Emulating or simulating an alkaline 9V battery

Thread Starter

Woodyd

Joined Apr 14, 2015
7
I have just started learning electronics and I am learning about Ohm's law. I have some simple experiments to do using an LED in series with a resistor. The experiments use a 9V pp3 alkaline battery. I have acquired a lab power supply and want to use the power supply instead of the 9v battery. I just don't know what current to set the power supply to so that I can get the same results as if I used the battery. I would appreciate some concept help please.
 

Kermit2

Joined Feb 5, 2010
4,162
Depending on the load.

It will handle most any load of 100mA or so for a short time.
It's terminal voltage will drop to a little above 8 volts, then the slow drop off begins.

Plot a load vs voltage chart using a few resistors. 100 ohms should give you 90mA. What voltage does the battery have when you place that resistance across it?
 

EM Fields

Joined Jun 8, 2016
583
I have just started learning electronics and I am learning about Ohm's law. I have some simple experiments to do using an LED in series with a resistor. The experiments use a 9V pp3 alkaline battery. I have acquired a lab power supply and want to use the power supply instead of the 9v battery. I just don't know what current to set the power supply to so that I can get the same results as if I used the battery. I would appreciate some concept help please.
If you take a typical red LED with 20mA of forward current (If) through it, it'll drop about 2 volts(Vf), so if you're driving the LED with a 9 volt battery you'll need the series resistor to drop the 7 volts the LED doesn't need with 20mA through both the resistor and the LED, since currents in a series circuit are everywhere the same.

The way to determine the value of the resistor is by writing R = Vs -Vf/If = 9V-2V/0.02A = 350 ohms

Note that whether you use a 9 volt supply capable of supplying several amperes or a 9volt battery capable of supplying only a much smaller current, the formula remains the same since the ballast resistor is doing the current limiting. Consequently, where you set the supply's current limit will be unimportant as long as it can supply the 20mA the LED needs in order to light up.
 
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GopherT

Joined Nov 23, 2012
8,009
I have just started learning electronics and I am learning about Ohm's law. I have some simple experiments to do using an LED in series with a resistor. The experiments use a 9V pp3 alkaline battery. I have acquired a lab power supply and want to use the power supply instead of the 9v battery. I just don't know what current to set the power supply to so that I can get the same results as if I used the battery. I would appreciate some concept help please.
9V on a power supply is the same as 9V battery. The only difference is that:
a good power supply will have
- higher maximum current output (but ohms law will still apply)
- stay at 9 v no matter the load (up to the limit of the supply)
- some supplies can also have a control to automatically lower the voltage if a preset current value is reached.

A real 9V battery will:
- start at 9.3V (approx for new) and decrease in value as the battery is discharged
- have "internal resistance" that increases from about 5 - 10 (new) to 100 volts (or more) as the battery is consumed.
- higher internal resistance means the voltage across the terminals drops as the load increases.


If your circuit is assembled correctly, the current setting of your power supply does not matter. Ohms law will apply.
The supply cannot "push" more current into the circuit than ohms law will tell you the current is. Use EM Fields rules to calculate the resistor you need to limit current to your LEDs (or to calculate the current to your LEDs if you are using designs from a textbook).
 

Thread Starter

Woodyd

Joined Apr 14, 2015
7
If you take a typical red LED with 20mA (If) through it it'll drop about 2 volts(Vf), so if you're driving the LED with a 9 volt battery you'll need the series resistor to drop the 7 volts the LED doesn't need with 20mA through both the resistor and the LED at 20mA.

The way to determine the value of the resistor is by writing R = Vs -Vf/If = 9V-2V/0.02A = 350 ohms

Note that whether you use a 9 volt supply capable of supplying several amperes or a 9volt battery capable of supplying only a much smaller current, the formula remains the same since the ballast resistor is doing the current limiting. Consequently, where you set the supply's current limit will be unimportant as long as it can supply the 20mA the LED needs in order to light up.
Thank you all very much for the input. Much appreciated. Getting clearer for me.
 
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Thread Starter

Woodyd

Joined Apr 14, 2015
7
9V on a power supply is the same as 9V battery. The only difference is that:
a good power supply will have
- higher maximum current output (but ohms law will still apply)
- stay at 9 v no matter the load (up to the limit of the supply)
- some supplies can also have a control to automatically lower the voltage if a preset current value is reached.

A real 9V battery will:
- start at 9.3V (approx for new) and decrease in value as the battery is discharged
- have "internal resistance" that increases from about 5 - 10 (new) to 100 volts (or more) as the battery is consumed.
- higher internal resistance means the voltage across the terminals drops as the load increases.


If your circuit is assembled correctly, the current setting of your power supply does not matter. Ohms law will apply.
The supply cannot "push" more current into the circuit than ohms law will tell you the current is. Use EM Fields rules to calculate the resistor you need to limit current to your LEDs (or to calculate the current to your LEDs if you are using designs from a textbook).
Based on Ohm's law and disregarding the internal resistance of the battery, if I dead-shorted the battery i.e., no resistance, then I=V/R and then I=9/0 = 9 amps. Is this correct? Could a 9v battery produce 9 amps with a zero load?
 

Kermit2

Joined Feb 5, 2010
4,162
You would divide the 9 volts by the INTERNAL resistance of the battery, not by zero.
You never divide by zero...
So a short across the terminals would deliver the current allowed by that resistance.
 

MrChips

Joined Oct 2, 2009
30,706
Based on Ohm's law and disregarding the internal resistance of the battery, if I dead-shorted the battery i.e., no resistance, then I=V/R and then I=9/0 = 9 amps. Is this correct? Could a 9v battery produce 9 amps with a zero load?
Basic math, anything divided by zero gives infinity, or more correctly an undefined result.
 

Thread Starter

Woodyd

Joined Apr 14, 2015
7
You would divide the 9 volts by the INTERNAL resistance of the battery, not by zero.
You never divide by zero...
So a short across the terminals would deliver the current allowed by that resistance.
Well taking into account the battery's internal resistance, that would be close to 9 amps wouldn't it? What I'm trying to find out, is the maximum current that can be drawn from a 9V battery.
 

crutschow

Joined Mar 14, 2008
34,280
According to this, the internal resistance of a fresh 9V alkaline battery is 1 to 2 ohms.
So if you shorted across the terminals you would get 4.5A to 9A, but only for a very short (pun) time.
 

dl324

Joined Mar 30, 2015
16,839
What I'm trying to find out, is the maximum current that can be drawn from a 9V battery.
This information is available from the battery manufacturers. This is from Energizer:
upload_2017-2-24_8-9-28.png
The way I read this is that the maximum current they recommend is 500mA and the battery will last about 30 minutes and the ending voltage will be 4.8V.

That type of battery isn't intended for any "high" power load.
 

dl324

Joined Mar 30, 2015
16,839
PSA: If you try checking Eveready batteries, remember that there's only one 'r' in eveready.com. Two r's will take you to a bad website that will tell you that you have a virus and need to call Microsoft. Kill your browser; don't click on the popup. It will likely give you a virus...

BTW, it appears that Energizer acquired Eveready.
 

GopherT

Joined Nov 23, 2012
8,009
Most 9V batteries should not be expected to run continuously over 100 mA or short duration at 500mA. Very short pulses (fractions of a second) at 1 amp ir more and low duty cycle.

There are some (at lease one) lithium-based 9-v that can do over an amp for extended operation.
 

GopherT

Joined Nov 23, 2012
8,009
PSA: If you try checking Eveready batteries, remember that there's only one 'r' in eveready.com. Two r's will take you to a bad website that will tell you that you have a virus and need to call Microsoft. Kill your browser; don't click on the popup. It will likely give you a virus...

BTW, it appears that Energizer acquired Eveready.
No, premium batteries were introduced as "Eveready Energizer".

When Ralston Purina pet foods company spun off Eveready in 2000, the new company was rebranded under the name Energizer Holdings.
 

EM Fields

Joined Jun 8, 2016
583
Based on Ohm's law and disregarding the internal resistance of the battery, if I dead-shorted the battery i.e., no resistance, then I=V/R and then I=9/0 = 9 amps. Is this correct?
No.
Consider: If you had a 9 volt battery with an internal resistance of 1 ohm and your external short had a resistance of zero ohms, when you placed the zero ohm shunt across the battery the current through the shunt would be determined by the battery's internal voltage and its internal resistance, and the current through the short would be I = E/R.

For a 9 volt battery with an internal resistance of 1 ohm, that current would be I = E/R = 9V/1Ω = 9 amperes.

Looking at a second case and supposing that the 9 volt battery had an internal resistance of 0.5 ohms, the current it could output into a zero ohm short would be 18 amperes.

Then, picturing the battery's internal resistance as getting smaller and smaller, the current into the shunt would get larger and larger until, in the limit, when the battery's resistance dropped to zero ohms, the current it would deliver into the zero ohm shunt would be infinite and our local universe would blow up.

Could a 9v battery produce 9 amps with a zero load?
No.

With a zero load the battery will only self-discharge.
 
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Thread Starter

Woodyd

Joined Apr 14, 2015
7
No.
Consider: If you had a 9 volt battery with an internal resistance of 1 ohm and your external short had a resistance of zero ohms, when you placed the zero ohm shunt across the battery the current through the shunt would be determined by the battery's internal voltage and its internal resistance, and the current through the short would be I = E/R.

For a 9 volt battery with an internal resistance of 1 ohm, that current would be I = E/R = 9V/1Ω = 9 amperes.

Looking at a second case and supposing that the 9 volt battery had an internal resistance of 0.5 ohms, the current it could output into a zero ohm short would be 18 amperes.




Then, picturing the battery's internal resistance as getting smaller and smaller, the current into the shunt would get larger and larger until, in the limit, when the battery's resistance dropped to zero ohms, the current it would deliver into the zero ohm shunt would be infinite and our local universe would blow up.



No.

With a zero load the battery will only self-discharge.


Thank you E.M. Fileds. So there must be a limit as to how much current can actually be drawn from a 9V battery. If a battery is rated say at 550 mAh until it drops down to 4.8V, how can I figure out what the maximum current would be if I discharged the battery across a very small load of say 0.50 ohms for 30 seconds? Manufacturers give the mAh ratings but I don't see maximum current ratings. Any light on this?
 

EM Fields

Joined Jun 8, 2016
583
Thank you E.M. Fileds. So there must be a limit as to how much current can actually be drawn from a 9V battery. If a battery is rated say at 550 mAh until it drops down to 4.8V, how can I figure out what the maximum current would be if I discharged the battery across a very small load of say 0.50 ohms for 30 seconds? Manufacturers give the mAh ratings but I don't see maximum current ratings. Any light on this?
If you model a battery as a perfect voltage source in series with an internal resistance, then from Ohm's law, the maximum current that can be taken from the battery will be I = E/(Rs+RL) , where I is the current in amperes, E is the source voltage in volts, Rs is the battery's internal resistance, and RL is the load resistance, both in ohms.

For example, if a [perfect] 9 volt battery with an internal resistance of 1 ohm was externally shorted with a very low resistance, say 10 milliohms, we'd get: I =9V/(1Ω+0.01Ω) ≈ 8.911 amperes out of it.

Bear in mind that this is an unreal expectation in the real world.

But... just for grins, I measured the resistance of a DMM and its leads on its 20A range and got about 40 milliohms, then I connected the leads across a brand new alkaline 9V Energizer and got 4 amperes out of it for a second or two before I chickened out and disconnected it.

Since, from Ohm's law, R = E/I, in this case we have R = 9V/4A = 2.25 ohms. Then, since the meter and lead resistances amounted to 40 milliohms, the battery's internal resistance must have been 2.25Ω - 0.04Ω = 2.21Ω.
 

MrChips

Joined Oct 2, 2009
30,706
I did the same on a fresh 9V battery and got 10A for less than 1 second then the current dropped very rapidly. Did you notice that the battery got very warm?
 
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