EEVblog Electronics Community Forum
Electronics => Beginners => Topic started by: javadesigner on April 02, 2021, 04:16:58 pm
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Hi - I'm a n00b so this is a n00b question.
I'm poking around at Signal/function generators (both separate units or oscilloscopes built in ones).
I see they all specify a max Voltage (peak to peak) and an impedance (say 50 ohms).
But for some reason, I do not see a power/current rating on any of them. Why is this ?
For example, unless I am missing something obvious, none of these specify a POWER rating at all:
1) Siglent: https://siglentna.com/resources/documents/waveform-generators/ (https://siglentna.com/resources/documents/waveform-generators/)
2) Rigol: https://www.rigolna.com/products/waveform-generators/dg900/ (https://www.rigolna.com/products/waveform-generators/dg900/)
3) R&S: https://www.rohde-schwarz.com/fi/knowledge-center/videos/r-s-rtb2000-waveform-and-pattern-generator-video-detailpage_251220-505224.html (https://www.rohde-schwarz.com/fi/knowledge-center/videos/r-s-rtb2000-waveform-and-pattern-generator-video-detailpage_251220-505224.html)
You could generate 100,000 Vpp if you want right, with a the power rating near zero ? (like a van de graaf generator for example).
Or you can have generate 1Vpp and a power rating of 100 A.
How do I know what the max current (AMPS) a signal generator can output ? Is the A/C power rating intrinsically tied to Vpp and only Vpp (with no other current/amps spec needed)?
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The key is the 50 ohm output impedance. If you set the generator for 1 volt output and then short it, you'll have 20mA (1/50) of current. Change the numbers however you like and get the answer for that setting.
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I see they all specify a max Voltage (peak to peak) and an impedance (say 50 ohms).
But for some reason, I do not see a power/current rating on any of them. Why is this ?
Because it is tacitly assumed the load will always be 50 ohms (which is not always true, can also be something else, but to keep the things simple). The generator has internally an ideal voltage source series with an internal 50 ohms resistor, then it comes your load/circuit in series.
That ideal voltage source inside the generator is adjustable, at the given Vpp you want (with a max possible Vpp), and it always has that internally 50 ohms series resistor. That series resistor is tricky in the sense that the Vpp you see in your circuit might NOT be the one you set !!!, for example:
- you set 10 Vpp and measure with an open circuit, and find there are 20 Vpp in fact.
- then with the same 10Vpp and short circuit on the generator (it's safe), the output will be zero Vpp.
- when you put a 100 ohms load resistor on the generator's output and measure again you find 6.66Vpp, WTF!
- only when the load is the usual one, a 50 ohms resistor, only then the 10Vpp you set on the generator's front panel will show as many Vpp on the load resistor, too.
Note that some generators have a setting to specify a load other than 50 ohms, but the resistor inside generator does not changes, only that it recalculate (i.e. if you set the generator expected load to 100 ohms, connect an 100 ohms load, set the generator to 10Vpp, then only then you'll measure indeed 10Vpp on the 100 ohms load, too).
Another (different) reason is that in AC, the voltage and the current can appear delay/shifted (with a delay, a phase shift), and then the power is not V*I any more.
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The key is the 50 ohm output impedance. If you set the generator for 1 volt output and then short it, you'll have 20mA (1/50) of current. Change the numbers however you like and get the answer for that setting.
From one of the funtion generator spec sheets:
2mVpp~10Vpp(50Ω,≤10MHz)
4mVpp ~ 20 Vpp(High impedance, <10MHz)
What is the maximum current (amps) this generator can output on "High impedance" (whatever that may be) ?
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The key is the 50 ohm output impedance. If you set the generator for 1 volt output and then short it, you'll have 20mA (1/50) of current. Change the numbers however you like and get the answer for that setting.
From one of the funtion generator spec sheets:
2mVpp~10Vpp(50Ω,≤10MHz)
4mVpp ~ 20 Vpp(High impedance, <10MHz)
What is the maximum current (amps) this generator can output on "High impedance" (whatever that may be) ?
The 50R or Hi-Z selection does not affect the actual operation of the signal generator, it just changes what it displays--in this case it divides the displayed voltage by 2 when you select 50R.
20 Vp-p is the open-circuit voltage, a.k.a. Hi-Z. This is what you would measure on the outputs with a typical high impedance device like a DMM or an oscilloscope (1M or 10M probe, not 50R of course). If the generator is used this way, the current will be very, very small and can be calculated as V/R where R = the measuring device impedance + 50 ohms. At 10MHz, the actual impedance will be a lot lower than the specified resistance, but let's ignore that for now.
There is a 50 ohm resistor in series with the output. So if you connect something with a resistance of 50R, like a scope with 50R input or a DMM using a 50R terminator, then the 20Vp-p will be dropped across two 50 ohm impedances--the internal one and the external device. Thus each sees half the voltage, thus the 10Vp-p rating--this is what the external device would see. In this case the current Ip-p would be 20 volts divided by 100 ohms, or 200mA.
Now if you short the generator, the 20Vp-p would be dropped entirely over the 50R internal resistor and the current would be V/R, 20/50 or 400mAp-p.
Note that peak-to-peak current can be a misleading concept because there is no time when 400mA of current is actually flowing in this case. Vrms and Irms would be easier to understand.
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2mVpp~10Vpp(50Ω,≤10MHz)
4mVpp ~ 20 Vpp(High impedance, <10MHz)
What is the maximum current (amps) this generator can output on "High impedance" (whatever that may be) ?
I think the max current on "High impedance" mode is the same as 50 R mode, because it can't be larger.
10Vpp/(2*1.41) = 3.54 V RMS for sine signal. And with only 50 R at its output the current is 3.54V/50R = 71 mA RMS for sine signal (or 100 mA for square-wave).
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20 Vp-p is the open-circuit voltage, a.k.a. Hi-Z
[..]
There is a 50 ohm resistor in series with the output. So if you connect something with a resistance of 50R, like a scope with 50R input or a DMM using a 50R terminator, then the 20Vp-p will be dropped across two 50 ohm impedances--the internal one and the external device. Thus each sees half the voltage, thus the 10Vp-p rating--this is what the external device would see. In this case the current Ip-p would be 20 volts divided by 100 ohms, or 200mA.
Now if you short the generator, the 20Vp-p would be dropped entirely over the 50R internal resistor and the current would be V/R, 20/50 or 400mAp-p.
Really appreciate the long and informative reply bdunham7 :-+
So, just to be clear (and ignorning rms numbers to keep it simple), the max current the device is capable of is:
High-Z (open) voltage / internal resistance
In this case, like you said, 400mA.
This is the correct way to calculate power rating on these devices when no explicit power rating is given ?
Now, as an aside, I'd like to drive something like this (for a personal hobby physics experiment):
https://www.pasco.com/products/lab-apparatus/waves-and-sound/ripple-tank-and-standing-waves/sf-9324#documents-panel (https://www.pasco.com/products/lab-apparatus/waves-and-sound/ripple-tank-and-standing-waves/sf-9324#documents-panel)
This is essentially a speaker with 8 ohms impedance and max current rating upto 1 A.
If I plug this device into the this particular function generator, I would then be able to output:
20 High-Z / ( 50 [internal]+ 8 [device] ) = 345 mA ?
If this is indeed the case, and I want to drive this device at 800 mA, I would need a "power" function generator instead ?
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I think the max current on "High impedance" mode is the same as 50 R mode, because it can't be larger.
10Vpp/(2*1.41) = 3.54 V RMS for sine signal. And with only 50 R at its output the current is 3.54V/50R = 71 mA RMS for sine signal (or 100 mA for square-wave).
Thanks! But based on the prior post, the High-Z and 50R max current were different (200mA vs 400mA) - could you please clarify a bit more ?
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Signal generators (as opposed to other kinds of power amplifiers) usually have an actual 50 ohm (or 600 ohm, or whatever the spec is) output resistance, comprising a physical resistor after a low-impedance source followed by an output attenuator that maintains the 50 ohm output impedance. As shown above, the generator will deliver twice the voltage into a high impedance load that it delivers into a 50 ohm load. Into a short circuit (which should not damage the generator, don't do this with a power amplifier), it will deliver a current equal to (high-Z output voltage) / (50 ohm output impedance). This follows directly from the Thévenin and Norton equivalent circuits (q.v.).
For other systems, it is important not to confuse the output impedance with the load impedance, but signal generators usually have an output impedance equal to the expected load impedance (50 ohms typical for RF generators). The maximum power will be delivered into a load resistance equal to the output impedance. Confusion results from power amplifiers and other power devices, where the circuit is designed for an optimal load resistance, but the source resistance (output impedance) is not directly related to the optimal load.
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Selecting Hi-Z or 50 Ohms makes no difference to the internals of the generator, it still has, and always will have, a 50 Ohm series resistor. Among other things, it helps with short-circuit protection.
What you are doing is telling the generator that the load is High-Z or 50 Ohms.
If you select 50 Ohms then the generator knows there is a series voltage divider and does the math accordingly. The actual output is 1/2 of what the generator actually produces internally (as measured open-circuit) but the generator compensates for that.
If you select High-Z, the generator has no idea what the load impedance is and it simply ignores the math involved with correcting the voltage. As a user, you need to figure out how the voltage divider is affecting things. As an example, if the load impedance is 10k, the 50 Ohm internal resistor just doesn't matter, the current is limited by the 10k impedance.
In terms of power, it's still E2/R or I2*R or I*E;. You probably want to use RMS values for power calculations.
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If I plug this device into the this particular function generator, I would then be able to output:
20 High-Z / ( 50 [internal]+ 8 [device] ) = 345 mA ?
If this is indeed the case, and I want to drive this device at 800 mA, I would need a "power" function generator instead ?
You really need to convert everything to RMS, because as I predicted, you will confuse yourself with peak-to-peak numbers. Your maximum current will be nowhere near 345mA in this case. 20Vp-p is about 7Vrms.
If you want 800mArms into a 8 ohm device, that is 6.4Vrms and 5.12 watts, way more than any standard 50R function generator will give you. There are specific power amplifiers intended to be used with function generators for this purpose. Cheapies from FeelTech and Juntek are $20-50 on ebay or Aliexpress. Note that these will have a very low output impedance, not 50R.
https://www.ebay.com/itm/FeelTech-DC-power-amplifier-module-amp-driver-For-DDS-Function-signal-Generator/132059122895?hash=item1ebf564ccf:g:DosAAOSw5cNYcywe (https://www.ebay.com/itm/FeelTech-DC-power-amplifier-module-amp-driver-For-DDS-Function-signal-Generator/132059122895?hash=item1ebf564ccf:g:DosAAOSw5cNYcywe)
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If you want 800mArms into a 8 ohm device, that is 6.4Vrms and 5.12 watts, way more than any standard 50R function generator will give you. There are specific power amplifiers intended to be used with function generators for this purpose. Cheapies from FeelTech and Juntek are $20-50 on ebay or Aliexpress. Note that these will have a very low output impedance, not 50R.
Thanks again for the further clarification!
This is slightly off-topic here but was wondering: since you have experience with this (and that ebay link above looks good), do you have any other recommendations for "better" quality power amps made to be used with function generators? They would take BNC/Banana IN and BNC/Banana OUT.
Price could be a upto a few hundred dollars. I have a +/- 5Vpp, 50R function generator BNC out from my oscilloscope already. I'd like to drive this signal as upto 1 amp into 8 Ohms. (Edit: upto about 5Khz)
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The answer depends on the signal frequency range.
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If I plug this device into the this particular function generator, I would then be able to output:
20 High-Z / ( 50 [internal]+ 8 [device] ) = 345 mA ?
yes. But if you're talking about 20 Vpp (peak-to-peak voltage) then it will be 345 mA peak-to-peak current.
If you're needs a peak (amplitude) current, then you're needs to use peak voltage, so you will need to divide it by two:
(20 / 2) / (50 + 8) = 10 / 58 = 172.4 mA peak
If this is indeed the case, and I want to drive this device at 800 mA, I would need a "power" function generator instead ?
no, if you want to get more power, you can buy power amplifier for the generator.
I'd like to drive this signal as upto 1 amp into 8 Ohms. (Edit: upto about 5Khz)
This is 8 Watt.
For the frequency below 20 kHz you can use usual 15-20 Watt audio power amplifier.
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do you have any other recommendations for "better" quality power amps made to be used with function generators? They would take BNC/Banana IN and BNC/Banana OUT.
I've no direct experience with any of them. I just use an audio amplifier like this:
https://www.ebay.com/itm/Marantz-Mono-Block-Amplifier-Model-MA700-Amp-Serial-MZ009731050249-MA700U/114747299537?epid=110959053&hash=item1ab778d6d1:g:yDEAAOSwAgRgY7xO (https://www.ebay.com/itm/Marantz-Mono-Block-Amplifier-Model-MA700-Amp-Serial-MZ009731050249-MA700U/114747299537?epid=110959053&hash=item1ab778d6d1:g:yDEAAOSwAgRgY7xO)
Of course that's massive overkill for what you want.
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With an outboard power amplifier, capable of the current you need, be careful about the actual load that the amplifier sees: the amplifier might be unhappy with a short circuit, and an old-fashioned vacuum-tube amplifier would also be unhappy with an open circuit. You will waste lots of power this way, but you could combine a shunt resistance across the output terminals with a series resistance to your load, and that would keep the actual load resistance between two calculable levels, regardless of your load. The calculation is left as an exercise for the reader.
If you know that your load impedance is close to 8 ohms, then a normal audio amplifier should be happy driving it. The mean power required is the (rms current)2 x (Rload). Some people call that "rms power", but that is incorrect.
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Hi - I'm a n00b so this is a n00b question.
I'm poking around at Signal/function generators (both separate units or oscilloscopes built in ones).
I see they all specify a max Voltage (peak to peak) and an impedance (say 50 ohms).
But for some reason, I do not see a power/current rating on any of them. Why is this ?
For example, unless I am missing something obvious, none of these specify a POWER rating at all:
1) Siglent: https://siglentna.com/resources/documents/waveform-generators/ (https://siglentna.com/resources/documents/waveform-generators/)
Typically 200mA, the same as their DC output rating....that is they can also be used a low power DC supply.
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Typically 200mA, the same as their DC output rating....that is they can also be used a low power DC supply.
Very low power, certainly not 200mA--that would be the short circuit current. The 50R output impedance makes for pretty bad load regulation--you lose a volt for every 20mA.