MAX868 issue

Thread Starter

atlas513

Joined Nov 12, 2012
12
C9 is there for extra security persay. It doesnt affect the output but can regulate the current in the circuit.

As for changing 4.7uF to 10uF. The main reason I do not have it as 10uF is because it doesnt give me -10V as an output. Instead it gives me -3V if 10uF. I only get -10V when that cap is 4.7uF.

Edit: This works if I use a 1.48nF cap instead of 4.7uF.
 
Last edited:

tshuck

Joined Oct 18, 2012
3,534
C9 is there for extra security persay. It doesnt affect the output but can regulate the current in the circuit.
But it will affect the output... this is the feedback path.. You are adding a low pass filter to it... you are forcing the device to work much harder as the capacitor resists a change in voltage...the exact thing this device uses to regulate the output....

The output capacitor is there for smoothing the output and providing instantaneous current capability....You should keep it...otherwise, you'll see lots of ramping on your voltage when any switching occurs...

The fact that the 10uF cap doesn't give you the desired output should tell you that your circuit isn't working as planned... Look at the Choosing the Output Capacitor section on page 7...


C9 is there for extra security persay.
I'm sorry to nitpick, but I hate this phrase and it's misuse in modern language.... it is Latin.. per se, which means in itself, or intrinsically... to say that a capacitor is there for extra security in itself, doesn't make much sense....a capacitor is not a security device. It can be used for security, but this is not its function...

Okay, rant over.... back to your regularly scheduled programming...:p
 

Thread Starter

atlas513

Joined Nov 12, 2012
12
Well I will be adding some more load to the output and see what the response from the device will be. I think that is the only way to see if the voltage will carry through the system instead of spending time to figure out why the schematic in the datasheet isnt working for me.
 

tshuck

Joined Oct 18, 2012
3,534
Well I will be adding some more load to the output and see what the response from the device will be. I think that is the only way to see if the voltage will carry through the system instead of spending time to figure out why the schematic in the datasheet isnt working for me.
I have a tendency to trust the people that design/make the chips that I use. Their circuits are usually fairly good guides as to how to make a circuit that is similar to theirs, but that's just me. There are other ways to circumvent the original intent and use it otherwise, but they know something we don't about the implementation of the device, so I tend to want to trust them when there are multiple options, but again that's just me...

I hope your circuit works, though, common sense says, it was smoking => it is damaged and will probably not work properly...

I hope you have plenty more components on hand...
 

Audioguru

Joined Dec 20, 2007
11,248
The IC operates at a frequency that is too high for a breadboard. Did you make it on a breadboard?
Is the 4.7uF output capacitor a ceramic type?
 

bountyhunter

Joined Sep 7, 2009
2,512
I think the iC was zapped the moment you tried it without the important input capacitor.
The reason that is very likely is:

High frequency switch devices have to draw very fast currents from somewhere.

Without a very low impedance cap right at the input pin, it is drawing it from the source down relatively high inductance leads which can create voltage ringing/spiking (both positive and negative). That's what can kill an IC.
 

Thread Starter

atlas513

Joined Nov 12, 2012
12
The IC operates at a frequency that is too high for a breadboard. Did you make it on a breadboard?
Is the 4.7uF output capacitor a ceramic type?
Yes I redid this on a breadboard and yes it is a ceramic type.

As of the voltage carrying through. It seems that adding more resistance to the output still keeps the voltage intact. I will have to add more components to see if it can work in a bigger system.
 
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