Sometimes when experimenting with a transistor amplifier or a LM386 which have a DC bias on the output I put a cap on the output to block the DC and I forget to put a load on the cap. When I measure the output I find my AC test signal and a DC bias. The cap is there and I would think it would block the DC. I get it, I need to tie the cap to ground but is there a physical interpretation for this? I mean, the scope is referenced to ground. It seems one plate is charged up and without a circuit reference the other plate follows.

 

hi pete,
Large electrolytics do have an internal leakage path resistance.
So using a high impedance voltmeter you will measure a DC value.
E

 

Sometimes when experimenting with a transistor amplifier or a LM386 which have a DC bias on the output I put a cap on the output to block the DC and I forget to put a load on the cap. When I measure the output I find my AC test signal and a DC bias. The cap is there and I would think it would block the DC.

It can only "block" the DC bias after it has charged up to the value of the DC bias and cancels it out; and it can only do that if the circuit is completed by connecting a load. Otherwise no current flows through the cap, and with no current to charge it up it remains in its discharged state and so when you measure the output voltage you'll see the DC voltage as well as the AC.

An interesting experiment: if you leave your scope or DMM connected to the output for a while (with load disconnected), you should see the DC voltage gradually decrease. This could take many seconds or minutes, or even hours. What you're seeing is the cap gradually charging up through the high-resistance "load" presented by the scope or DMM.

 

It can only "block" the DC bias after it has charged up to the value of the DC bias and cancels it out; and it can only do that if the circuit is completed by connecting a load. Otherwise no current flows through the cap, and with no current to charge it up it remains in its discharged state and so when you measure the output voltage you'll see the DC voltage as well as the AC.

An interesting experiment: if you leave your scope or DMM connected to the output for a while (with load disconnected), you should see the DC voltage gradually decrease. This could take many seconds or minutes, or even hours. What you're seeing is the cap gradually charging up through the high-resistance "load" presented by the scope or DMM.

So the cap has to be charged up to block DC. I did try attaching a cap to a power supply and looking at the negative terminal with a scope. Except for a small initial charge that was still on the cap that dissipated I do read 0 volts is what I expected I'm going to try that again and meanwhile think about the charging up.

Thanks

 

hi pete,
Use a larger electrolytic cap, say 1000uF and a low voltage DC source say 12Vdc or 24Vdc.
If the impedance of your voltmeter is high ,say 1meg to 10meg Ohms, the charge indication on your meter will last longer.

Lets know what you see.
E

 

hi pete,
Use a larger electrolytic cap, say 1000uF and a low voltage DC source say 12Vdc or 24Vdc.
If the impedance of your voltmeter is high ,say 1meg to 10meg Ohms, the charge indication on your meter will last longer.

Lets know what you see.
E

I rechecked my set up and had the scope on ac coupling from last night. Correcting that, a voltage on the negative terminal appears. That's in agreement. It's just natural for me to think that the dielectric would separate the plates' charges when one terminal isn't connected. But in this case my thinking is wrong, or at least incomplete for now.

 

hi,
Its the effect OBW posted about, the cap is charging up via your meters internal resistance, so you measure a neg value.

If you leave it connected long enough, the current will not go to zero due to the caps internal leakage resistance.

E

 

Put a bleeder resistor from the output of the cap to GND.

 

If there is no path for current in the circuit, there will be no mechanism for change in charge on the capacitor. If the capacitor was discharged to begin with, connecting one electrode to a zillion volts and leaving the other open will not put any charge on the capacitor. You simply have one piece of metal electrically insulated from another. Once you introduce a current path the capacitor will begin to charge.

Ignoring leakage current (which you can't if the cap is electrolytic), a large capacitance with the input resistance of a 10 megohm attenuator probe can make a very long times constant. Even with a 10 µF capacitor, the settling time to get within 1% of final value, which will look stable on a scope, is over 8 minutes. 100 nF will settle to the same error in about 5 seconds. It takes about 5 time constants, that being the product of the capacitance in farads times the resistance in ohms, with the result in seconds, for a capacitor to charge to within about 1% of the source voltage or the current to drop to about 1% of its initial value. Capacitors for DC blocking are normally chosen so that the time constant formed by the capacitance and the effective resistance is long relative to the period of the lowest frequency of interest so as not to attenuate the signal.

Leakage current (DC current which flows through the cap even when it is nominally fully charged) in electrolytic capacitors is rarely specified to be less than about 3 µA except for special low-leakage types, which aren't very common.. It can be much more and in practice can be less. 3 µA of current into a 10 meghohm load (meter or scope input) equates to 30 volts! Of course you can't get 30 V if the supply voltage is less than that, but the leakage current can still lead to a high voltage reading if the only "load" is a meter or scope, even after the capacitor has had a very long time to charge. All film type capacitors and ceramic types have much much lower leakage current.

 

If there is no path for current in the circuit, there will be no mechanism for change in charge on the capacitor. If the capacitor was discharged to begin with, connecting one electrode to a zillion volts and leaving the other open will not put any charge on the capacitor. You simply have one piece of metal electrically insulated from another. Once you introduce a current path the capacitor will begin to charge.

Ignoring leakage current (which you can't if the cap is electrolytic), a large capacitance with the input resistance of a 10 megohm attenuator probe can make a very long times constant. Even with a 10 µF capacitor, the settling time to get within 1% of final value, which will look stable on a scope, is over 8 minutes. 100 nF will settle to the same error in about 5 seconds. It takes about 5 time constants, that being the product of the capacitance in farads times the resistance in ohms, with the result in seconds, for a capacitor to charge to within about 1% of the source voltage or the current to drop to about 1% of its initial value. Capacitors for DC blocking are normally chosen so that the time constant formed by the capacitance and the effective resistance is long relative to the period of the lowest frequency of interest so as not to attenuate the signal.

Leakage current (DC current which flows through the cap even when it is nominally fully charged) in electrolytic capacitors is rarely specified to be less than about 3 µA except for special low-leakage types, which aren't very common.. It can be much more and in practice can be less. 3 µA of current into a 10 meghohm load (meter or scope input) equates to 30 volts! Of course you can't get 30 V if the supply voltage is less than that, but the leakage current can still lead to a high voltage reading if the only "load" is a meter or scope, even after the capacitor has had a very long time to charge. All film type capacitors and ceramic types have much much lower leakage current.

 

Thank you. I did find by experiment that simply attaching one lead of a capacitor to a power supply voltage and measuring the the other I do get a DC voltage at the 2nd terminal which is correct.
It raises a further question to me of the mechanism. If on any given circuit diagram I attach a component at some point there is a voltage and leave the other end unattached the voltage measured will be the voltage at the node. This easy to see with a resistor or an inductor. But with a capacitor with 2 plates how the voltage at the node transfers through the dielectric to the other plate is a mystery to me. I accept it, the voltage at a node is the voltage at a node but even with an air capacitor, which I don't have any of, it's hard to see how the voltage on a plate would occur on the other.

 

hi.
You should think in terms of electron charge on one plate relative to the other plate.
If the two plates are connected via an external conductive path, excess electrons on one plate will move in direction in the conductor to the plate that has fewer electrons, until the two plates have the same net electron charge, ie: voltage

This is a very rough description, but it should help you understand what is happening regarding the capacitor.

 

hi.
You should think in terms of electron charge on one plate relative to the other plate.
If the two plates are connected via an external conductive path, excess electrons on one plate will move in direction in the conductor to the plate that has fewer electrons, until the two plates have the same net electron charge, ie: voltage

This is a very rough description, but it should help you understand what is happening regarding the capacitor.

 

But I was thinking of perhaps something else and wasn't clear. With one end of a capacitor attached to a voltage and the other end sticking in air the voltage shows up on the other plate. There is no conductive path

 

But I was thinking of perhaps something else and wasn't clear. With one end of a capacitor attached to a voltage and the other end sticking in air the voltage shows up on the other plate. There is no conductive path

You need to understand that you are not measuring actual voltage when you use a mutlimeter or oscilloscope or any other instrument. What you are actually measuring is the loss or drop or difference in voltage between 2 points (your probes) required to bridge a resistance. If the resistance is an open circuit (like a capacitor), then the value will be the maximum possible by any related source(s).

In fact, truly understanding this concept will enable you to understand and resolve both voltage and current issues almost intuitively in circuits far easier than people who don't understand this.

 

There is no conductive path

hi pete,
So how are you measuring this voltage.???

E

 

hi pete,
So how are you measuring this voltage.???

E

He's probably creating a path by connecting his meter to the floating end of the capacitor and the other end to ground.

 

hi Bob,
I know that , I was asking him the question, so he might have a Eureka moment.
Eric

 

hi Bob,
I know that , I was asking him the question, so he might have a Eureka moment.
Eric

 

MOD:
Pete, why do you post the Quote on its own in a post.?
You should add your new text under the Quote in the same post.
E