some electronic questions

Thread Starter

s3b4k

Joined Feb 15, 2010
38
I don't really know where to find these answers

1. In an oscilloscope, what is the purpose for AC-DC-GND button.
2. Will an oscilloscope interfere with a circuit under test?
3. For a dual power supply, show the necessary connection to obtain -5V and -8V with respect to a common(chassis) ground. Repeat for -5V and +8V.
 

PRS

Joined Aug 24, 2008
989
I don't really know where to find these answers
LOL! I hope you came to the right place! ;)

1. In an oscilloscope, what is the purpose for AC-DC-GND button.
That button, if Ac is selected, puts a capacitor at the input of the scope so that it blocks dc. Only a varying signal will be input to the scope.

If Dc is selected, the scope reads the dc bias voltage as well. There is no capacitor at the input to block dc. The waveform will show the ac plus the dc bias. I wish I could show you this.

2. Will an oscilloscope interfere with a circuit under test?
It can. The probe has a resistance and a capacitance. The probe has a switch to compensate for this. On x1 it will usually show 1 MegOhm resistance (no problem usually, but sometimes it is a problem) and about 20 pF of capacitance. This capacitance is a problem when measuring circuits such as radios have. By switching to the x10, the probe will have about 2pF of capacitance and an input resistance of 10 Mhohms. This is to keep the probe from affecting the parameters of the circuit.

3. For a dual power supply, show the necessary connection to obtain -5V and -8V with respect to a common(chassis) ground. Repeat for -5V and +8V.
I need graphics, which I don't have to show you this. But, if you insist, I will elaborate.
 

Thread Starter

s3b4k

Joined Feb 15, 2010
38
I need graphics, which I don't have to show you this. But, if you insist, I will elaborate.
Yeah i needed help with this one, don't even know what it was asking.
And also if you can please answer the following.
1. what is the purpose of the current limiting feature on a power supply
2. when a power supply is shorted what happens to the output current and voltage.

in question two, doesn't the current become an open circuit?
 

PRS

Joined Aug 24, 2008
989
Yeah i needed help with this one, don't even know what it was asking.
And also if you can please answer the following.
1. what is the purpose of the current limiting feature on a power supply
2. when a power supply is shorted what happens to the output current and voltage.

in question two, doesn't the current become an open circuit?
I attached two supply setups. The first puts out -5 and -8 volts, the second puts out -5 and +8 volts. Note that the symbol for ground in this case has nothing to do with earth ground, but is, rather, a common connection point and is called ground.

To answer your quesions: 1. A current-limitting feature limits the amount of current that can be drawn from the supply. It is needed in case someone shorts the ground to the positive terminal.

This brings up 2: When a power supply is shorted it trys to draw an infinite amount of current. The supply will get hot and quickly burn up.
 

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Thread Starter

s3b4k

Joined Feb 15, 2010
38
@ PRS

When the supply is shorted then what will happen to the actual output of the current and voltage values? you said the supply will burn up, but won't there be a value for the output voltage but 0 for the current?

@ Joe
thanks for showing me the diagrams for the AC/DC/GND it really cleared that up

sorry to bother you but i also needed help with this question?
http://img844.imageshack.us/img844/2087/elecquest.jpg
I believe that for the AC setting it will only read the sine wave, While when it is on DC, it will read the DC bias and the AC part. but i am confused about the rms value. Do i calculate the rms for both DC and AC parts and how would that be Done?? I know the one of the RMS values is 3/2^1/2 but what is the other.
 
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PRS

Joined Aug 24, 2008
989
When a voltage supply is looking at a resistance, R, then the current is given by ohm's law as I = V/R. When shorted, R is equal to zero. This would be I=V/0 which is infinity. Of course infinite amperage is not available, but the supply tries to comply and supplies what current is available. Without short circuit protection, this maximum current produces a large amount of heat due to the collisions of electrons (and/or holes in a semiconductor) and this heat destroys the supply.

As for your question about the DMM reading: When you set the DMM to AC any DC is blocked by a capacitor. Therefore the 3Vp-p sinewave is the only voltage read, and in RMS this is 3/(2*sqrt2) = 1.06 volts RMS.

When you switch the DMM to DC, the 6 Volt DC voltage upon which the AC is riding, is the value that will show up on your DMM. To see this think about 3V-p-p sinewave riding on a dc bias of zero. The average of 1.5 volts and -1.5 volts is zero and therefore the DMM will not read any voltage with respect to the AC. But be aware that if the AC was larger then 6 volts peak (12 V p-p) some of the AC would be included in the RMS voltage. I can draw this if you want.
 

Thread Starter

s3b4k

Joined Feb 15, 2010
38
When a voltage supply is looking at a resistance, R, then the current is given by ohm's law as I = V/R. When shorted, R is equal to zero. This would be I=V/0 which is infinity. Of course infinite amperage is not available, but the supply tries to comply and supplies what current is available. Without short circuit protection, this maximum current produces a large amount of heat due to the collisions of electrons (and/or holes in a semiconductor) and this heat destroys the supply.

As for your question about the DMM reading: When you set the DMM to AC any DC is blocked by a capacitor. Therefore the 3Vp-p sinewave is the only voltage read, and in RMS this is 3/(2*sqrt2) = 1.06 volts RMS.

When you switch the DMM to DC, the 6 Volt DC voltage upon which the AC is riding, is the value that will show up on your DMM. To see this think about 3V-p-p sinewave riding on a dc bias of zero. The average of 1.5 volts and -1.5 volts is zero and therefore the DMM will not read any voltage with respect to the AC. But be aware that if the AC was larger then 6 volts peak (12 V p-p) some of the AC would be included in the RMS voltage. I can draw this if you want.
No i understood for the graph thanks a lot. And for the Power supply being shorted, the final voltage will be zero right?
 

BillB3857

Joined Feb 28, 2009
2,570
Yeah i needed help with this one, don't even know what it was asking.
And also if you can please answer the following.
1. what is the purpose of the current limiting feature on a power supply
2. when a power supply is shorted what happens to the output current and voltage.

in question two, doesn't the current become an open circuit?

The one mentioned in item #1........ It is a knob you turn to set the max current before the power supply starts to automatically change voltage to maintain the PRESET current with a variable load. Like this one...... http://www.mpja.com/prodinfo.asp?number=9600+PS
 

R!f@@

Joined Apr 2, 2009
9,918
Hmmm that I know of.... didn't u know that a current preset is varying the internal resistance of a PSU.
The post was at standard point of view.
Not every supply has current limiting or Short circuit protection
 

BillB3857

Joined Feb 28, 2009
2,570
If regulating the PWM of a switching supply is causing the internal resistance to vary, I must yield. It must have been an assumption that the supply in question had a current limit function since the question was as to what was it's function.
 
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BillB3857

Joined Feb 28, 2009
2,570
We were originally talking about a homework question the OP posted that asked, "1. what is the purpose of the current limiting feature on a power supply?" The correct answer to that question would, to most people familiar with modern power supply technology would be, "To fold back the voltage output of the supply in order to prevent the output current from exceeding the limit that had been established, no matter what the load." From the simple supplies you are talking about, they may have a current limit, after which they will be overloaded, but not a current limit FEATURE. A supply with a true current limit FEATURE, can operate at any current setting (below it's max available) all day long, even into a shorted output. As an example, output voltage is desired to be 5.0 volts, with 1 amp limit. First, with the voltage section set for very low output, short the supply and set the Current limit to 1 amp. Then open the output and set the Voltage output to the desired 5 volts. Any time the current attempts to exceed 1 amp, the voltage will be reduced to a point that the current will maintain 1 amp and no more.

Page 2 of this pdf also explains it well.......rfic.eecs.berkeley.edu/ee100/man/Power%20Supply%20E3631A.pdf
 

Wendy

Joined Mar 24, 2008
23,421
There is more than one form of current limiting. The standard method was current foldback. If you tried to force more current than the power supply was capable of it would drop the voltage to very low values until the "short" was removed.

Modern power supplies use constant current to limit current. This has both good and bad sides. The good side is constant current is darn useful in its own right, while the bad side is it can still cause undesirable heating of bad circuits (usually near the failure location).

Both are very different, but do the same thing in protecting the power supply and the circuit being powered.
 

BillB3857

Joined Feb 28, 2009
2,570
T the bad side is it can still cause undesirable heating of bad circuits (usually near the failure location).

.

That feature alone can be very useful in a bench supply. The time spent finding a shorted bus cap where many are distributed throughout a circuit board is greatly reduced. By limiting the current to safe levels for the traces, enough current can be forced through the bad component that may have been overloading the normal power supply for the unit resulting in full foldback. Just follow the heat!.
 
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