Hi,

Could someone set me up with some 'general rules' around source impedance/resistance, and why it is important? I mean in relation to power supplies such as a battery. I understand that any device's output is going to have an associated impedance, even a power supply, but do not 'get' what sort of impedance say, a 9V battery or a transformer, is going to offer to the following circuit. So I don't know what to 'assume' the impedance of an AC source or a battery should be in circuit simulation, etc., and how it affects current flow to the device it's powering. It's easy to figure out what a transistor's output Z will be, but a power supply??

It's not like you can measure a battery to find its internal impedance
For AC, is the DC resistance of the secondary sufficient to assume its impedance? Or would one have to go thru the steps of finding L and C reactance and actually do the impedance calculation? Could be difficult if the L and C values aren't known...how's that done?

Thanks in advance for your input!

~Mike

 

Hi,
...It's not like you can measure a battery to find its internal impedance
...

Sure you can. Hook a DVM to the battery. While watching the meter, momentarily connect a 1K resistor to the battery. Note the drop in voltage.

Now do it again with a 100Ω resistor.

Now try a 10Ω resistor.

Plot the voltage drop (ΔV) vs battery current (which is just the voltage measured with the load applied divided by the resistor value). The slope of the line is the battery impedance.

 

Thanks Mike! I DO understand voltage drop/sag (at least mechanically)...so the drop is a consequence of the internal resistance? Ability of the source to provide power.... This works with all power sources, and changes with load, eh?
So at a given 'place' on the 'load line', Y=MX+B will equal the impedance?
Cool.

 

Components are never ideal. They always come with extra capacitance or inductance or resistance etc. which means the real life component usually will perform slightly differently from that predicted.

Batteries have an internal resistance, equivalent to a small internal resistor in series with an ideal voltage source.

The value of this internal resistance [Ri] can be measured because it is in series with the load, so the same current flows through the Ri as the external load. If you know the battery terminal voltage without a load [V1], then you put a known load (known current ) on the battery and measure the now lower terminal voltage [V2], the difference is the voltage that has been dropped across the Ri.

So using ohms law, Ri = (V1-V2)/I

A batteries internal resistance increases as it runs down, so a partially exhausted battery may have the same unloaded terminal voltage as a fresh battery, but the voltage will drop more when the load is applied. This is a good way to measure remaining charge in a battery.

You can use the same technique to measure the source resistance of any power supply.

 

Thank you, Boriz! It is something everyone has seen in action, but I didn't have any mathematical model to get semi-quantitative with ('home-schooled electronics guy' here).

These source resistances show up in simulations I work with from time to time (mainly tube amp power supplies). It will be nice to be able to include that info for my own junkbox stuff by doing the quick measurement, just to get a rough idea of what to expect when building a circuit! If you're going to include tube Zin and Zout, might as well consider the source Z, also...

 

Components are never ideal. They always come with extra capacitance or inductance or resistance etc. which means the real life component usually will perform slightly differently from that predicted.

Batteries have an internal resistance, equivalent to a small internal resistor in series with an ideal voltage source.

The value of this internal resistance [Ri] can be measured because it is in series with the load, so the same current flows through the Ri as the external load. If you know the battery terminal voltage without a load [V1], then you put a known load (known current ) on the battery and measure the now lower terminal voltage [V2], the difference is the voltage that has been dropped across the Ri.

So using ohms law, Ri = (V1-V2)/I

A batteries internal resistance increases as it runs down, so a partially exhausted battery may have the same unloaded terminal voltage as a fresh battery, but the voltage will drop more when the load is applied. This is a good way to measure remaining charge in a battery.

You can use the same technique to measure the source resistance of any power supply.

Yip. If it weren't for internal resistance, you could get a million watts from a nine volt radio battery.


Maximum power transfer occurs when the load resistance is equal to the internal resistance of a source. If reactance is involved, the maxiumum power transfer occurs when the load is the complex conjugate of the source impedance.

eric

 

There is always more to uncover, ha ha ;o) Good to know; this is a nice explanation of how 'sag' is actually operating! Good luck on my finding the complex conjugate of a transformer, though! But I doubt I'll need to get THAT involved for sim purposes; creating a 'load line' of sorts for a power supply would probably do the job for what I deal with.

I am mainly interested in how to get a handle on how your voltages will sag as you design gain stages (you know, how that painstakingly designed V1 anode voltage will drop as you mess around with V2's voltages), since there is no perfect power supply. Thanks for the info, really helpful!

~Mike