driving the opamp out of its linear region.

3. ### The Engineer Thread Starter Member

Sep 3, 2014
35
1
Thankyou for your reply, this is taking me further into my assignment in an hour than I would have done in a day otherwise..

I am starting to understand what you mean although it is still fairly fuzzy as this is new to me !
I can see that rail voltage is supply voltage, however what do you mean by
" So you are starting out with the opamp railed so it will hold the voltage at the load to about 0.7V below the saturation voltage." ?

also would the voltage decrease at 7Ω be due to the fact that according to the datasheet the maximum value for VCE before breakdown is 10mA ?
and this is pushing through 1400mA.

Collector-Emitter
Breakdown Voltage
(IB = 0)
IC = 10 mA
I have tried finding the information regarding :

"A look at the data sheet shows that the output short circuit current (i.e., when the output has been pulled all the way down) is typically only 25mA but can be as little as 10mA"

however I have not been able to find the information on datasheets, would you be able to clarify what you mean by this ?
sorry about being such a dumbass !
If you know of any useful links online to save time for yourself that would be fantastic, I am happy to do as much reading as it takes to gain a full understanding of the subject.

4. ### #12 Expert

Nov 30, 2010
16,668
7,314
You are finding the limits of the op-amp. The 741 can barely get to 10.7 volts of output with a 12 volt supply and you have increased the load until you found out how much current the 741 can supply at that voltage. Change the zener diode to 3 volts and watch a "better" performance result.

5. ### crutschow Expert

Mar 14, 2008
13,479
3,367
How will that help?

It's the output voltage that's the problem, not the zener voltage.

6. ### WBahn Moderator

Mar 31, 2012
18,085
4,917
Reducing the zener voltage will give the opamp room to work with since the opamp output voltage needs to be about (2*Vz+0.7V).

Adding a transistor to make the output transistor a Darlington will also help, as will replacing the 2N2222 with a transistor better suited to the current levels the OP is trying to get out of it.

7. ### The Engineer Thread Starter Member

Sep 3, 2014
35
1
Are you sure that the sudden decrease in voltage at 1.4A is due to the limits of the op amp, or could it be the transistor ? allthough 2n2222a transistors max amp rating is 800mA so i would expect the voltage decrease to be sooner if it was the transistor ?

8. ### #12 Expert

Nov 30, 2010
16,668
7,314
This is the Homework section. It's the lack of understanding that's the problem. The question was about why this circuit fails to maintain its voltage output as the current increases. One of the options is that the op-amp has hit its limit for positive voltage and current output with a 12 volt supply and a 10 volt target. Reducing the target voltage will betray something about the output characteristics of the op-amp model in this simulator. I could just as well have told the OP to add 1 ohm between the op-amp and the transistor and measure the current available. That wouldn't fix the output either, but it would provide information about the behavior of the 741 and the transistor.

Last edited: Jan 20, 2015
9. ### #12 Expert

Nov 30, 2010
16,668
7,314
Am I sure exactly which one is the limiter? Not without measurements of voltage and current.
This being YOUR homework, why do you think an 800 ma rated transistor doesn't quite work perfectly when you ask it to deliver 1400 ma? Does the gain of a transistor diminish when you exceed the current it was designed for? Did it melt when you asked it to dissipate 2.8 watts?

Last edited: Jan 20, 2015
10. ### WBahn Moderator

Mar 31, 2012
18,085
4,917
Probably not in simulation.

Which is one of the reasons that I don't like the move toward almost completely simulation based "labs" -- there is a LOT to be learned from burned fingers.

11. ### WBahn Moderator

Mar 31, 2012
18,085
4,917
Well, what kind of tests could you do to determine which was which?

If you knew what the output characteristic (the current vs. voltage) curve of the opamp was when it is being overdriven (i.e., the inputs are such as to drive the output as high as it can go), would that shed some light on things? If so, then how could you construct a circuit to test that out?

12. ### #12 Expert

Nov 30, 2010
16,668
7,314
Thanks. I was beginning to wonder if one of the rules was, "Don't change any parts and don't measure anything, just stand there and guess."

13. ### The Engineer Thread Starter Member

Sep 3, 2014
35
1
The voltage at the load does not begin to diminish until it reaches 1400mA, although it does not show on the graph the voltage remains constant until it reaches 1400mA. this is why I am having trouble figuring out where this sudden drop has came from. I have tested the current gain of the transistor, and have got the following results.
The current gain will not have any effect on the output anywhere above a gain of 2, this will be because the output is 10.2V. I know that the op amp cannot give an output ah high as the supply voltage, due to it needing its own voltage to operate. however have not been able to find a definition of max output on datasheets (only found min output ). the collector to emitter current flattens off at 1.41A.

Im sorry about my lack of understanding ! and have only got limited access to the lab, but still can not figure why this the load voltage suddenly decreases at 7ohms.

14. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,990
1,115
1 . Do you understand how non-inverting amp work?
2. Do you know how emitter follower work?

As for the op amp output voltage the answer is maximum output voltage swing.
Maximum output voltage swing is the range of voltage that an op amp physically provide at its output.
With a power supply of +15V/-15V, and an ideal amplifier that can swing its output voltage just as far as power supply rail voltages (+15V/-15V).
But most real world op-amps cannot swing their output voltages exactly to +V/-V power supply rail voltages.
For example the LM741 with a split power supply of +15/-15 volts, a 741 op-amp's output may go as high as +14 volts or as low as -13 volts, but no further. And these two voltage limits are known as the positive saturation voltage and negative saturation voltage respectively.
The output voltage range (or output voltage swing/capability) is commonly specified relative to the rail voltages.
For example:

In this case, (V-)+2V to (V+)–1V. So in our example the positive saturation voltage 1V and negative saturation voltage is 2V.
And the maximum output voltage swing is +14V - (-13V) = 27V or 30V - (1 + 2) = 27V

At the beginning let as assume that BJT β/Hfe is constant and equal to 49, also Vbe is constant and equal to 0.7V. Also op amp output short circuit current is Iop_max = 20mA.

Case 1 - RL = 100Ω
The load current is IL = 10V/10Ω = 0.1A = 100mA. The base current is equal to
Ib = IL/(Hfe + 1) = 100mA/50 = 2mA.

So for RL = 100Ω op amp output current is equal to Iop = 2mA and the op amp output voltage is Vop = Vout + Vbe = 10V+0.7V = 10.7V

Case 2
- RL = 50Ω

IL = 10V/50Ω = 0.2A and Iop = 0.2A/50 = 4mA

Case 3 - RL = 10Ω

IL = 10V/10Ω = 1A and Iop = 1A/50 = 20mA
Notice that op amp current reach the limit, so op amp current cannot be higher than 20mA. And this means that for RL<10Ω op amp cannot provide enough current to keep Vout at constant value (10V). Or BJT do not have enough current gain or both.

Case 4 - RL = 5Ω

IL = 10V/5Ω= 2A and Iop = 2A/50 = 40mA

But since Iop > Iop_max the IL current cannot be greater than IL_max = Iop_max * Hfe = 20mA * 50 = 1A

And this means that for RL = 5Ω Vout is no longer equal to 10V. The new Vout values is 1A*5Ω = 5V

I hope that now you see why the load voltage suddenly starts to drop. Also my simplified analysis do not include all factors that affect the suddenly drop in Vout.
Because in real life Hfe is not constant, the higher the Ic current, the smaller the Hfe.
Vbe voltage also increases with the current and op amp saturation voltage also increases with the current.

Last edited: Jan 22, 2015
15. ### WBahn Moderator

Mar 31, 2012
18,085
4,917
As suggested previously, measure the output characteristic of the opamp. It's a simply thing to do and I suspect it will answer your question almost immediately.

16. ### The Engineer Thread Starter Member

Sep 3, 2014
35
1
I realise now that as the load resistance is decreased , more current will flow, however at a certain point the op amps max current output combined with the transistors Hfe with will reach its current gain limitations for flow to the load, therefore when the MAX current has been reached (and the resistance continues to lower) the high pressure (voltage) to push the current through is no longer required as much and also begins to decrease.

So glad that this has finally begun to sink in.
This has been a great help, Thankyou !

17. ### WBahn Moderator

Mar 31, 2012
18,085
4,917
Good. Now I recommend going back and re-reading Post #2 (and some of the others) in light of your dawning awareness and see if they make more sense to you now.

18. ### #12 Expert

Nov 30, 2010
16,668
7,314
Page 4, top left graph.

Decreasing gain above a certain current is the usual behavior for bipolar transistors.

File size:
308.1 KB
Views:
11
19. ### The Engineer Thread Starter Member

Sep 3, 2014
35
1

i can see that the op amp max current output is around 25mA, and the Hfe of the 2n2222a is reading at 58 in the simulation, this calculates to be an output of around 1.45 through to the load, therefore when the load is decreased to 7ohms the voltage at the load becomes (1.45 * 7) = 10.15, and will continue to decrease in this manner. also to clarify whats going on the zener will peg the voltage from going any higher, however when the load resistance is decreased enough to voltage will decrease regardless of the zener.
This stuff may seem simple once its known but until this point i have been very confused . Thankyou for your help and for staying cool when faced up with someone who has alot to learn .
All the best !

20. ### WBahn Moderator

Mar 31, 2012
18,085
4,917
Since you're doing this in simulation, it's not surprising that the op amp limits at 25mA since this is the typical value given in the data sheet.

No problems with helping you work through things -- we've all been there (and are all still there, just on different topics).