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Help me to understand the circuit

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renzen

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See the attached file

schem1.jpg

I have only basic electronic. Help me to understand the circuit.
1. The circuit is a constant current or a basic led circuit(DC converter + resistor + led)? If the circuit is a constant current, why? It looks like DC converter + resistor + led.
2. The 1st circuit doesn't have output capacitor and it work(tested by the circuit creator). Which one is better, using and not using output capacitor?
3. The 2nd circuit: 'current setting with R1: by the bias of the FB connection with D1 and D2 is the typical voltage drop of 0.2 R1'(using Google translate).
I=(1.23-0.6-0.6)/R1 => I=0.03/R1 is this correct?
4. The 1st circuit using 1K resistor and 2nd circuit using 47K(R2). How to calculate the resistor?

Thank you :)
 
The FB pin on the lm2575-adj is made to work at 1.23 volts. If this voltage is below 1.23 the lm2575 tries to output more power. If the voltage is above 1.23 the lm2575 stops sending power. Normally there is a voltage divider here. (two resistors) For 5 volts output you make a divider that takes 5V and output 1.23V.

This circuit is a constant current LED driver. The LED current is measured in a 0.7 ohm resistor. If the FB pin was connected directly to the 0.7 ohm resistor then the regulator will try to make 1.23 volts across the resistor. A diode and pull up resistor was added so.....The FB pin is at 1.23, the bottom of the diode is at 1.23=0.7= 0.53 volts (approx). The regulator is trying to make about 1/2 volt across 0.7 ohms. This current passes through the LEDs.

Why no cap? I would use a small cap. A capacitor filters out noise. It will help average out the voltage. BUT The inductor is also a filter. It averages out the voltage from the IC to make a DC current. The thought is that the inductor is enough of a filter.

Hope this helps.
 
The hand-written circuit has two flaws:

First, removing the output cap. The LEDs will see a lot of high frequency ripple current. This causes unnecessary heating on the LEDs.

Second, it removed one of the two biasing diodes. Because now it can't reach the proper ffedback voltage, he had to significantly reduce R2 from 47k to 1k. The resulting voltage divider (with respect to the input) will have the effect that the output current will now have dependency with the input voltage.

Why ruin a circuit just to save a pair of inexpensive components? This is a very poor design practice.
 
putting diodes in the feed back path is not good. the diode voltave changes with temperature. There are ICs with the FB voltage at a fraction of a volt. (made to do this job)
 
I don't have a spice model for the lm2575. I have used a LT1616 which has the same FB pin.
Here is a way to get the PWM to respond at 0.25 volts. When you are making a current controlled PWM you don't want 1.25 volts across the current scene resistor.
Here I use two 2N2222 transistors to amplify the voltage FB voltage from 0.25 to 1.25 volts. This circuit is more temperature stable. Transistor 1 and transistor 2 temperature compensate each other.
upload_2014-7-11_19-56-10.png
 

Attachments

  • 1616 LED driver.asc
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If C2 goes to ground then C4 = 0.01uf. The transistor amp is very fast. (too fast)
If C2 goes to the current sense resistor then the transistor amp can have less slow down capacitor.

I built the PWM (in spice) and know there would be a problem. It wants to oscillate at 1/3 the switching frequency. I can not get to the error amp. I slowed down my transistor amp (added phase shaft) for the frequency the thing wanted to scintillate at. It is good to have much gain at DC. At some high frequency point adding more gain will cause problems. At that point (frequency) I want the gain of the transistor amp to drop to one.
 
Thank you all for the explanation .

In the 2nd schematic, the circuit use diode(D4) near the input. What is the use of the diode(D4)?
Since the diode(in the feed back) voltage changes with temperature, what if I use transistor as a diode?

Thank you Ron for the schematic, it's a new 'homework' for me :D
 
In the 2nd schematic, the circuit use diode(D4) near the input. What is the use of the diode(D4)?

D4 it there for reverse polarity protection.
Sometimes referred to as an "idiot diode".
If some fool connects the supply the wrong way round, the diode will conduct and blow the fuse in the supply, so protecting the rest of the circuit.
Always assuming that there is a fuse in the supply! I there is no fuse, the rectifier diode will turn into a Light and Smoke Emiting Diode. :eek::D

JimB
 
Since the diode(in the feed back) voltage changes with temperature, what if I use transistor as a diode?
The diode in a transistor is also temper not-stable.
I used two transistors so one has a negative and the other has a positive temper effect. Thus they cancel out. Do you need more on how that works?
 
Q1 is a common base amplifier. (non-inverting)
Often the Base is on ground. In this case the base is about 1 volt positive.
Usually there is a emitter resistor. If the collector resistor is 1ok (R1) and the emitter resistor is 1k the amp will have a gain of 10. If you put 10mV across the Emitter resistor then you will get 100mV across the Collector resistor.

If the emitter voltage is high then there will be little emitter current (maybe no current) so there will be little to no collector cur4rent so the voltage is high.
If the emitter voltage is low, in out case 0V then there is a large current in the B-E so large current in C-E and the C is at a low voltage.

Look at R4 & R2. The 3.3V point is not exactly 3.3 but reasonably stable. Most of the 3.3V is across R4, About 0.65 volts are in B-E of the Q2, and the rest of the voltage is on R2. About 0.25 volts on R2. So the Q2 emitter is at 0.25V. The base is one diode drop above that. Q1 Base is at the same voltage and its emitter is one diode drop below. So the two emitters are at the same voltage. If the temperature changes it should change in both diodes by the same amount.

You could use a diode in place of Q2. It will be close. Not the best. You could use just the B-E of Q2. That will be better. I used Q2 in what should be about the same way as Q1. If I can get aQ1 and Q2 to have almost the same amount of base current and the same collector current then the two Vbe will batch and the two emitter voltages will be the same. There will be no temperature problem based on Vbe.

How does Q2 work? Current flows through R4 into B of Q2. Q2 turns on and the collector turns on. The collector can pull below the base voltage. What happens is that the C sends most of the current to ground leaving a small amount of current for the base. Current is 2.3V/10k. Most of that current goes to the C-E and to ground via R2. The base current is just enough to get the C turned on for that current.

Q1 current is 2.3V/10k. Q2 current is 2.0/10k. About the same thing.
If the transistors have a current gain of 100, (at that current):
Q1 base current is 1/100 of its collector current. Q2 base current is 1/100 of its collector current. About the same thing.
The Vbe is dependent on current. Because Q1 = Q2 in many ways, the Vbe should be the same. Both Bases are 1 diode drop above the voltage on R2.

We have a high gain amplifier, (Q1). It thinks any voltage below 0.24V is low. Any voltage above 0.26 is high. There is a small linear region at about 0.25V.

The IC is looking to see if the voltage on FB is 1.25 volts. Because of Q1 this is translated to (is the voltage across the 0.7 ohms equal to 0.25 volts)? Current=350mA


upload_2014-7-11_19-56-10-png.87317
 
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