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Designing a Charger ? help

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Hi MrAl,

I've run the simulation. Voltage across R2 is approximately 5.25 volts and current is about 52.5 mA.
 
Hi MrAl,

I've run the simulation. Voltage across R2 is approximately 5.25 volts and current is about 52.5 mA.

hi Squire,
Your sim runs OK, I would recommend that you reduce the Base resistor from 10K to say 1K.

Sir K. :D
 
hi Squire,
Your sim runs OK, I would recommend that you reduce the Base resistor from 10K to say 1K.

Sir K. :D

Changed it Knight E.
Zener current now at 3.75mA. :D

SORRY. Many years ago (about 53 years ago) I also had NO knowledge of electronics (but I was very involved with nice young ladys). Then I learned about electronics.

Then...what happened to the nice young ladies after you learned about electronics? :D
 
Then...what happened to the nice young ladies after you learned about electronics? :D
I played with many nice voung ladies then I married the nicest (sexiest) one! Now I have 3 grown up kids and I am a grandfather.
 
Hi again,

3 or 4ma sounds pretty good, maybe 5ma optimum, have to read the zener spec sheet.

I did a few runs myself and found that as the resistor gets larger one of the interesting things is that since the gain of the transistor is (somewhat) fixed, the lower current means less emitter current which means the output cap charges slower which in turn means the startup takes longer. With 1k under 1us, but with 10k up to 5us startup time. Startup time here is defined as the time it takes the output to ramp up to the final output voltage from zero volts.


aguru:
So the young ladies found out you were a nerd (after you got into electronics) and dropped you like a pair of wet gloves in the middle of winter :)
 
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Hi,

Hi again,

3 or 4ma sounds pretty good, maybe 5ma optimum, have to read the zener spec sheet.

I did a few runs myself and found that as the resistor gets larger one of the interesting things is that since the gain of the transistor is (somewhat) fixed, the lower current means less emitter current which means the output cap charges slower which in turn means the startup takes longer. With 1k under 1us, but with 10k up to 5us startup time. Startup time here is defined as the time it takes the output to ramp up to the final output voltage from zero volts.

I'm not getting the same startup time sir. I see about 10us for the startup time with 1k. I am measuring the voltage on the junction of C2, emitter, and R2.
 
Hi again,


You may have to decrease your maximum step value to maybe 1ns to see fast times like this.
 
I may not be doing it correctly. :p

Here's a picture of what I have.
 
Hi,

Yes something is wrong.

If we do a quick sanity check (in my sim), the current in the cap is measured at 0.58 amps and because we have a cap 0.1e-6 that means in 1us it can charge by a voltage equal to ten times the current, which is 5.8 volts. Thus, the output cap can charge by 5.8v in 1us meaning it should get up to the operating point in approximately 1us. That's what my sim is actually doing too, so i assume it is correct.

What you should check:
First reduce your max step time to 1e-8.
Check the model of the transistor see that Bf is equal to 300.
Check the model of the zener see that the capacitance is less than 200pf.
Check the version number of your LT Spice simulator software, see that it is 4.15w or better (Help/About LT Spice).
 
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Have you got the sim command set to 'Start external DC supply voltages at 0V' ? If so, that slows the rise time.
 
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Hi,

What you should check:
First reduce your max step time to 1e-8.
Check the model of the transistor see that Bf is equal to 300.
Check the model of the zener see that the capacitance is less than 200pf.
Check the version number of your LT Spice simulator software, see that it is 4.15w or better (Help/About LT Spice).

I did what is stated above sir.
Bf is 300. Capacitance for the zener is 185pf.
I am using 4.15w.

I am getting the same waveform as before.

alec_t
Have you got the sim command set to 'Start external DC supply voltages at 0V' ? If so, that slows the rise time.

Yes, I checked the box containing 'Start external DC supply voltages at 0V'.
 
I did what is stated above sir.
Bf is 300. Capacitance for the zener is 185pf.
I am using 4.15w.

I am getting the same waveform as before.



Yes, I checked the box containing 'Start external DC supply voltages at 0V'.

Hi again,

I checked this with another simulator and it did 'almost' the same. I say 'almost' because the model they use for the transistor is slightly different, so it gave me a slightly different rise time but only by a factor of about 50 percent. This is interesting in itself, but what you should do next is post another copy of your schematic *exactly* how it is now, and i'll take a look.

The things that can make the rise time slower are:
Cap larger than indicated, transistor gain too low, sim settings not the same.

You can also check your Simulate/Control Panel, and click "Reset to Default Values"
 
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.... or you can just uncheck the sim command box containing 'Start external DC supply voltages at 0V' and you'll get the same pulse rise time as MrAl.
 
it worked for me. you can edit
C:\Program Files\LTC\LTspiceIV\lib\cmp\standard.dio
if you need different or additional diode
 
The .asc file would open but not run until I saved the .asc file, opened it then it ran fine.
 
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Hi,

Here's my simulation.


Hello again,


Ok you have the simulation options set wrong. Instead of checking "Skip initial operating point solution" you have checked "Sart external DC supply voltages at 0v"
So simply uncheck that and check the one that says "Skip initial operating point solution".

This is similar to what alec mentioned.
 
Hi,

I have fixed it already.

Q: What is the option "Start external DC supply voltages at 0v" for?
 
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