# Designing a Charger ? help

Discussion in 'Homework Help' started by Freeza, Sep 7, 2012.

1. ### meowth08Member

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Hi MrAl,

I think it would be easier for you sir to explain if you know how I understand the circuit so you could point out what am I not understanding well.

1. The voltage across the capacitor, C1, is the same as the input voltage.
2. The voltage across the resistor, R, is the total input voltage minus the zener diode voltage rating.
3. The voltage that can be measured at V+ is the zener voltage minus the base emitter voltage of the transistor.

Questions:
1. Isn't it that the input voltage here is also the power supply voltage?
2.
How can we see a voltage gain if we'd end up with the full power supply voltage? If the input voltage is the same with that of the output voltage, then I am thinking sir that there is no gain.

meowth08

Last edited: Sep 12, 2012
2. ### audioguruWell-Known MemberMost Helpful Member

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The transistor is an emitter-follower that has no voltage gain. A transistor in a different circuit that uses its collector as the output can have voltage gain.

3. ### MrAlWell-Known MemberMost Helpful Member

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Hi again,

The voltage gain YOU are talking about is the ratio of the output voltage to the power supply voltage Vo/Vs..
The voltage gain that I am talking about is the ratio of the output voltage to the zener voltage Vo/Vz.

The voltage gain from the standpoint of the transistor alone is based on the voltage gain looking at it as an amplifier. The common collector configuration has a gain of slightly less than 1, we'll call it 1. If you have 10v power supply and 5v output, that's a gain (according to your intended view) of only 1/2, yet the gain of a voltage follower circuit has a gain of 1 (common collector). So you see when we talk about the voltage gain of the transistor we talk about the input of the transistor, not the input power supply. The input power supply is not the input it is just the power supply. It could be 10v, 15v, 20v, 100v, etc., but we'd still see a voltage gain of 1 because the zener is (close to) 5v and the output is 5v.
The input power supply is not the input, it's just the power supply, just like any other transistor amplifier, and in the voltage regulator the theory of operation is such that the transistor acts as an amplifier. In common collector config, it operates as a current amplifier so there is current gain usually greater than 1, but the voltage gain is 1 or a little less.
In common emitter config the voltage gain is usually more than 1 and so is the current gain, so we have both voltage and current gain. In the common collector config we just have current gain and no voltage gain (we all a voltage gain of 1 "no" voltage gain).

Another example where we see the property of a voltage gain of 1 is with an op amp voltage follower. We use a very small input current and get a larger output current but the output voltage is the same as the input voltage so we only see a voltage gain of 1. The power supply just supplies the power to the circuit that's it.

In a voltage regulator we usually talk about other properties like 'regulation' with input line change, and 'regulation' with output load change, like that. But for the transistor itself in the circuit, we often talk about the gain of the transistor itself as being either a voltage gain or a current gain or both. It's a property of the transistor itself and it's configuration and doesnt matter what the input supply voltage is.

Sometimes we talk about a regulator circuit as having a certain "input voltage", where we mean that we use a 12v input supply and even what the minimum "input" voltage should be (referring to the input power supply voltage) but that's not the same as when we view the transistor itself and the transistor configuration. So when we talk about the voltage 'gain', we are not talking about the input power supply voltage to output voltage we're talking about how the transistor itself is behaving.

So when we talk about the whole power supply we say the 'input' voltage is the power supply input, but then we dont talk about any gain. When we talk about the transistor itself and the way it is used, we talk about the gain of the transistor itself and not the whole power regulator.

So when i say the gain is 1, it means the transistor gain is 1, not the whole circuit gain.

Last edited: Sep 12, 2012

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5. ### meowth08Member

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Yes. It is exactly what I was thinking.

I have read again your previous post and it was so clear that you were talking about the voltage on top of the zener as the input voltage.

It is now clear to me. Thank you for the replies.

meowth08

Is it a good idea to replace the transistor with a darlington pair?

Last edited: Sep 12, 2012
6. ### MrAlWell-Known MemberMost Helpful Member

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Hi again,

You might want to do that if you needed more output current than you can get with only one transistor.
With one transistor we can get a certain current gain (the transistor itself only has a certain max beta) and we only have a certain base current due to the lowest input power supply voltage and the choice of zener bias resistor and zener power rating. To use a small zener we have to use a small base current, so two transistors would effectively increase the output current ability quite a bit.

The downside to this is then we'd have to take into consideration the drop of TWO base emitter diodes instead of just one. This means we would get more variation in the output voltage set point using a darlington instead of just one transistor (if it was available) with the same current gain as the two. So it would probably be a good idea to set up a simulation and then see what happens as the load current is increased. If the voltage starts out at 5.1v say and then decreases to 4.9v at full load it might be ok for the intended application. If it is not then we might have to add an op amp to enhance the voltage regulation ability.

The circuit i posted in this thread with the NPN and PNP transistor would also increase the output current ability of the circuit. We would have to investigate stability a little better first though. The nice thing about doing it that way is that we only have to worry about one diode drop again even though we get an increase in output current capability.

7. ### meowth08Member

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I would take advantage to learn everything I can in this thread

I have LT spice but I don't actually know how to use it. I tried before but I was not so successful. I guess this simple circuit would be a good start if you help me

I would want to set up the circuit now but I don't know what values or components to use. Thanks in advance.

8. ### alec_tWell-Known MemberMost Helpful Member

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Keep practising . It's well worth getting familiar with it. Use the Help (F1 key) to get you started.

9. ### MrAlWell-Known MemberMost Helpful Member

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Hi,

I could help you get started with LT Spice yes. Do you have it up and running so you can draw a schematic?
Like how far did you get so far?

Last edited: Sep 14, 2012
10. ### meowth08Member

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I have drawn the schematic sir. However, I have not yet placed any value for the components.
I already have the .asc file.

11. ### MrAlWell-Known MemberMost Helpful Member

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Hi,

Ok, now you have to right click on each component and set their values. Try that next.

12. ### meowth08Member

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Here's the schematic with values already sir.

13. ### MrAlWell-Known MemberMost Helpful Member

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Hi,

1. Reverse the diode D1, it is backwards.
2. Add some output resistance emitter to ground, 100 ohms.
3. Go to menu: Simulate/EditSimulationCmd, enter:
Stop Time 500m
Time to Start Saving Data 0
Maximum Timestep 1e-5
Check the checkbox "Skip Initial operating point solution".
Click 'OK' and then you'll get a rectangle that contains text,
you drop that rectangle onto the schematic near the bottom.
5. Wait a few seconds, then move the mouse to the top of the
new output resistor, you'll see a little probe appear.
If you see a clamp on ammeter, dont click, but when you
see a little probe that looks like a pencil then right
click. The waveform will appear.
6. You can then try changing some things, then do a 'Run'
again. If you add parts you may have to right click
with the probe again though, or else attach a text label
to the output using that little "A" enclosed in a square
on the tool bar.

14. ### meowth08Member

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Hi,

I'm a bit sleepy. Ph is GMT+8. The zener diode used as a regulator should be operated in reversed bias. The way I put it, it acts as a normal diode.

Thank you so much MrAl for your effort and time. This is a good start for me. I will try changing the values and see the effect on the waveform.

meowth08

15. ### audioguruWell-Known MemberMost Helpful Member

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Why is the zener diode upside down?

16. ### MrAlWell-Known MemberMost Helpful Member

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Hi,

He's fixing that.

17. ### meowth08Member

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Hi,

Sorry I forgot to post the correct shematic in .asc.
I believe it's always a good idea to finish a thread with everything posted correctly.
Other members (newbies like me ) may also want to learn from it.

18. ### audioguruWell-Known MemberMost Helpful Member

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Your zener diode does not have enough current. Look at its datasheet.
Your zener diode current is about 0.4mA but its 6.2V (plus or minus 7%) rating is when its current is 5mA. So your output voltage might be too low.

(Why don't nOObs NEVER look at datasheets??)

Last edited: Sep 16, 2012
19. ### MrAlWell-Known MemberMost Helpful Member

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Hi,

Ok great. Did you try the run command yet?
Yes the resistor should be lowered.

Last edited: Sep 16, 2012
20. ### alec_tWell-Known MemberMost Helpful Member

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Most datasheets assume a fair knowledge of electronics to comprehend them. It's a classic Catch-22 situation

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