Input 24Vac OutPut 40Vdc /1A circuits -issues

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The MUR420 is 4A 200V fast diode. It will work good.
i tried above schematic............
but the voltage in no load condition rises above 100V
Click to expand...
You NEVER use a current source with no load. The voltage will get very high. I am not certain what schematic but a "boost" power supply will make very high voltage is there is no limit.

I added a 1k resistor in the feedback path. It will have no effect.
I added a high voltage Zener diode from "OUT" to "FB". If there is no load (diode was not in circuit) then a small amount of current will flow in FB and cause the LT1070 to limit the voltage to about 57 volts. The value of the diode is no important. It needs to be more than the 38 volts of the diode and less than the maximum voltage of the IC and D1 and C1.
 

We tried above schematic voltage in no load condition is solved but output current is too low. The Report of above schematic is

Input voltage 24vac
Input current 0.586mA
Watts consumed 9.450W

Output Voltage 34.41Vdc
Output current 0.232mA
Watts 7.9W
power factor 0.673 should be above 0.9
efficiency 83%

But what we need in output voltage and current is 38V/1A ..With 4 to 6W driver loss is applicable...
 
If we remove 1k from Feed back , these are the following report..


Input voltage 24vac
Input current 1.7062A
Watts consumed 31.02W

Output Voltage 37.74Vdc
Output current 0.678mA
Watts 25.58W
power factor 0.771 should be above 0.9
efficiency 82%
 
With out 1K:
output current 0.678mA? Why not 0.83?

With 1k:
This gives me the idea the Zener has 500uA of current. Change the 1k to 100 or find a zener with less current. You need a resistor there!

power factor should be 0.9
Click to expand...
You should have said this in post #1. This changes everything! You need a power factor corrected design.
 
We tried above schematic voltage in no load condition is solved but output current is too low
Click to expand...
Simulation shows the schematic with Ronsimpson's helpful mod (post #21) should produce an output current of 0.83A.

Input voltage 24vac, Input current 0.586mA, Watts consumed 9.450W
Click to expand...
Even assuming you mean 0.586A those figures don't make sense to me. With more than 0.5A flowing the power would be more than 24 x 0.5 = 12W (regardless of whether we're talking RMS values or not). What was your actual load for that test?
 
actual load for the above test is Single CREE LED 40V /1A... we use zener 51V ...i will try with less voltage and give the feed back
 
we use zener 51V ...i will try with less voltage and give the feed back
Click to expand...
The voltage is not so important. It is to keep the IC from braking if the voltage is too high.
You can change the 1k to 220 or 100. That will help the leaky zener problem.

Back to the power factor question:
You need PF=0.9.
Where is this power coming from? 110vac or 220vac?
Do you have a 220 to 24V transformer.
This requires a different design!
 
ronsimpson said:
Where is this power coming from? 110vac or 220vac?
Click to expand...

The client have a direct 24Vac source , so transformer not required .... we kept burning from yesterday driver loss is 5w ..input watt is 25 and out put is 20W... we want increase the current alone in output around 0.900mA or 1A so we get the exact output.
 
The input Watts are variously stated to be 9.45 (post #22), 31.02 (post #23) and 25 (post #28). Why the big variation? What conditions are you changing?
 
Output current 0.678mA
Click to expand...
Why not 0.833A? Maybe you are not regulation. What is the voltage on "FB"? It should be 1.25V.
What inductor are you using? What is the saturation current?
What type of resistor are you using for the 1.5 ohm resistor?
Can you send pictures? We need pictures.

Zener problem: I can see the zener is send too much current into the FB pin. This new circuit will send the zener current to ground through a 1K resistor. Only when the current reaches 1.5mA then the zener will pull up on FB and shut down the IC.

input watt is 25 and out put is 20W
Click to expand...
You are 80%. Where is the 5 watts going. Something must be hot. What is hot? IC? Diode? What efficiency do you want? 90%?
 
How are you measuring the input current? A normal DMM will give misleading current readings because the current is pulsed and non-sinusoidal.

Edit:
I don't think the feedback pin current is a problem, Ron. Simulation shows that it actually goes close to zero when the zener begins to conduct if the LED is disconnected.


Edit2:
When the LED is passing 0.83A the peak current in the inductor is approaching 3A, so the inductor must be rated to handle at least 3A without saturating.
 
Last edited:
Alec_t,
From #22 & #23, when the Zener is in place, with a good LED, the current is 1/3. That tells me that his Zener pass 500uA to 700uA. The high voltage Zeners pass some current below their turn on voltage. The circuit in #30 will probably solve this problem. We are looking for a zener that will not effect FB at 40V but will turn on just below 60V. (FB current is very small, the problem is zener current at 37V)

It looks like he has a good current meter. He can measure PF and peaks current.

He has some problem we can't see, like inductor saturation.
 
It looks like he has a good current meter. He can measure PF and peaks current.
Click to expand...
Considering the disparities between the measurements of input current, who knows? If the current measurements are doubtful then won't the PF values be too?
 
Today i will update you all ...sorry i was in vacation ..

we used this type inductor

744732121 Size 8075; L = 120 µH ±10%; IR = 1 A; Isat = 1.1 A; RDC = 0.38 Ω; Shrink Tube
attached the datasheet for reference ...

Actual inductor spec is
**broken link removed**


i have to increase the inductor current to 5A and re-test the circuit ..
 

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