DC converter

How does that 6V look? Lots of ripple? Depending on your load current

60Hz requires a few hundred uF. So if other than the 10uF there's nothing on the input side that would be your problem :-)

--
Regards, Joerg

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On a sunny day (Thu, 18 Feb 2010 12:16:57 -0800 (PST)) it happened john wrote in :

Without looking at your diagram, did you take into account the voltage drop of 2 diodes in the bridge?

2 x 0.7 = 1.4 5 - 1.4 = 3.6

A function generator is _feeding_ the bridge?

Jon

Hi, Yes the function generator is feeding the bridge. I increased the input voltage frequency upto 5000Hz and 50khz and amplitude to 24 volts peak to peak. And also took the full bridge out and replace with one diode and connect a 20kohm as a load and 220uF 25 volts capacitor in parallel with the load. I can see the output of 9 volts (volt meter mesurement)

So, now I have half bridge rectifier output but still I am not getting the 5 volts output. As soon as I connect the output of the single diode bridge to the DC to DC converter the output of the single diode bridge falls from 9 volts to 3.5 volts and the out put of the DC to DC converter is 3.5 volts not 5 volts. The ground of the fumction generator is connected to ground of the dc to dc converter

John

SO, the function generator has output of 50 ohm impedance. My be it can drive the bridge or a single diode.

Hmm, Maybe you are loading down the signal generator? Does the output change when you attach the DC-DC converter? How much current is the DC-DC drawing.

George H.

Let's take this step by step. A couple of questions before we attempt to solve the 'mystery': Did you measure the output voltage from the function generator before or after you connect the bridge rectifier and the DC-DC converter to it? Did you connect any load at the output of the DC-DC converter?

john wrote:

Your posts are confusing to me. If you have the generator ground connected to the DC-DC ground along with the negative terminal of the rectifier bridge, that's wrong... the generator ground should be isolated from the rectifier negative, otherwise, you have a half-wave rectifier. Your hookup should look like the diagram below. Is this what you have?

Please view in a fixed-width font such as Courier.

Generator

+------+ DC-DC Conv | | +-----------+ | | FWBridge | | | | +----+ | | | |Out AC| |+ | | | +-----------+ +-----+-------------|Vin | | | | | | | | | | | | | | | | | | | |+ | | | | | | -+- | | | | | | -+- | | | | | | | | | | | | | | | | | |Gnd AC| |- | | | | +-----------+ +-----+ | | | | | | | | | | | +----+ | | | | | | | | | | | | | +------+ | | | | | | | | Gnd | | +--+--------+ | | +----------------+ | | ---+--- --- -

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Dave M
dgminala at mediacombb dot net

--- First off, your function generator has an output impedance of 50 ohms, which means that it looks like a voltage source with a 50 ohm resistor connected in series with it, the unconnected end of the resistor being presented to the outside world as the output of the generator, like this: (View in Courier)

E1 / +---[50R]----->E2 | [E] | +------------->GND

With a load connected , the circuit becomes a voltage divider, so we have:

E1 / R1 +---[50R]--+--->E2 | | [E] [xR] | | +----------+--->GND

Now, with the circuit you've described, a half-wave rectifier followed by a reservoir capacitor and a load,:

E1 E2 / R1 / +---[50R]-+-[DIODE>]--+------+ | |C1 |R2 [E] [220µF] [20k] | | | +---------------------+------+ \ GND

We'd expect, to a first approximation, that:

(E1 * R2) 12V * 20kR E2 = ----------- = ------------ = 11.97 volts R1 + R2 50R + 20kR

Subtracting about 0.7 volts for the drop across the diode would make E2 equal to about 11.3 volts, but I suspect that the reverse capacitance of the rectifier at the frequency you're driving it at is sucking charge out of the cap, making its terminal voltage fall.

Drop the frequency down to around 60Hz and you should see the voltage across the cap rise to around 11 volts.

---

--- Remembering that there's a 50 ohm resistor in series with the output of your function generator, whatever you load it with will cause its output voltage to drop and, eventually, to starve your load if the current your load draws causes the voltage dropped across the generator's output resistance to increase to the point where there's not enough left to supply the load with what it needs.

JF

This is an analog signal generator with an output impedance of 50 ohms, referenced to ground. If you ground an output from the bridge with a scope probe, you change the schematic to a half-wave rectifier, giving 16V/2 - 2 x diode drops.

RL

Your function genarator even if you hook everything up correctly as others have suggested likely is'nt up to the task you are useing it for.

When you use a bridge cap input the cap gets recharged in current pulses which your FG is probably crapping out on.

Assumunig your buck is the 1.75w Buck in the AN your FG would have to supply takeing into consideration rectifier and SMPS conversion losses about 2.1W. It would have to supply larger peak power to charge the caps which its likely not able to do.The more current it provides the less the output voltage would be.Plus the other problems people have mentioned.

OK. Coming out of the gate you have introduced major problems by using=20 a laboratory signal source as a power source. Get a decent audio power=20 amplifier (50 W should do it), feed it from the 4011, and use that=20 combination as the substitute for a transformer.

This is caused by the internal impedance of the 4011. You should see=20 about 14 V peak. and read about 12 with an average reading voltmeter=20 (both cases without filter cap).