Power good signal

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# Power good signal

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(@shar)
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Joined: 1 month ago
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In one of a reference board this switching regulator is there which is stepping down 3.3 v to 1.8v. The power good signal of the 1.8volt domain has to be sent to the CPLD, which they have sent the way shown in down side of the figure. What I understand is that this is a voltage translation as CPLD works on 3.3v logic. So I am stuck that how this circuit is designed...How resistance values selected and why 1.8v has been given to the npn bjt using voltage divider and not directly.

(@davee)
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Joined: 3 years ago
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Hi @shar,

Note that 'power good' signals are not usually intended as precision measurement signals to detect a substandard power source with a slightly out of spec output voltage, but instead as a sequencing mechanism, so that parts of the circuit are held in a reset state until the power voltages applied to those parts have reached a high enough value to enable the circuit to begin to function correctly.

Hence, a 'power good' signal might be asserted 'true' when the voltage is still rising, but close enough to the final value, assuming it will reach the final expected voltage in a very short time after the 'power good' signal begins to change from false to true.

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why 1.8v has been given to the npn bjt using voltage divider and not directly

PQ162 is a mmbt3904ltig, and its datasheet at

implying that for saturation (transistor to be fully turned on, and hence collector pulled to the lowest voltage) typically requires the base-emitter voltage to be 0.65V, but could be up to 0.85V, assuming collector current of 10mA, and base current of 1mA.

As maximum collector current (i.e. saturation) will be (slightly less than) 3.3V/(PR168) 10k = 0.33mA, a much smaller base current than the stated value will be sufficient, so we may guess saturation will occur when base-emitter voltage is slightly lower than the 0.65-0.85V range.

Thus, to act as a 'power good' signal, assume it will be necessary to provide the base of the transistor with at least 0.85V, when the 1.8V power line is actually reaches 1.8V. In practice, it is likely the transistor will be switched on just before the voltage actually reaches 1.8V, say at 1.6V, but the difference in timing will probably not be significant.

How resistance values selected

Simplistically, the circuit must ensure the transistor is saturated when the 1.8V rail voltage is actually 1.8V, and the previous paragraph suggested the voltage available should exceed 0.85V. If the effect of current passing through PR 160 is ignored, then PR179/181 will act as a divide by two, voltage divider, so that the voltage at the midpoint will be 0.9V, when the supply voltage is 1.8V. No value is specified for PR160, but I might (widely) guess it will be something like 10k, so that when the supply voltage is 1.8V, the base current will be small enough to make little change to the voltage divider ratio, but sufficient to saturate the transistor, given the collector current only needs to be 0.3mA and the minimum current gain of the transistor is 100.

(PR160 value should be calculated and checked more carefully than my wild guess, before a production run is attempted.)

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The above discussion is simplified, compared to a full 'theoretical' analysis, but in practice, the variation in part characteristics due to process tolerances, will probably exceed the effects of the simplifications. A common design approach might start with a simplified approach, like that shown here, and then carefully examine each part of the circuit, using simulations and prototypes, adjusting the values of any components that appear critical, to produce the most robust design.

I hope this helps. Best wishes, Dave

(@shar)
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@davee Thank you so much for your answer. I hadn't expected that this forum has people like you who take out time to answer others' queries that too in detail . Its going to be a learning and exciting journey on this forum.

One thing I would like to mention that you said since collector current is small around 0.33 mA, a value of voltage less than 0.65-0.85 voltage range would be enough to saturate the transistor. But in the next paragraph you said that it will be saturated when it exceeds 0.85volt. Just a little confusion here.

And I think the base resistance value they have taken is 316 ohms ... I think its written in blue just beside the resistor. This would work fine I guess because if we take 0.85v also Ib= (0.9-0.85)/316= 0.158 mA which is more than (Ic/Beta i.e 0.33mA/100 = 0.0033mA) enough to saturate the transistor. Am I correct here?

Once again thank you Dave, looking forward to have more informative and knowledgeable conversation with you.

Shardul

(@davee)
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Joined: 3 years ago
Posts: 1851

Hi @shar,

With regard to the base-emitter voltage confusion:

The 'first priority' of the that part of circuit is to make sure that when the 1.8V line is actually at 1.8V, that the npn transistor is 'fully switched on' or 'saturated', so that the collector voltage also becomes very low (say ~0.5V), and hence controls the FET.

The data sheet range of 0.65V-0.85V base-emitter voltage is a guide to the minimum voltage to achieve that. If you had several of this type of transistor, possibly from different production batches, and tested each of them in the circuit, you would find the actual base-emitter voltages when the 1.8V rail was at 1.8V, would vary, possibly from about 0.65V to 0.85V. This is because the base-emitter junction is basically the same as a silicon diode, the curve of base current versus base-emitter voltage has a sharp 'knee' in this range, in which a tiny voltage increase will result in a large current increase.

For illustration, see the following plot of a silicon diode curve from:

The 'knee' is marked on the right hand side.

Applying a higher voltage than 0.85V, through a 'suitable value' base resistor, will aim to ensure that the actual base-emitter voltage is sufficient for that particular device. Oversimplifying, if a voltage of (say) 0.9V appears at the junction of the potential divider PR179/181, then this will be enough to cause the base-emitter junction to start conducting, and the voltage across that junction will probably be somewhere near 0.7V, with the remaining 0.2V dropped across the base resistor. i.e. the base-emitter voltage will be roughly determined by the characteristics of the particular device, and the current flow will automatically be 'adjusted' to produce this voltage drop, given the values of the resistors.

I hope this helps a little. I probably haven't explained it that clearly, but give it a little time, and you will find it becomes clearer.

Best wishes, Dave

(@shar)
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Joined: 1 month ago
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Topic starter

@davee Thanks Davee.. Got your point. Had been off the forum for some time.

DaveE reacted