I am still trying to work my way around exercise 1, sigh, the simple zener diode exercise. My task this time was to put together a simple circuit so that I could get 5v, and 5ma to the load.
I did not post my schematic as it is the same one as on my other threads and in the tutorials.
Source v = 8.85
Zener resistor = 655 ohms, voltage drop = 2.90v
Zener = BZX55C6V2, 6.2 V
Load resistor = 1320 ohms
Load voltage drop = 5.62

Load current = 12.13ma ?? Huh

Given Ohms law, I=E/R,
I=5.62/1350 = 4.16ma close enough to 5

So why the heck am I reading 12.13 ma ?

And thanks again.

 

because you are trying to measure current by connecting multimeter in parallel with the load.
your multimeter is about 75 Ohm when on mA range.

655+75 = R

8.85V/R = 12.13mA

 

Zener = BZX55C6V2, 6.2 V

If you want 5V, why are you using a 6.2V zener?

 

because you are trying to measure current by connecting multimeter in parallel with the load.
your multimeter is about 75 Ohm when on mA range.

655+75 = R

8.85V/R = 12.13mA

Thank you panic and that means that I have no idea where to take the measurements.
I need to find me a young EE to look over my shoulder as I wire this booger and watch me as I measure stuff,
It's apparent that I have no idea what in an doing. And to think that computer programming comes easy to me.
Let me see if I can draw my circuit with reference points so it might be easier to pound it into my head.
I understand your explanation but don't know why the formula used the source voltage, 9, rather than the voltage at the load, 5.62.
I will try and be back later if I can make it clearer.

 

Thank you panic and that means that I have no idea where to take the measurements.
I need to find me a young EE to look over my shoulder as I wire this booger and watch me as I measure stuff,
It's apparent that I have no idea what in an doing. And to think that computer programming comes easy to me.
Let me see if I can draw my circuit with reference points so it might be easier to pound it into my head.
I understand your explanation but don't know why the formula used the source voltage, 9, rather than the voltage at the load, 5.62.
I will try and be back later if I can make it clearer.

In order to measure current with an MM, you have to make that current go through the MM. That is, connect the MM in series with the load. So you have to, for instance, connect the zener to the voltage source, then connect the other end of the zener to the resistor, then connect one probe of the MM to the other lead of the resistor, and then the other probe of the MM to the circuit's ground. Just make sure you've set your MM to read current when you do this.

 

If you want 5V, why are you using a 6.2V zener?

Thank you Albert and I have no clue. I seem to remember reading something about the zener value being > than the desired voltage.
I really have no idea how to size one, I am firstly trying to get to the point where I can predict the results. So, is that how it's sized ?
Whatever voltage I need, that's the zener to select? Other values being relevant as well which is beyond me.
I glanced at the data sheet and mostly it's indecipherable except for min, max. I remember also that the source needs to be > than the zener min,( I think ).
Mostly they are a mystery.

 

Does his help?

As for a zener, just think of them as a weir on a waterway. The water level stays about the same.

Figure out your load current. Then select a zener that can carry that load current plus a bit (in case the load is disconnected the zener has to carry the lot) based on the zener power rating, then size the series resistor to allow that current, load + a bit for the zener, to flow.

 

In order to measure current with an MM, you have to make that current go through the MM. That is, connect the MM in series with the load. So you have to, for instance, connect the zener to the voltage source, then connect the other end of the zener to the resistor, then connect one probe of the MM to the other lead of the resistor, and then the other probe of the MM to the circuit's ground. Just make sure you've set your MM to read current when you do this.

!!!!! Hallaleu brethren !!!!!!
By gosh, by golly, that was it. I knew it had to do something to do with how I was doing the measuring.
Finally !
Cmartinez, you are my new god. Can I send you a tip ?

 

View attachment 138794
Does his help?

As for a zener, just think of them as a weir on a waterway. The water level stays about the same.

Figure out your load current. Then select a zener that can carry that load current plus a bit (in case the load is disconnected the zener has to carry the lot) based on the zener power rating, then size the series resistor to allow that current, load + a bit for the zener, to flow.

Dendad, yes that did help,,,, a lot.
Thanks, I think that I made this more difficult than it actually is. It says a lot for in person education.

 

!!!!! Hallaleu brethren !!!!!!
By gosh, by golly, that was it. I knew it had to do something to do with how I was doing the measuring.
Finally !
Cmartinez, you are my new god. Can I send you a tip ?

Thanks, but I have a no-tips policy... but I'll let you buy me a beer... something like this will do:

​

Glad to be of help.

 

Thanks, but I have a no-tips policy... but I'll let you buy me a beer... something like this will do:

View attachment 138798​

Glad to be of help.

My word but I really must give that a try. 27%!!