# RL time constant

Discussion in 'Homework Help' started by suzuki, Sep 15, 2011.

1. ### suzuki Thread Starter Member

Aug 10, 2011
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0
Hi all,

this seems to be a fairly simple question but i can't quite get what im looking for. hopefully someone can help me spot what is probably an obvious error.

I am trying to find the RL time constant of the following circuit. I turn off the switch at t = 0.5s and measure the time it takes for the inductor voltage to go from its high to "zero". I believe we can approximate the time constant of the circuit by taking that time (5.03e-3 in this case) and then dividing by 5 to get the value of a single time constant. This value is about 1000μs.

However, my calculation for this circuit uses time constant = L/R. using the values for my circuit, that gives me 10m/101, which is about 99μs.

thanks in advance for all replies.

2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
782
The FET in your circuit is not really turning off the current. The anti-parallel diode will still conduct for half the AC cycle. In effect the response you are seeing is a driven response rather than the natural transient response.

3. ### suzuki Thread Starter Member

Aug 10, 2011
119
0
ah ok. In that case, do you have a tip on how to measure physically measure the time constant in the lab? Would I just switch off the power supply? but i'm not sure that the oscilloscope would be able to pick up that signal in time for me to measure the decay.

4. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
782
One issue with the series LR circuit is that if you simply open circuit the power source (by whatever means) there is no path for non-zero current to continue to flow and there is a large self-induced emf arising in the inductor.

So to measure the time constant in an LR circuit you need to organize the circuit layout such that current will continue to flow when the power source is removed or alternatively you measure the time constant when the energy source is connected (rather than disconnected). It's simpler in principle & practice to use a fixed DC source rather than an AC source.

Using a CRO to measure a short duration transient is easier if the scope is a storage type and you take a stored single shot measurement. Triggering is an issue if you want to catch the start of the transient so a pre-triggering feature on the CRO is useful (but not always available). Alternatively you would generate a CRO trigger signal coupled to the control circuit which turns the source on or off.

I'm not sure why you are attempting to do the measurement with an AC source other than the advantage of having a cyclic source at line frequency which is usually compatible with the (often standard) line trigger feature on a CRO. You'd need a more complex on-off control to synchronize the turn off time so that the transient is reproduced consistently and the display appears stable on the CRO.

One idea attached for a single shot turn off condition .... there are obviously many options.

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5. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
782
Another option for a line frequency based repetitive version.

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6. ### suzuki Thread Starter Member

Aug 10, 2011
119
0
ah sorry, t_n_k. ive been away for a last few days so i didnt get the chance to see your replies.

for other members future reference, i found that the simplest solution was to actually remove the switching element and then replace the ac voltage source with a step function or a square wave. that way you can measure the voltage decay after the voltage source has gone from high to low state.