Beginner: measuring maximum load current

[nyoo]

A very junior question.

Often I'd like to know the maximum load current of a simple circuit. After the curcuit is built, I can of course measure this with a multimeter. But for the circuit on paper, I just don't know. And the web won't help me.

The usual comparator circuit might make the problem more concrete. See attached. The relay has a nominal coil current of 60 mA.

How do I measure Imax? Is it the 60 mA coil current, plus the 16 mA Isink of the LM2903? Plus the approximate 1.2 mA for the thermistors? The 2N3904 has a "collector current - continuous" of 200 mA, should I include that? And the diode's 30 uA? Is the capacitor by its nature not included?

Thanks for your help.

 

[Wade Hassler]

A very junior question.

Often I'd like to know the maximum load current of a simple circuit. After the curcuit is built, I can of course measure this with a multimeter. But for the circuit on paper, I just don't know. And the web won't help me.

The usual comparator circuit might make the problem more concrete. See attached. The relay has a nominal coil current of 60 mA.

How do I measure Imax? Is it the 60 mA coil current, plus the 16 mA Isink of the LM2903? Plus the approximate 1.2 mA for the thermistors? The 2N3904 has a "collector current - continuous" of 200 mA, should I include that? And the diode's 30 uA? Is the capacitor by its nature not included?

Thanks for your help.

First of all: the LM2903 has an open-collector output, so you'll need a pull-up resistor to the 2N2904's base.

After that, the current will be
LM2903 supply-current 2.5 mA Abs Max
Relay 60 mA
Thermistors 12v/20k .6 mA
Pot 12v/10k 1.2 mA
base-current 2.6 mA depends on the pull-up resistor

As you said, the capacitor doesn't figure into it, and the diode leakage is to small to worry about.
The transistor's collector-current in this case is only what the relay wants, and the comparator only sinks
the current drawn by the pull-up resistor.

 

[MikeMl]

That circuit will barely work! The naive designer thinks that 220K will supply enough current into the base of the 3904 to properly switch it on. There needs to be a pull-up resistor from the output of the comparator to 12Vdc.

About your question: Ignoring the current for the sensors and opamp, the relay current is determined by subtracting the VceSat of the 3904 (~0.3V) from 12V, and dividing by the relay coil resistance. A first order approximation is that the circuit will require 12*0.06= 0.72W

 

[nyoo]

Thanks very much. I now know (more of) who the actual current users are. I learned something here.

It's a little off the subject of counting current, but who knows? I could use clarification on MikeML's statement about the 220K resistor (for hysteresis) and the transistor. I know I'm pretty poor at reading the datasheets, but the 2N3904 looks like it will switch on with 1 mA and 0.85 V. The LM2903 will sink 16 mA at more than 10 V, depending on the pull-up resistor. What am I missing?

 

[Wade Hassler]

Thanks very much. I now know (more of) who the actual current users are. I learned something here.

It's a little off the subject of counting current, but who knows? I could use clarification on MikeML's statement about the 220K resistor (for hysteresis) and the transistor. I know I'm pretty poor at reading the datasheets, but the 2N3904 looks like it will switch on with 1 mA and 0.85 V. The LM2903 will sink 16 mA at more than 10 V, depending on the pull-up resistor. What am I missing?

The comparator's sink current can only pull the transistor's base down, cutting the transistor off.
When the LM2903 is off, it's as if it were disconnected from the circuit and the pull-up resistor drives current into the base.
And, as Mike pointed out, that current must be, when multiplied by the transistor beta, equal to or greater than what the relay needs.

 

[nyoo]

Aha. Learned a little more.

Thanks.

 

[nyoo]

So, taking the advice of a 4.7K pull-up resistor. And mentally removing the LM2903 from the circuit, for a moment.

Current through the 4.7K = 12/4700 = 2.55 mA.
H(FE) for the 2N3904 = 40. Is Beta also known as H(FE)?
If so, current through the transistor = 40 * 2.55 = 102 mA. And this is more than the 60 mA that the relay needs.
Is this what you meant?

Did you choose a 4.7K pull-up resistor (rather than, say, a 3.3K resistor) to meet the relay's requirement, with the minimum mA to spare?

 

[MikeMl]

Actually, a transistor used as a saturated switch should have a much higher base current injected into the base-emitter junction than implied by the simple Hfe calculation above. A rule of thumb is Ib <= (Ic/20), so about 3mA to 6mA. That would require the pull-up to be (12-0.6)/0.006 ~= 2K.

 

[audioguru]

The datasheet for the 2N3904 transistor and for most other transistors lists its saturation voltage loss when its base current is 1/10th its collector current, not 1/20th.

The datasheet for the LM2903 and LM393 shows that its minimum output current is 6mA when its saturation voltage is 1.5V which is much too high so some of them certainly will not turn off a transistor.

 

[Wade Hassler]

He's right: the 2n3904's Hfe has a variation over current and temperature and voltage

... and it looks like using Hfe=20 will all but guarantee operation for you

 

[MikeMl]

The datasheet for the 2N3904 transistor and for most other transistors lists its saturation voltage loss when its base current is 1/10th its collector current, not 1/20th.

As a practical matter (we are not designing spacecraft or war-machines here), 1/20 is more than good enough!

 

[audioguru]

Why not use the values listed on the datasheets? Then every passable transistor will work well, not just some of them.
I have always used minimum and maximum values from datasheets so that every circuit I designed worked perfectly (like spacecraft or war-machines).
Maybe if I cheated a little on the spec's (like you imply) then my products would cost less but then some won't work and will need to be fixed or replaced.

 

[MikeMl]

Why not use the values listed on the datasheets?

Show me on the data sheet where it says that to switch 60mA at the collector I have to inject 6mA into the base!

Even if the the actual Vce was slightly more than the spec'ed VceSat, say it was 1V instead of 0.3V, which is the difference between injecting 1mA vs 6ma, then the dissipation in the transistor would be 60mW instead of 18mw, and the relay would have 11V across it instead of 11.7V.

BFD

 

[nyoo]

We're still measuring current, yay! And I'm grateful for your time and patience.

I see the LM2903's minimum I(sink) of 6.0 mA. Is this an outlier condition: I(sink) is normally 16 mA, but some poorly made LM2903 might only give 6 mA?

Let's see if I'm still keeping up with you all. Is it something like this: if some comparator sinks only 6 mA and if the transistor requires 6 mA, then there is a chance the pull-up will always choose the transistor over the comparator output, even in its On state?

By the way, I guess the 120 VAC gives away the secret that this is not attached to a spacecraft, at least yet.

 

[audioguru]

ICs with one part number are not all exactly the same. There are excellent ones and there are pretty poor ones. Manufacturers sell most of them.
When you buy them you don't know if they are excellent or pretty poor.

If you design a circuit using the guaranteed minimum and maximum spec's listed on the datasheets then every circuit will work perfectly.
But if you gamble that the transistors are all "typical" then some of your circuits will not work.

A transistor needs an input of 0.5V or less to turn off. But the LM2903 has a minimum output current of 6mA and its saturation voltage has a maximum of 1.5V which will not turn off a transistor.

 

[MikeMl]

...
Let's see if I'm still keeping up with you all. Is it something like this: if some comparator sinks only 6 mA and if the transistor requires 6 mA, then there is a chance the pull-up will always choose the transistor over the comparator output, even in its On state?...

The comparator must sink whatever current the pull-up resistor can source AND at the same time get the base voltage of the NPN low enough so that it doesn't begin to turn on. If we use 3mA or less for the base current, then the Vo of the comparator when it is sinking that 3mA is under 0.3V, so it will work at your house, but not in space

 

[audioguru]

Again you show the graph of a typical device. But you can't buy a typical device, you get whatever they have which might have minimum spec's.
Don't you want every single circuit to work perfectly or do you want to test every one then throw away the ones that don't work?

Oh. Maybe you want your customers to test them then throw away your reputation with the ones that don't work. (wink)

 

[Wade Hassler]

The LM2903 will turn off the transistor by pulling its base-voltage too low to turn on the base-emitter junction. The minimum value of Vbe(sat) is about .65 Volts.
To do so, the 2903 must be able to pull enough current through the pull-down resistor that the voltage is below .65, keeping in mind that the 2903's output saturation voltage will rise with current:


Notice that a Vsat=.4 volts can be attained at worst case if the LM2903's output-current is in the neighborhood of 4 mA. That will dictate the pull-up's value:
Rpu=(12-Vsat)/.004 --> 2.9K

That pull-up will provide approximately the same current into the transistor-base (the two Vsat values are both swamped by the +12v supply,) so the minimum transistor beta that will suit this is
60mA/4mA = 15... should work

The transistor could be replaced by a MOSFET and reduce the calculating considerably, but you'd still have to calculate

 

[MikeMl]

Again you show the graph of a typical device. But you can't buy a typical device, you get whatever they have which might have minimum spec's.
Don't you want every single circuit to work perfectly or do you want to test every one then throw away the ones that don't work?

Oh. Maybe you want your customers to test them then throw away your reputation with the ones that don't work. (wink)

And you always have to have the last word, don't you!

 

[nyoo]

The transistor could be replaced by a MOSFET and reduce the calculating considerably, but you'd still have to calculate

What you have given me here, are some tools and worked examples to understand the balance of voltage and current in bipolar transistors. Only you all of this forum have the depth of skill to simplify it for folks like me.

Thank you. It is always entertaining, usually enlightening.

I will face MOSFETs eventually, have already used two without understanding them. But for now I'll calculate max load current. Maybe to size a zener diode voltage regulator. And I'll look more carefully at these newly explained chararcteristics on the datasheets, trying to match comparator with transistor.