FM transmitter (mod4)

[audioguru]

If you can't get a National Semiconductor LM1117, then maybe the NCP1117 is a copy-cat that is sold over where you are. Here is the datasheet for the LM1117:
**broken link removed**

Here is its typical (some will have a higher or lower voltage) dropout voltage:

 

[bananasiong]

If you can't get a National Semiconductor LM1117, then maybe the NCP1117 is a copy-cat that is sold over where you are. Here is the datasheet for the LM1117:
**broken link removed**

Here is its typical (some will have a higher or lower voltage) dropout voltage:

Yea, what I found is almost similar to your attachment, but I've forgotten which semiconductor company I was searching to.

Is there any method to calculate the voltage drop or ripple voltage when a filter capacitor is added after the rectifier circuit?

Thanks

 

[audioguru]

Is there any method to calculate the voltage drop or ripple voltage when a filter capacitor is added after the rectifier circuit?

I have this graph for my 120Hz ripple frequency. The scale on its left side is the amount of peak-to-peak ripple voltage.

Note that big filter capacitors have a wide tolerance that could be -20%, +80%.

 

[bananasiong]

Oh, after the full wave rectifier and the filter capacitor, the ripple frequency is twice of the original supply frequency, 60Hz. Is the x-axis the output current? As you said before, the ripple voltage becomes higher when the current increases.

Note that big filter capacitors have a wide tolerance that could be -20%, +80%

The tolerance can affect the ripple voltage?

Thanks

 

[Hero999]

The datasheet list a condition of having Vin - Vout of 3V and of 5V for some of its important regulating spec ratings.

Perhaps for worst case the values on the graph should be increased by a third then.

 

[Hero999]

Obviously, suppose a 1000:mu: capacitor has a tollerance of -20%, +80%, there's a chance it might really be a 800:mu: capacitor which would give you more ripple.

Use the following formula to calculate the minimum capacitor value with a fullwave rectifier:

[latex]C = \frac{I{LOAD}}{2FV_{RIPPLE}}[/latex]
Where:
Vripple = maximum allowable ripple, the peak transformer output voltage (1.4 times RMS) minus the diode drops and regulator droput.

F = Line frequency, 50/60Hz.

 

[bananasiong]

Use the following formula to calculate the minimum capacitor value with a fullwave rectifier:

C = \frac{I{LOAD}}{2FV_{RIPPLE}}
Where:
Vripple = maximum allowable ripple, the peak transformer output voltage (1.4 times RMS) minus the diode drops and regulator dropout.

F = Line frequency, 50/60Hz.

For drop out voltage, it depends on how much is the load current right?
Where did you get this equation??

Thanks

 

[Hero999]

1. Yes, look at the datasheets. Don't get me wrong asking questions is good but most of the time you're right in your assumptions. You know more than you think you do, try to be more confident. I'm not having a go by the way.
2. I defined that formula myself, a linear regulator is a constant current load. The capacitor discharges linearly between AC cycles so the minimum capacitor size is proportional to the load current and inversely proportional to the frequency (which is double, hence the *2 term), maximum allowable ripple. I have tested this formula both with simulation software and in practice and it generates consistant results, if you see it in a book or on another websit then please tell me, I'm interested.

 

[bananasiong]

1. Yes, look at the datasheets. Don't get me wrong asking questions is good but most of the time you're right in your assumptions. You know more than you think you do, try to be more confident. I'm not having a go by the way.
2. I defined that formula myself, a linear regulator is a constant current load. The capacitor discharges linearly between AC cycles so the minimum capacitor size is proportional to the load current and inversely proportional to the frequency (which is double, hence the *2 term), maximum allowable ripple. I have tested this formula both with simulation software and in practice and it generates consistant results, if you see it in a book or on another websit then please tell me, I'm interested.

Yea, I know that I'm not confident enough, and sometimes I couldn't understand well as my English is so poor. That's why I'm asking so much
I've tried a 470uF, I can't see the waveform from CRO. It seems the capacitor is discharging very slow.

 

[audioguru]

I can't see the waveform from CRO. It seems the capacitor is discharging very slow.

The capacitor has 12VDC on it so the trace on the CRO is off scale, so you need to turn down the trace position control or reduce the amplitude setting so the trace can be seen. Maybe it needs to be switched to AC input to be seen.

The voltage will be higher without a load and then the capacitor will not discharge.

 

[Hero999]

Yea, I know that I'm not confident enough, and sometimes I couldn't understand well as my English is so poor. That's why I'm asking so much

You seem to be doing very well. I didn't know English wasn't your first language

I've tried a 470uF, I can't see the waveform from CRO. It seems the capacitor is discharging very slow.

What was the load current?

If no load was connected then you won't see any ripple. A 470:mu:F capacitor will give 1V of ripple at 47mA if your power frequency is 50Hz and 833mV if it's 60Hz. Of course you sould be prepared for the capacitor bieng 376:mu:F which would give a 1.25V ripple @50Hz.

 

[bananasiong]

You seem to be doing very well. I didn't know English wasn't your first language

Thanks

The capacitor has 12VDC on it so the trace on the CRO is off scale, so you need to turn down the trace position control or reduce the amplitude setting so the trace can be seen. Maybe it needs to be switched to AC input to be seen.

The voltage will be higher without a load and then the capacitor will not discharge.

What was the load current?

If no load was connected then you won't see any ripple. A 470F capacitor will give 1V of ripple at 47mA if your power frequency is 50Hz and 833mV if it's 60Hz. Of course you sould be prepared for the capacitor bieng 376F which would give a 1.25V ripple @50Hz.

The waveform I've seen is just a flat line (but not nothing as I mentioned previously). I didn't put any load to it, I thought I can see the charging and discharging waveform. A load can be just a resistor right? Then the load current should be V/R.
I'll try it tomorrow, the lab is not opening just for me on Sunday

Thanks

 

[bananasiong]

Hi,
I found this. Get 13.4 volts from 7812. Just because of 2 diodes? How come 2 diodes can create 1.4 volts as reference to the ground pin? I thought supply voltage is needed.
If it is working, how if I put many diodes in series at the ground pin? For example 10 diodes, the output should have 19 volts when 21 volts supply to the 7812.

Thanks

 

[Hero999]

The two diodes, lift the ground pin to 1.4V, because the regulator keeps the output 12V above the ground terminal, the output is 12 + 1.4V = 13.4V above 0V. I personally wouldn't bother with this circuit, I'd use an LM317 with the correct resistors values to give 13.4V.

The waveform I've seen is just a flat line (but not nothing as I mentioned previously). I didn't put any load to it, I thought I can see the charging and discharging waveform.

How can you see a charging and discharging waveform with no load?

There is nothing to discharge the capacitor, so all you'll see is the charging waveform, consisting of a fairly fast voltage increase followed by a straight DC line, you probably only noticed the latter because the first event was too fast for you to see.

A load can be just a resistor right? Then the load current should be V/R.

A resistor will create less ripple than a regulator, because the current drawn will reduce when the voltage falls while a linear regulator draws the same current for a given load regardless of the input voltage.

The best dummy load for your rectifier is an LM317 constant current source, the circuit is on the datasheet.

I'll try it tomorrow, the lab is not opening just for me on Sunday

You have access to a lab?

Do you do your experiments at work, like I do?

 

[audioguru]

Don't ruin the design of an IC regulator by adding diodes in series with the ground pin. Diodes change their voltage drop when their current and their temperature changes.

 

[bananasiong]

A resistor will create less ripple than a regulator, because the current drawn will reduce when the voltage falls while a linear regulator draws the same current for a given load regardless of the input voltage.

The best dummy load for your rectifier is an LM317 constant current source, the circuit is on the datasheet.

I see there is a transistor and a few resistors. Do I need to put any load for the regulator?

You have access to a lab?

Do you do your experiments at work, like I do?

I go to the lab of my school, I'm still a student, and work as a part time too.

Don't ruin the design of an IC regulator by adding diodes in series with the ground pin. Diodes change their voltage drop when their current and their temperature changes.

Oh I see. I won't use it, since LM317 has adjustable output voltage. I just want to make sure that what I saw is correct

 

[Hero999]

I see there is a transistor and a few resistors.

Not there should be just one LM317, and one resistor - even simpler than the voltage regulator.

Do I need to put any load for the regulator?

You don't need to, it's just a method I suggested of testing the rectifier and smoothing capacitor ripple under a constant current load.

 

[bananasiong]

Not there should be just one LM317, and one resistor - even simpler than the voltage regulator.

You don't need to, it's just a method I suggested of testing the rectifier and smoothing capacitor ripple under a constant current load.

One LM317 and a resistor? I see a current regulator there. Somewhere has to be connected to the ground right? Otherwise it acts as opened circuit, like no load.

Thanks

 

[Hero999]

Obviously it needs a ground connection.

Go on, post the circuit taken from the circuit and annotate it showing the +V and 0V connections. I'm sure you can guess.

 

[bananasiong]

I guess this. If the resistor is 12hm:, the load current is around 100mA. Right?